In this post, All matrices are assumed to be in $M_n(\mathbb{C}),$ the ring of $n \times n$ matrices with complex entries. Also, $e_{ij} \in M_n(\mathbb{C})$ is the matrix with $(i,j)$-entry $1$ and all other entries $0.$

Recall that a matrix $A=[a_{ij}]$ is said to be upper triangular if $a_{ij}=0$ for all $i > j.$ Similarly, $A$ is lower triangular if $a_{ij}=0$ for all $i < j.$ If $A$ is either upper triangular or lower triangular, then $A$ is said to be triangular. If $A$ is triangular and also $a_{ii}=0$ for all $i,$ then we say that $A$ is strictly triangular.

Problem 1. Let $A=[a_{ij}]$ be an $n \times n$ matrix with the adjugate $\text{adj}(A).$ Show that

i) if $A$ is strictly upper triangular, then $\text{adj}(A)=(-1)^{n+1}a_{12}a_{23} \cdots a_{n-1,n}e_{1n}$

ii) if $A$ is strictly lower triangular, then $\text{adj}(A)=(-1)^{n+1}a_{21}a_{32} \cdots a_{n,n-1}e_{n1}.$

Solution. I only prove i) and leave the proof of ii), which is similar to the proof of i), as an easy exercise.
Let $M_{ij}$ be the $(i,j)$-minor of $A,$ i.e. the $(n-1) \times (n-1)$ matrix that results from deleting the $i$-th row and the $j$-th column of $A.$ Let $d_{ij}=\det M_{ij}.$
If $(i,j) \ne (n,1),$ then, since both the first column and the last row of $A$ are zero, either the first column or the last row or both the first column and the last row of $M_{ij}$ is zero and hence $d_{ij}=0$ in this case. But if $(i,j)=(n,1),$ then $M_{ij}$ is just an upper triangular matrix with the diagonal entries $a_{12}, a_{23}, \cdots , a_{n-1,n}$ and hence $d_{n1}=a_{12}a_{21} \cdots a_{n-1,n}.$ Thus

$\text{adj}(A)=[(-1)^{i+j}d_{ij}]^T=(-1)^{n+1}d_{n1}e_{1n}=(-1)^{n+1}a_{12}a_{23} \cdots a_{n-1,n}e_{1n}. \ \Box$

Problem 2. Let $A=[a_{ij}]$ be a strictly triangular $n \times n$ matrix with the adjugate $\text{adj}(A).$ Show that if $n > 2$ and $\text{adj}(A)=cA$ for some scalar $c \ne 0,$ then $A=0.$ Also, show that the result does not always hold if $n=2.$

Solution. I’ll assume that $A$ is strictly upper triangular; the argument for strictly lower triangular is similar. So we are given that $cA=\text{adj}(A)= (-1)^{n+1}a_{12}a_{23} \cdots a_{n-1,n}e_{1n},$ by Problem 1, and hence $a_{ij}=0$ for all $(i,j) \ne (1,n)$ and $ca_{1n}=(-1)^{n+1}a_{12}a_{23} \cdots a_{n-1,n}.$ Thus if $n > 2,$ then, since $a_{23}=0,$ we have $a_{12}a_{23} \cdots a_{n-1,n}=0$ and hence $ca_{1n}=0$ implying $a_{1n}=0.$ So $a_{ij}=0$ for all $i,j,$ and hence $A=0.$
The result does not always hold if $n=2$ because, for example, $\text{adj}(e_{12})=-e_{12}$ but $e_{12} \ne 0. \ \Box$

Remark. In the solution of the next problem, we are going to use this important fact that if $A,B \in M_n(\mathbb{C})$ and $AB=BA,$ then $A,B$ are simultaneously triangularizable, i.e. there exists an invertible matrix $P \in M_n(\mathbb{C})$ and upper triangular matrices $U,V \in M_n(\mathbb{C})$ such that $A=PUP^{-1}$ and $B=PVP^{-1}.$

Problem 3. Let $A$ be an $n \times n, \ n > 2,$ matrix and suppose that there exists a scalar $c \ne 0$ such that

$A^2=(\text{adj}(A))^2=\text{adj}(A)+cA=0.$

Show that $A=0.$

Solution. It follows from the relations given in the problem that $\text{adj}(A)A=A\text{adj}(A)=0$ and hence, by the above Remark, there exists an invertible matrix $P$ and upper triangular matrices $U,V$ such that $A=PUP^{-1}$ and $\text{adj}(A)=PVP^{-1}.$ Since $A^2=0,$ the eigenvalues of $A$ are all zero and hence $U$ is strictly upper triangular. Also

$\text{adj}(A)=PVP^{-1}=\text{adj}(PUP^{-1})=\text{adj}(P)^{-1} \text{adj}(U) \text{adj}(P)=P \text{adj}(U)P^{-1}$

and thus $V=\text{adj}(U),$ which gives

$0=\text{adj}(A)+cA=P(V+cU)P^{-1}=P(\text{adj}(U)+cU)P^{-1}.$

Hence $\text{adj}(U)+cU=0$ and therefore, by Problem 2, $U=0$ giving $A=PUP^{-1}=0. \ \Box$

Let’s begin this post with a well-known result about the normality of subgroups of prime index.

Problem 1. Let $G$ be a finite group and let $p$ be the smallest prime divisor of $|G|.$ Suppose that $G$ has a subgroup $H$ such that $[G:H]=p.$ Show that $H$ is normal in $G.$

Solution. See Problem 2 in this post. $\Box$

A trivial consequence of Problem 1 is that in finite groups, every subgroup of index $2$ is normal. In fact, subgroups of index $2$ are always normal, whether the group is finite or infinite. Let’s prove this little result.

Problem 2. Let $G$ be a group and suppose that $H$ is a subgroup of $G$ such that $[G:H]=2.$ Show that $H$ is normal in $G.$

Solution. Choose $x \in G \setminus H.$ Since $[G:H]=2,$ we have $G=H \cup xH.$ Now, suppose, to the contrary, that $H$ is not normal. So there exist $g \in G, \ h \in H$ such that $ghg^{-1} \notin H.$ So $ghg^{-1} \in xH$ and $g \in xH.$ Therefore there exist $h_1,h_2 \in H$ such that $g=xh_1$ and $ghg^{-1}=xh_2.$ But then, eliminating $x,$ we get $g=h_2^{-1}h_1h \in H,$ contradiction. $\Box$

Are subgroups of prime index always normal? Of course not. For example, let $p > 2$ be any prime number, and let $G$ be the dihedral group of order $2p,$ i.e. $G$ is generated by two elements $a,b,$ such that $o(a)=o(ab)=2, \ o(b)=p.$ Then the subgroup $H:=\langle a \rangle =\{1,a\}$ has index $p$ but $H$ is not normal because $bab^{-1} \notin H.$ On the other hand, the subgroup $\langle b \rangle$ has index $2$ and so it’s normal, by Problem 2 or directly: $ab^ka^{-1}=(aba^{-1})^k=(aba)^k=b^{-k} \in \langle b \rangle.$

Problem 3 (T. Y. Lam). Let $G$ be a group, and let $p$ be a prime number. Suppose that $G$ has a subgroup $H$ such that $[G:H]=p.$ Show that the following statements are equivalent.

i) $H$ is normal in $G.$

ii) $g^p \in H$ for all $g \in G.$

iii) If $g \in G \setminus H,$ then $g^k \notin H$ for all positive integers $k < p.$

Solution. i) $\Longrightarrow$ ii). Since $H$ is normal and $[G:H]=p,$ the set $G/H$ is a group of order $p$ and hence $x^p=1_{G/H}=H$ for all $x \in G/H,$ i.e. $g^pH=(gH)^p=H$ for all $g \in G$ and so $g^p \in H.$

ii) $\Longrightarrow$ iii). Let $g \in G \setminus H$ and suppose, to the contrary, that $g^k \in H$ for some integer $1 \le k < p.$ Then, since $p$ is prime, $\gcd(k,p)=1$ and hence there exist integers $r,s$ such that $rk+sp=1.$ But then we get $g=g^{rk+sp}=(g^k)^r(g^p)^s \in H,$ contradiction.

iii) $\Longrightarrow$ i). Suppose, to the contrary, that $H$ is not normal. Hence there exist $x \in G$ and $h \in H$ such that $g:=xhx^{-1} \notin H.$ Thus, by iii), $g^k \notin H$ for all integers $1 \le k < p,$ and hence we have $p$ distinct cosets $H, gH, \cdots , g^{p-1}H.$ Thus, since $[G:H]=p,$ we get $G=\bigcup_{k=0}^{p-1}g^kH.$ So $x \in \bigcup_{k=0}^{p-1}g^kH$ and hence $x=g^kh_1$ for some $h_1 \in H$ and some integer $0 \le k \le p-1.$ Therefore

$x=g^kh_1=(xhx^{-1})^kh_1=xh^kx^{-1}h_1,$

which gives $x=h_1h^k \in H$ and hence $g=xhx^{-1} \in H,$ contradiction. $\Box$

Remark 1. Note that, in the solution of Problem 3, we used the hypothesis “$p$ is prime” only once and that was to prove that ii) $\Longrightarrow$ iii).

Let’s finish this post with a nice but, surprisingly, not well-known result about the normality of subgroups of prime index.

Problem 4 (J.C. Bioch & R.W. van der Waall). Let $p$ be a prime number. Show that if $G$ is a finite group and $H$ is a subgroup of $G$ such that $[G:H]=p$ and $\gcd(|H|,p-1)=1,$ then $H$ is normal in $G.$

Solution. Proof by induction over $|G|.$ There’s nothing to prove if $|G|=p.$ Now suppose that $G$ is a finite group that satisfies the conditions given in the problem, i.e. $G$ has a subgroup $H$ such that $[G:H]=p$ and $\gcd(|H|,p-1)=1.$ By Problem 1 in this post, there exists a normal subgroup $N$ of $G$ such that $N \subseteq H$ and $[G:N] \mid p!.$ Now, we consider two cases.

Case 1: $|N| > 1.$ In this case, $|G/N| < |G|$ and clearly $[G/N:H/N]=[G:H]=p.$ Also, since $|H/N|$ divides $|H|$ and $\gcd(|H|,p-1)=1,$ we also have $\gcd(|H/N|, p-1)=1.$ Thus, by our induction hypothesis, $H/N$ is normal in $G/N$ and hence $H$ is normal in $G.$

Case 2: $|N|=1.$ In this case, $|G|=[G:N] \mid p!$ and thus, since $p \mid |G|,$ we get that $|G|=pm$ for some integer $m$ with $\gcd(m,p)=1.$ Since $[G:H]=p,$ we have $|H|=m$ and also

$\gcd(|G|,p-1)=\gcd(pm,p-1)=\gcd(m,p-1)=\gcd(|H|,p-1)=1. \ \ \ \ \ \ \ \ \ \ \ (1)$

Now let $P$ be the Sylow $p$-subgroup of $G.$ So $|P|=p.$ Let $C(P), N(P)$ be, respectively, the centralizer and the normalizer of $P$ in $G.$ By the lemma in this post, $N(P)/C(P)$ is isomorphic to a subgroup of $\text{Aut}(P).$ But since $P \cong \mathbb{Z}_p,$ we have $|\text{Aut}(P)|=p-1$ and so $|N(P)/C(P)|$ must divide $p-1.$ Thus, since $|N(P)/C(P)|$ also divides $|G|,$ we get $|N(P)/C(P)|=1,$ by $(1).$ So $N(P)=C(P)$ and hence, by Burnside’s Normal Complement Theorem, there exists a normal subgroup $Q$ of $G$ such that $P \cap Q=\{1\}$ and $G=PQ.$ Thus $|Q|=m=|H|$ and so

$\displaystyle |HQ|=\frac{|H| \cdot |Q|}{|H \cap Q|}=\frac{m^2}{|H \cap Q|}. \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Since $Q$ is normal, $HQ$ is a subgroup of $G$ and thus, by $\displaystyle (2), \ \frac{m^2}{|H \cap Q|}$ must divide $|G|=pm$ and that is possible only if $|H \cap Q|=m.$ Thus $|H \cap Q|=|H|$ and hence $H \cap Q=H,$ which gives $H=Q$ because $|H|=|Q|.$ So $H$ is normal in $G$ and the solution is complete. $\Box$

Remark 2. Note that Problem 1 follows from Problem 4.

## Kernel of the commutator of two involutions

Posted: March 23, 2021 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , , ,

Problem. Let $V$ be a vector space over a field of characteristic $\neq 2.$ Let $f,g: V \to V$ be two linear transformations such that $f^2=g^2=I,$ where $I : V \to V$ is the identity map. Show that

$\ker(fg-fg)=\ker(f+g) \oplus \ker(f-g).$

Solution. We need to prove three things.

i) $\ker(f+g) \cap \ker (f-g)=(0).$

Proof. If $x \in \ker(f+g) \cap \ker (f-g),$ then

$2f(x)=(f+g)(x)+(f-g)(x)=0$

and so $f(x)=0$ hence $x=f^2(x)=0.$

ii) $\ker(f+g) + \ker(f-g) \subseteq \ker(fg-gf).$

Proof. We have

$fg-gf=(f-g)(f+g)=-(f+g)(f-g), \ \ \ \ \ \ \ \ \ \ \ (1)$

which gives $\ker(f+g) \subseteq \ker(fg-gf)$ and $\ker(f-g) \subseteq \ker(fg-gf).$

iii) $\ker(fg-gf) \subseteq \ker(f+g) + \ker(f-g).$

Proof. Let $x \in \ker(fg-gf).$ So $fg(x)=gf(x)$ and hence $gfg(x)=f(x), \ fgf(x)=g(x),$ which give

$(fg-gf)g(x)=(fg-gf)f(x)=0. \ \ \ \ \ \ \ \ \ \ \ (2)$

We also have

$\displaystyle x=y+z, \ \ \ \ \ \ \ y:=\frac{1}{2}(f-g)f(x), \ \ \ \ z:=\frac{1}{2}(f+g)g(x)$

and, by $(1),(2),$

$(f+g)(y)=(f-g)(z)=0,$

i.e. $y \in \ker(f+g)$ and $z \in \ker(f-g).$ Hence $x \in \ker(f+g) + \ker(f-g). \ \Box$

## A torsion-free group with a cyclic subgroup of finite index is cyclic

Posted: March 20, 2021 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , , , ,

Recall that an element of a group is called torsion if it has a finite order and a group is called torsion if every element of the group is torsion. A (non-trivial) group is called torsion-free if the only torsion element of the group is the identity element. Clearly every finite group is torsion and every torsion-free group is infinite.

Problem. Let $G$ be a torsion-free group and suppose that $G$ has a cyclic subgroup of finite index. Show that $G$ is cyclic.

Solution. The heart of the solution is the following theorem.

Theorem (Schur). Let $G$ be a group with the center $Z(G)$ and the commutator subgroup $G'.$ If $G/Z(G)$ is finite, then $G'$ is finite too.

Proof. See this post!

Corollary. Let $G$ be a torsion-free group. If $G/Z(G)$ is finite, then $G$ is abelian.

Proof. By Schur’s theorem, $G'$ is finite and so torsion. But $G$ is torsion-free and so $G'=(1),$ i.e. $G$ is abelian.

Now let’s solve the problem. Let $H$ be a cyclic subgroup of finite index in $G.$ It’s clear that $H$ is infinite because $G$ is torsion-free, hence infinite, and $[G:H]$ is finite. I give the solution in several steps.

Claim 1. We may (and will) assume that $H$ is normal in $G.$ Also, $H \cong \mathbb{Z}.$

Proof. Since $H$ is cyclic and infinite, $H \cong \mathbb{Z}.$ Also, by Cayley’s theorem (see Problem 1 in this post!), $H$ contains a subgroup which is both normal and finite index in $G.$ So, by replacing $H$ with that normal subgroup, we may assume that $H$ is normal in $G.$

Claim 2. If $G$ is abelian, then $G$ is cyclic.

Proof. Let $[G:H]=n.$ Then the map $f: G \to H$ defined by $f(x)=x^n$ is a group homomorphism and, since $G$ is torsion-free, $f$ is injective. Thus $G$ is isomorphic to a subgroup of $H$ and hence it is cyclic because $H$ is cyclic.

Claim 3. Let $C$ be the centralizer of $H$ in $G.$ Then $C$ is cyclic.

Proof. Let $Z(C)$ be the center of $C.$ Clearly $H \subseteq Z(C) \subseteq C \subseteq G$ and hence $C/Z(C)$ is finite (because we’re assuming that latex [G:H]$is finite). Thus by the above Corollary, $C$ is abelian and hence cyclic, by Claim 2. Claim 4. If $G$ has a cyclic subgroup of index $2,$ then $G$ is cyclic. Proof. Let$K=\langle a \rangle\$ be a cyclic subgroup of $G$ with $[G:K]=2.$ Then $K$ is normal in $G$ and choosing $x \in G \setminus K,$ we have $G=K \cup Kx.$ Now, $x^2 \in K$ and so $x^2=a^p$ for some integer $p.$ We also have $xax^{-1}=a^q$ for some integer $q.$ Thus

$a^p=x^2=xx^2x^{-1}=xa^px^{-1}=(xax^{-1})^p=a^{pq}$

and so $p=pq$ because $G$ is torsion-free. Thus either $p=0$ or $q=1.$ If $p=0,$ then $x^2=1,$ which gives $x=1 \in K,$ contradiction. Hence $q=1$ and therefore $xa=ax,$ i.e. $G$ is abelian hence cyclic, by Claim 2.

Claim 5. $G$ is cyclic.

Proof. Let $C$ be the centralizer of $H$ in $G.$ Define the map $f: G \to \text{Aut}(H)$ by $f(x)(h)=xhx^{-1}$ (don’t forget that, by Claim 1, we’re assuming that $H$ is normal in $G$). It’s clear that $f$ is a well-defined group homomorphism and $\ker f=C.$ So $G/C$ is isomorphic to a subgroup of $\text{Aut}(H) \cong \mathbb{Z}_2.$ Thus either $G=C$ or $[G:C]=2.$ If $G=C,$ then $G$ is cyclic, by Claim 3. If $[G:C]=2,$ then again $G$ is cyclic, by Claim 3 and Claim 4. The proof of the claim and the solution is now complete. $\Box$

## Solving a functional equation for matrices

Posted: February 20, 2021 in Elementary Algebra; Problems & Solutions, Linear Algebra

Problem (V. Brayman). Find all functions $f: \mathbb{R} \to M_3(\mathbb{R})$ that satisfy the following functional equation

$\displaystyle f(x)f(y)f(z)=\begin{pmatrix}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{pmatrix}, \ \ \ \ \ \ \ \forall x,y,z \in \mathbb{R}. \ \ \ \ \ \ \ \ \ \ \ (*)$

Solution. I show that the answer is all functions in the following form

$\displaystyle f(x)=\begin{pmatrix}0 & 0 & ax \\ b & 0 & 0 \\ 0 & c & 0\end{pmatrix}, \ \ \ \ \ \ \ a,b,c \in \mathbb{R}, \ \ abc=1.$

First see that such functions satisfy $(*).$ Now suppose that $f$ satisfies $(*)$ and let

$\displaystyle f(1)=\begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix}.$

By $(*),$ we have $(f(1))^3=I,$ the identity matrix, and so

$\displaystyle f(x)=(f(1))^3f(x)=f(1)f(1)f(1)f(x)=f(1)\begin{pmatrix}1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & x \end{pmatrix},$

and

$\displaystyle f(x)=f(x)(f(1))^3=f(x)f(1)f(1)f(1)=\begin{pmatrix}x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}f(1).$

Thus

$\displaystyle f(x)=f(1)\begin{pmatrix}1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & x \end{pmatrix}=\begin{pmatrix}x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}f(1),$

which gives $a_{11}=a_{12}=a_{23}=a_{33}=0$ and

$\displaystyle f(x)=\begin{pmatrix}0 & 0 & a_{13}x \\ a_{21} & a_{22} & 0 \\ a_{31} & a_{32} & 0\end{pmatrix}.$

Put the above in $(f(x))^3=xI,$ which holds by $(*),$ to get

$a_{13}a_{32}a_{21}=1, \ \ \ a_{22}=a_{31}=0,$

and that completes the solution. $\Box$

## Rings with only finitely many zero divisors

Posted: January 25, 2021 in Elementary Algebra; Problems & Solutions, Finite Rings, Rings and Modules
Tags: , ,

Recall that in a ring $R$ an element $z \in R$ is said to be a left zero divisor if $zr=0$ for some non-zero element $r \in R.$ Similarly, $z \in R$ is a right zero divisor if $rz=0$ for some non-zero element $r \in R.$ A zero divisor of $R$ is an element which is both a left and a right zero divisor.

Proposition 1. Let $R$ be a ring, and let $Z$ be the set of left zero-divisors of $R.$ If $Z$ is finite and $|Z| > 1,$ then $R$ is finite and $|R| \le |Z|^2.$

Proof. Since $|Z| > 1,$ there exists $z \in Z$ such that $z \ne 0.$ Let

$I:=\{r \in R: \ \ rz=0\}.$

If $r \in I,$ then $rz=0$ and so $r \in Z$ because $z \ne 0.$ Thus $I \subseteq Z$ and hence

$|I| \le |Z|. \ \ \ \ \ \ \ \ \ \ \ (1)$

Now clearly $I$ is a subgroup of the abelian group $(R,+)$ and so we may consider the group $R/I$ and the map

$f: R/I \longrightarrow Z, \ \ \ \ \ f(r+I)=rz, \ \ \ \forall r \in R.$

Then $f$ is well-defined and one-to-one because clearly $rz \in Z$ for all $r \in R$ and if $r_1,r_2 \in R,$ then $f(r_1+I)=f(r_2+I)$ if and only if $r_1z=r_2z=0$ if and only if $(r_1-r_2)z=0$ if and only if $r_1-r_2 \in I$ if and only if $r_1+I=r_2+I.$ Thus

$|R/I| \le |Z|. \ \ \ \ \ \ \ \ \ \ (2)$

It now follows from $(1),(2)$ that

$|R|=|I| \cdot |R/I| \le |Z|^2. \ \Box$

Example 1. The equality in the above proposition is possible, i.e. we can have $|R|=|Z|^2.$ To see that, consider the commutative ring $R:=\mathbb{Z}/p^2\mathbb{Z},$ where $p$ is a prime. It’s easy to see that the set of zero-divisors of $R$ is $\{np+p^2\mathbb{Z}: \ \ n=0,1, \cdots , p-1\}.$ So $R$ has exactly $p$ zero divisors and clearly $|R|=p^2.$

Example 2. An infinite field has only one zero-divisor, which is $0.$ That means the condition $|Z| > 1$ in the above proposition cannot be omitted.

Proposition 2. Let $R$ be a finite ring with $1.$ Let $Z, U$ be, respectively, the set of left zero divisors and the set of units of $R.$ Then

i) $U=R \setminus Z$

ii) if $R$ is not a division ring, then $|U| \le |R|-\sqrt{|R|}.$

Proof. i) Let $r \in R \setminus Z.$ Then the map $f: R \longrightarrow rR$ defined by $f(x)=rx$ is one-to-one and hence $R=rR$ because $R$ is finite and $rR \subseteq R.$ Thus $rx=1$ for some $x \in R$ and hence $r(xr-1)=0,$ which gives $xr=1.$ So we have shown that $R \setminus Z \subseteq U.$ Thus $U=R \setminus Z$ because obviously $Z \cap U=\emptyset.$

ii) Since $R$ is not a division ring, $U(R) \ne R \setminus \{0\}$ and so $Z$ contains a non-zero element. Therefore, by Proposition 1, $|R| \le |Z|^2$ and hence $|Z| \ge \sqrt{|R|}.$ Thus, by i), $|U|=|R|-|Z| \le |R|-\sqrt{|R|}. \ \Box$

Remark. The results given in Proposition 1 and Proposition 2, i), still hold if we replace “left zero divisors” with “right zero-divisors”.

Example 3. Let $n > 1$ be an integer which is not prime. Let $\phi$ be the Euler’s totient function. We now use Proposition 2 to prove that $\phi(n) \le n-\sqrt{n},$ which is a well-known result in number theory. To do that, we consider the commutative ring $R:=\mathbb{Z}/n\mathbb{Z}.$ Let $U$ be the set of units of $R.$ It’s clear that $|U|=\phi(n).$ Thus, since $R$ is not a field (because $n$ is not prime), we have from Proposition 2, ii), that

$\phi(n)=|U| \le |R|-\sqrt{|R|}=n-\sqrt{n}.$

Example 4. Using Proposition 2, we show that for any prime $p$ and integers $m \ge 1, n > 1,$

$\displaystyle \prod_{i=1}^n(p^{mn}-p^{(i-1)m}) \le p^{mn^2}-p^{mn^2/2}.$

To prove the above, let $F$ be the finite field of order $p^m$ and consider the ring $R:=M_n(F),$ the ring of $n \times n$ matrices with entries from $F.$ Let $U$ be the set of units of $R.$ It’s clear that $|R|=|F|^{n^2}=p^{mn^2}$ and also, as we proved in Problem 3 in this post, $\displaystyle |U|=\prod_{i=1}^n(|F|^n-|F|^{i-1})=\prod_{i=1}^n(p^{mn}-p^{(i-1)m}).$ So, since $R$ is not a division ring (because $n > 1$), the result follows from Proposition 2, ii).

## Group actions: orbit, stabilizer, and the orbit-stabilizer theorem; some examples

Posted: January 17, 2021 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

See the first part of this post here and the second part here.

Example 1. By Example 1 in the first part of this post, $G$ acts on itself by left multiplication, i.e. $g_1 \cdot g_2=g_1g_2,$ for all $g_1,g_2 \in G,$ where $g_1g_2$ is the usual multiplication in $G.$ Now if $x \in G,$ then

$\text{orb}(x)=\{g \cdot x: \ \ g \in G\}=\{gx: \ \ g \in G\}=G$

and

$\text{stab}(x)=\{g \in G: \ \ g \cdot x=x\}=\{g \in G: \ \ gx=x\}=(1).$

So if $G$ is finite, then $\displaystyle |G|=|\text{orb}(x)|=\frac{|G|}{|\text{stab}(x)|},$ which is the orbit-stabilizer theorem (see the fourth part of the theorem in the second part of this post).

Example 2. Let $H$ be a normal subgroup of $G.$ By Example 2 in the first part of this post, $G$ acts on $H$ by conjugation, i.e. $g \cdot h=ghg^{-1}$ for all $g \in G, \ h \in H.$ Now let $h \in H$ and let $\text{Cl}(h), \ \text{C}(h)$ be the conjugacy class and the centralizer of $h$ in $G,$ respectively. Then

$\text{orb}(h)=\{g \cdot h: \ \ g \in G\}=\{ghg^{-1}: \ \ g \in G\}=\text{Cl}(h)$

and

\begin{aligned}\text{stab}(h)=\{g \in G: \ \ g \cdot h=h\}=\{g \in G: \ \ ghg^{-1}=h\}=\{g \in G: \ \ gh=hg\}=C(h).\end{aligned}

Now suppose that $G$ is finite. Then the orbit-stabilizer theorem gives $\displaystyle |\text{Cl}(h)|=\frac{|G|}{|C(h)|}.$ It is clear that $|\text{orb}(h)|=|\text{Cl}(h)|=1$ if and only if $|G|=|C(h)|$ if and only if $G=C(h)$ if and only if $h \in Z(G),$ the center of $G.$ Now, by the second part of the theorem in the second part of this post, $\{\text{orb}(h): \ \ \ h \in H\}$ is a partition of $H.$ So if $\text{orb}(h_1), \cdots , \text{orb}(h_n),$ where $h_1, \cdots , h_k \in Z(G),$ are all the distinct orbits, then

\displaystyle \begin{aligned}|H|=\sum_{j=1}^n|\text{orb}(h_j)|=\sum_{j=1}^n|\text{Cl}(h_j)|=\sum_{j=1}^n\frac{|G|}{|C(h_j)|}=\sum_{j=1}^k\frac{|G|}{|C(h_j)|}+\sum_{j=k+1}^n\frac{|G|}{|C(h_j)|}\end{aligned}

$\displaystyle =k+\sum_{j=k+1}^n\frac{|G|}{|C(h_j)|}=|H \cap Z(G)|+\sum_{j=k+1}^n\frac{|G|}{|C(h_j)|},$

which, as you know, is known as the class equation.

Example 3. Let $X$ be the set of all subgroups of $G.$ Then, by Example 4, i), in the first part of this post, $G$ acts on $X$ by conjugation. Let $H \in X$ and let $\text{Cl}(H)$ be the set of all subgroups of $G$ which are conjugate to $H.$ Let $N(H)$ be the normalizer of $H$ in $G.$ Then

$\text{orb}(H)=\{g \cdot H: \ \ g \in G\}=\{gHg^{-1}: \ \ g \in G\}=\text{Cl}(H)$

and

$\text{stab}(H)=\{g \in G: \ \ g \cdot H=H\}=\{g \in G: \ \ gHg^{-1}=H\}=N(H).$

Now suppose that $G$ is finite. Then the orbit-stabilizer theorem gives $\displaystyle |\text{Cl}(H)|=\frac{|G|}{|N(H)|}.$ It is clear that $|\text{orb}(H)|=1$ if and only if $|G|=|N(H)|$ if and only if $G=N(H)$ if and only if $H$ is normal in $G.$ Now, by the second part of the theorem in the second part of this post, $\{\text{orb}(H): \ \ \ H \in X\}$ is a partition of $X.$ So if $\text{orb}(H_1), \cdots , \text{orb}(H_n),$ where $\{H_1, \cdots , H_k\}$ is the set of normal subgroups of $G,$ are all the distinct orbits, then

\displaystyle \begin{aligned}|X|=\sum_{j=1}^n|\text{orb}(H_j)|=\sum_{j=1}^n|\text{Cl}(H_j)|=\sum_{j=1}^n\frac{|G|}{|N(H_j)|}=\sum_{j=1}^k\frac{|G|}{|N(H_j)|}+\sum_{j=k+1}^n\frac{|G|}{|N(H_j)|}\end{aligned}

$\displaystyle =k+\sum_{j=k+1}^n\frac{|G|}{|N(H_j)|}.$

Example 4. Let $p$ be a prime dividing $|G|$ and let $X:=\{P_1, \cdots , P_m\}$ be the set of all Sylow $p$-subgroups of $G.$ By Example 4, ii), in the first part of this post, $G$ acts on $X$ by conjugation. Now, we know that every two Sylow $p$-subgroups are conjugate. Thus $\text{orb}(P_1)=X$ and clearly $\text{stab}(P_1)=N(P_1),$ the normalizer of $P_1$ in $G.$ Thus, by the orbit-stabilizer theorem,

$\displaystyle m=|X|=\text{orb}(P_1)=\frac{|G|}{|N(P_1)|}.$

Example 5. Let $H,K$ be two subgroups of $G.$ By Example 9 in the first part of this post, the group $H \times K$ acts on $HK$ where the action is defined by $(h_1,k_1) \cdot (hk)=h_1hkk_1^{-1}$ for all $h,h_1 \in H, \ k,k_1 \in K.$ Now

$\text{orb}(1)=\{(h_1,k_1) \cdot 1: \ \ h_1 \in H, k_1 \in K\}=\{h_1k_1^{-1}: \ \ h_1 \in H, k_1 \in K\}=HK$

and

\begin{aligned}\text{stab}(1)=\{(h_1,k_1) \in H \times K: \ \ (h_1,k_1) \cdot 1=1\}=\{(h_1,k_1) \in H \times K: \ \ h_1k_1^{-1}=1\}\end{aligned}

$=\{(h_1,k_1) \in H \times K: \ \ h_1=k_1\}=\{(h,h): \ \ h \in H \cap K\}.$

So, if $H,K$ are both finite, then $|\text{orb}(1)|=|HK|$ and $|\text{stab}(1)|=|H \cap K|$ and hence, by the orbit-stabilizer theorem,

$\displaystyle |HK|=\frac{|H \times K|}{|H \cap K|}=\frac{|H| |K|}{|H \cap K|},$

which is a well-known result.

## Group actions: the orbit-stabilizer theorem

Posted: January 12, 2021 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , , ,

For the first part of this post, see here.

Throughout this post, $X$ is a set, $G$ is a group with identity element $1,$ and $G$ acts on $X.$

We now show that the action of $G$ on $X$ partitions $X$ into subsets called orbits and if $G$ is finite, then the number of elements of each orbit is finite and it divides the order of $G.$ But first a couple of definitions.

Definition. Let $x \in X.$ The orbit of $x,$ denoted $\text{orb}(x)$ or $G \cdot x,$ is defined by

$\text{orb}(x)=G \cdot x:=\{g \cdot x: \ \ \ g \in G\}$

and the stabilizer of $x,$ denoted $\text{stab}(x)$ or $G_x,$ is defined as follows

$\text{stab}(x)=G_x:=\{g \in G: \ \ \ g \cdot x=x\}.$

So, for any $x \in X, \ \text{orb}(x)$ is a subset of $X$ but $\text{stab}(x)$ is a subset of $G.$ In fact, $\text{stab}(x)$ is a subgroup of $G,$ as the third part of the following theorem shows.

Theorem. i) If $x_1,x_2 \in X$ and $\text{orb}(x_1) \cap \text{orb}(x_2) \neq \emptyset,$ then $\text{orb}(x_1) = \text{orb}(x_2).$

ii) The set $\{\text{orb}(x): \ \ x \in X\}$ is a partition of $X.$ In particular, if $X$ is finite and $\text{orb}(x_1), \cdots , \text{orb}(x_n)$ are all the distinct orbits, then $\displaystyle|X|=\sum_{j=1}^n|\text{orb}(x_j)|.$

iii) If $x \in X,$ then $\text{stab}(x)$ is a subgroup of $G.$

iv) (The Orbit-Stabilizer Theorem) Let $x \in X$ and put

$H:=\text{stab}(x), \ \ \ \ L=\{gH: \ \ g \in G\}.$

Then the map $\varphi : \text{orb}(x) \to L$ defined by $\varphi(g \cdot x)=gH$ is bijective. In particular, if $|L|=[G:H]$ is finite, then $\text{orb}(x)$ is finite too and $|\text{orb}(x)|=[G:H]$ and so if $G$ is finite, then $\displaystyle |\text{orb}(x)|=\frac{|G|}{|H|}.$

Proof. i) Let $x \in \text{orb}(x_1) \cap \text{orb}(x_2).$ Then $x=g_1 \cdot x_1=g_2 \cdot x_2$ for some $g_1,g_2 \in G$ and so, by the remark in the first part of this post, we have $x_1=(g_1^{-1}g_2) \cdot x_2$ and hence

$g \cdot x_1=g \cdot ((g_1^{-1}g_2) \cdot x_2)=(gg_1^{-1}g_2) \cdot x_2$

for all $g \in G.$ So $\text{orb}(x_1) \subseteq \text{orb}(x_2)$ and, by symmetry, we also have $\text{orb}(x_2) \subseteq \text{orb}(x_1),$ hence $\text{orb}(x_1) = \text{orb}(x_2).$

ii) Since $1 \cdot x=x,$ for all $x \in X,$ we have $x \in \text{orb}(x)$ for all $x \in X,$ and the result now follows from i).

iii) First notice that since $1 \cdot x=x,$ we have $1 \in \text{stab}(x)$ and so $\text{stab}(x) \neq \emptyset.$ Now let $g \in \text{stab}(x).$ Then, by the remark in the first part of this post, $g \cdot x=x$ gives $x=g^{-1} \cdot x$ and so $g^{-1} \in \text{stab}(x).$ Also, if $g_1,g_2 \in \text{stab}(x),$ then

$(g_1g_2) \cdot x=g_1 \cdot (g_2 \cdot x)=g_1 \cdot x=x$

and so $g_1g_2 \in \text{stab}(x),$ hence the result.

iv) For $g_1,g_2 \in G,$ we have, by the definition of $\varphi,$ that $\varphi(g_1 \cdot x)=\varphi(g_2 \cdot x)$ if and only if $g_1H=g_2H$ if and only if $g_1^{-1}g_2 \in H$ if and only if $g_1^{-1}g_2 \cdot x=x$ if and only if $g_1 \cdot x=g_2 \cdot x.$ So $\varphi$ is a well-defined injective map. Clearly $\varphi$ is surjective and so $\varphi$ is bijective. $\Box$

Remark. An important result of the orbit-stabilizer theorem, i.e. part iv) of the above theorem, is that if $G$ is finite and $x \in X,$ then $|\text{orb}(x)|$ divides $|G|.$ Also it follows from part ii) and part iv) of the above theorem that if $X,G$ are both finite and $\text{orb}(x_1), \cdots , \text{orb}(x_n)$ are all the distinct orbits, then

$\displaystyle |X|=\sum_{j=1}^n\frac{|G|}{|\text{stab}(x_j)|}.$

If $x \in X,$ then $\text{stab}(x)$ is always a subgroup of $G,$ by the third part of the above theorem. But when is $\text{stab}(x)$ normal in $G$ ? The following proposition answers this question.

Proposition. Let $x \in X$ and $H:=\text{stab}(x).$ Then

$\text{stab}(g \cdot x)=gHg^{-1}$

for all $g \in G$ and thus $H$ is normal in $G$ if and only if $H=\text{stab}(y)$ for all $y \in \text{orb}(x).$

Proof. Let $g \in G.$ Then $g_1 \in \text{stab}(g \cdot x)$ if and only if $g_1 \cdot (g \cdot x)=g \cdot x$ if and only if $(g_1g) \cdot x = g \cdot x$ if and only if $g^{-1} \cdot ((g_1g) \cdot x)=x$ if and only if $(g^{-1}g_1g) \cdot x=x$ if and only if $g^{-1}g_1g \in H$ if and only if $g_1 \in gHg^{-1}.$ So we have shown that $\text{stab}(g \cdot x)=gHg^{-1}$ for any $g \in G.$ Hence $H$ is normal in $G$ if and only if $H=gHg^{-1}=\text{stab}(g \cdot x)$ for all $g \in G.$ In other words, since $\text{orb}(x)=\{g \cdot x: \ \ \ g \in G\}, \ H$ is normal in $G$ if and only if $H=\text{stab}(y)$ for all $y \in \text{orb}(x). \ \Box$

## Group actions: definition & examples

Posted: January 12, 2021 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

Throughout this post, $X$ is a set and $G$ is a group with the identity element $1.$

Definition. An action of $G$ on $X$ is any group homomorphism $f: G \to \text{Sym}(X),$ where $\text{Sym}(X)$ is the group of permutations of $X.$ Then we say that $G$ acts on $X$ or $X$ is a $G$-set.

The following proposition says that $G$ acting on $X$ basically means that we can “multiply” any  element of $G$ by any element of $X$ and get an element of $X,$ of course provided that a couple of conditions are satisfied (just like rings and modules!).

Notation. For the sake of simplicity, from now on, given a map $G \times X \to X$ and $g \in G, \ x \in X,$ we denote the image of $(g,x)$ by $g \cdot x.$

Proposition. $G$ acts on $X$ if and only if there exists a map $G \times X \to X$ such that

i)   $1 \cdot x=x,$

ii)   $(g_1g_2) \cdot x=g_1 \cdot (g_2 \cdot x),$

for all $g_1,g_2 \in G, \ x \in X.$

Proof. Suppose first that $G$ acts on $X.$ So there exists a group homomorphism $f: G \to \text{Sym}(X).$ Now define a map $G \times X \to X$ by $g \cdot x=f(g)(x),$ for all $g \in G, \ x \in X.$ Then for all $x \in X$ and $g_1,g_2 \in G$ we have $1 \cdot x=f(1)(x)=\text{id}_X(x)=x$ and

$(g_1g_2) \cdot x=f(g_1g_2)(x)=(f(g_1)f(g_2))(x)=f(g_1)(f(g_2)(x))=g_1 \cdot (g_2 \cdot x).$

Conversely, suppose that there exists a map $G \times X \to X$ such that

$1 \cdot x=x, \ \ \ (g_1g_2) \cdot x=g_1 \cdot (g_2 \cdot x),$

for all $g_1,g_2 \in G, \ x \in X.$ Now define the map $f: G \to \text{Sym}(X)$ by $f(g)(x)=g \cdot x$ for all $g \in G$ and $x \in X.$ So we need to prove two things: $f$ is basically well-defined, i.e. $f(g) \in \text{Sym}(X)$ for all $g \in G,$ and $f$ is a group homomorphism.
Let $g \in G.$ We first show that $f(g)$ is injective. Suppose that $f(g)(x_1)=f(g)(x_2)$ for some $x_1,x_2 \in X.$ Then, by the definition of $f,$ we have $g \cdot x_1=g \cdot x_2$ and so

$x_1=1 \cdot x_1=(g^{-1}g) \cdot x_1=g^{-1} \cdot (g \cdot x_1)=g^{-1} \cdot (g \cdot x_2)=(g^{-1}g) \cdot x_2=1 \cdot x_2=x_2.$

So $f(g)$ is injective. Now we show that $f(g)$ is surjective. Let $x_2 \in X$ and put $x_1=g^{-1} \cdot x_2.$ Then

$f(g)(x_1)=g \cdot x_1=g \cdot (g^{-1} \cdot x_2)=(gg^{-1}) \cdot x_2=1 \cdot x_2=x_2$

and so $f(g)$ is surjective, hence $f \in \text{Sym}(X),$ i.e. $f$ is well-defined.
Finally, $f$ is a group homomorphism because for all $g_1,g_2 \in G$ and $x \in X$ we have

$f(g_1g_2)(x)=(g_1g_2) \cdot x=g_1 \cdot (g_2 \cdot x)=f(g_1)f(g_2)(x)$

and so $f(g_1g_2)=f(g_1)f(g_2). \ \Box$

Remark. If $G \times X \to X$ is an action of $G$ on $X,$ then $g_1 \cdot x_1=g_2 \cdot x_2$ for some $g_1,g_2 \in G, \ x_1,x_2 \in X$ if and only if $x_1=(g^{-1}g_2) \cdot x_2.$ That’s because, by the two properties of an action given in the above proposition, if $g_1 \cdot x_1=g_2 \cdot x_2,$ then

$x_1=1 \cdot x_1=(g_1^{-1}g_1) \cdot x_1=g_1^{-1} \cdot (g_1 \cdot x_1)=g_1^{-1} \cdot (g_2 \cdot x_2)=(g_1^{-1}g_2) \cdot x_2$

and if $x_1=(g_1^{-1}g_2) \cdot x_2,$ then

$g_1 \cdot x_1=g_1 \cdot ((g_1^{-1}g_2) \cdot x_2)=(g_1g_1^{-1}g_2) \cdot x_2=g_2 \cdot x_2. \ \Box$

Now we use the above proposition to give a few examples of group actions.

Example 1. $G$ acts on itself by (left) multiplication, i.e. the map $G \times G \to G$ defined by $g_1 \cdot g_2=g_1g_2,$ for all $g_1,g_2 \in G,$ where $g_1g_2$ is just the usual multiplication of elements $g_1,g_2 \in G,$ satisfies the two conditions given in the above proposition. That’s because, by the definition, $1 \cdot g=1g=g$ and

$(g_1g_2) \cdot g=g_1g_2g=g_1 \cdot (g_2g)=g_1 \cdot (g_2 \cdot g)$

for all $g,g_1,g_2 \in G.$

Example 2. Let $H$ be a normal subgroup of $G.$ Then $G$ acts on $H$ by conjugation, i.e. the map $G \times H \to H$ defined by $g \cdot h=ghg^{-1}$ for all $g \in G, \ h \in H,$ satisfies the two conditions given in the above proposition. To see  that, first note that the map is well-defined, i.e. $g \cdot h \in H,$ for all $g \in G, \ h \in H,$ because $H$ is normal. Also, we have $1 \cdot h=h$ and

$(g_1g_2) \cdot h=(g_1g_2)h(g_1g_2)^{-1}=g_1g_2hg_2^{-1}g_1^{-1}=g_1 \cdot (g_2hg_2^{-1})=g_1 \cdot (g_2 \cdot h),$

for all $g_1,g_2 \in G, \ h \in H.$

Example 3. Let $H, K$ be subgroups of $G$ and let $X:=\{aH: \ \ a \in G\}$ be the set of left cosets of $H$ in $G.$ Then $K$ clearly acts on $X$ by left multiplication, i.e. the map $K \times X \to X$ defined by $g \cdot aH=gaH,$ for all $g \in K, \ a \in G,$ satisfies the two conditions given in the above proposition.

Example 4. i) Let $X$ be the set of all subgroups of $G.$ Then $G$ acts on $X$ by conjugation, i.e. the map $G \times X \to X$ defined by $g \cdot H=gHg^{-1},$ for all $g \in G, \ H \in X,$ is an action of $G$ on $X.$

ii) Let $n \ge 1$ be a given integer and let $X$ be the set of all subgroups of order $n$ in $G.$ Then $G$ acts on $X$ by conjugation.

Example 5. Let $H$ be a characteristic subgroup of $G.$ Let $\text{Aut}(G)$ be the group of automorphisms of $G.$ Then the map $\text{Aut}(G) \times H \to H$ defined by $\sigma \cdot h=\sigma(h),$ where $\sigma \in \text{Aut}(G), \ h \in H,$ is clearly an action of $\text{Aut}(G)$ on $H.$ Note that the action is well-defined since $H$ is characteristic.

Example 6. Let $n$ be a positive integer, $k$ a field, $X$ the set of $n \times 1$ matrices with entries from $k,$ and $G$ the multiplicative group of $n \times n$ invertible matrices with entries from $k.$ Then $G$ acts on $X$ by left multiplication, i.e. the map $G \times X \to X$ defined by $g \cdot x=gx$ for all $g \in G, \ x \in X,$ where $gx$ is the usual matrix multiplication, satisfies the two conditions given in the above proposition.

Example 7. Let $n$ be a positive integer, $k$ a field, $X$ the set of $n \times 1$ matrices with entries from $k,$ and $G:=S_n,$ the permutation group of $\{1,2, \cdots ,n\}.$ Then $G$ acts on $X$ by permutation, i.e.

$\sigma \cdot [x_1, \cdots , x_n]^T=[x_{\sigma(1)}, \cdots, x_{\sigma(n)}]^T,$

for all $\sigma \in G, \ x_i \in k,$ defines an action of $G$ on $X.$

Example 8. i) If $G$ acts on $X_1,X_2,$ then $G$ also acts on $X_1 \times X_2$ if we define $g \cdot (x_1,x_2)=(g \cdot x_1, g \cdot x_2)$ for all $g \in G, \ x_1 \in X_1, \ x_2 \in X_2.$

ii) If $G$ acts on $X,$ then $G$ also acts on $X':=\{(x,y): X \times X: \ \ x \ne y\}$ if we define

$g \cdot (x,y)=(g \cdot x, g \cdot y)$

for all $g \in G, \ (x,y) \in X'.$ Note that the action is well-defined since $g \cdot x=g \cdot y$ if and only if $x=y$ (see the above Remark!).

Example 9. Let $H,K$ be subgroups of $G.$ The group $H \times K$ acts on $HK=\{hk: \ \ h \in H, \ k \in K\}$ where the action is defined by $(h_1,k_1) \cdot (hk)=h_1hkk_1^{-1}$ for all $h,h_1 \in H, \ k,k_1 \in K.$ To see that, we need to prove that the two conditions in the above proposition hold. Well, the first condition obviously holds because, by definition, $(1,1) \cdot (hk)=hk.$ We also have

$(h_1h_2,k_1k_2) \cdot (hk)=h_1h_2hk(k_1k_2)^{-1}=h_1h_2hkk_2^{-1}k_1^{-1}$

and

$(h_1,k_1) \cdot ((h_2,k_2) \cdot (hk))=(h_1,k_1) \cdot (h_2hkk_2^{-1})=h_1h_2hkk_2^{-1}k_1^{-1}.$

So $(h_1h_2,k_1k_2) \cdot (hk)=(h_1,k_1) \cdot ((h_2,k_2) \cdot (hk))$ for all $h,h_1,h_2 \in H, \ k,k_1,k_2 \in K$ and that proves of the second condition.

## The subsets {x^k: x \in G} of a finite group G

Posted: January 5, 2021 in Elementary Algebra; Problems & Solutions, Groups and Fields

Problem (Marian Andronache). For any finite group $G$ and positive integer $k,$ let

$P_k(G):=\{x^k: \ \ x \in G\}.$

Find all finite groups $G$ of order $\ge 4$ that satisfy the following condition

$|P_1(G)| > |P_2(G)| > \cdots > |P_{\lfloor |G|/2\rfloor}(G)|,$

where $\lfloor |G|/2 \rfloor$ is the integer part of $|G|/2.$

Solution. There are only three finite groups that satisfy the condition: $\mathbb{Z}_4, \ \mathbb{Z}_2 \times \mathbb{Z}_2,$ and $\mathbb{Z}_6.$ It’s easy to see that the groups $\mathbb{Z}_4, \ \mathbb{Z}_2 \times \mathbb{Z}_2$ satisfy the condition. So we may assume that $|G| \ge 5.$ But before we continue with the solution, we need the following simple result.

Claim. Let $G$ be a group of order $n.$ If $k \ge 1$ is an integer and $\gcd(k,n)=1,$ then $P_k(G)=P_1(G).$

Proof. Since $\gcd(k,n)=1,$ there exist integers $p,q$ such that $np+kq=1$ and so for any $x \in G,$ we have

$x=x^{np+kq}=x^{kq}=(x^q)^k \in P_k(G)$

and thus $G=P_1(G) \subseteq P_k(G) \subseteq G,$ which completes the proof.

Back to the problem, let $G$ be a group of order $n \ge 5$ that satisfies the condition in the problem. If $n$ is odd, then $\gcd(2,n)=1$ and hence $P_1(G)=P_2(G),$ by the above claim, and that contradicts the condition $|P_1(G)| > |P_2(G)|.$ So $n$ is an even number $\ge 6.$ Let $n=2^km,$ where $m$ is odd and $k \ge 1.$ If $k \ge 2,$ then $\gcd(2^{k-1}m-1,n)=1,$ and hence, by the above claim, $P_{2^{k-1}m-1}(G)=P_1(G),$ contradiction again. So $k=1$ and $n=2m.$ If $m > 3,$ then $\gcd(m-2,n)=1$ and so, by the above claim, $P_{m-2}(G)=P_1(G),$ contradiction. So $m=3$ and hence $n=6.$ There are only two groups of order $6$ : $\mathbb{Z}_6$ and $S_3.$ It’s easy to see that $\mathbb{Z}_6$ satisfies the condition and $S_3$ does not satisfy the condition. $\Box$