Maximum order of abelian subgroups in a symmetric group

Posted: April 21, 2012 in Elementary Algebra; Problems & Solutions, Groups and Fields
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For any set $X,$ we denote by ${\rm{Sym}}(X)$ the group of bijective maps from $X$ to $X.$

Problem. Let $\alpha : \mathbb{N} \longrightarrow [1, \infty)$ be any map which satisfies the following two conditions: $\alpha(p) \geq p$ and $\alpha(p+q) \geq \alpha(p) \alpha(q)$ for all $p,q \in \mathbb{N}.$ Let $X$ be a set with $|X|=n.$ Prove that if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq \alpha(n).$

Solution.  The proof is by induction on $n.$ If $n=1,$ then $|G|=1 \leq \alpha(1).$ Let $X$ be a set with $|X|= n \geq 2$ and suppose that the claim is true for any set of size $< n.$ Let $G$ be an abelian subgroup of ${\rm{Sym}}(X).$ Clearly $gx=g(x), \ g \in G, x \in X,$ defines an action of $G$ on $X.$ Let $X_1, \ldots , X_k$ be the orbits corresponding to this action and consider two cases.

Case 1. $k=1$: Fix an element $x_1 \in X.$ Then $X=X_1=Gx_1.$ Suppose that $g_1x_1=x_1$ for some $g_1 \in G$ and let $x \in X.$ Then $x=gx_1$ for some $g \in G.$ Thus, since $G$ is abelian, we have

$x=gx_1=gg_1x_1=g_1gx_1=g_1x.$

Hence $g_1x=x$ for all $x \in X$ and thus $g_1=1.$ So the stabilizer of $x_1$ is trivial and therefore, by the orbit-stabilizer theorem, $|G|=|X|=n \leq \alpha(n).$

Case 2$k \geq 2$: Let $|X_i|=n_i, \ i=1,2, \ldots, k.$ Clearly $\sum_{i=1}^k n_i=n$ and, since $k \geq 2,$ we have $n_i < n$ for all $i.$ For every $g \in G$ and $1 \leq i \leq k$ let $g_i=g|_{X_i},$ the restriction of $g$ to $X_i,$ and put

$G_i=\{g_i: \ g \in G\}.$

Then $g_i \in {\rm{Sym}}(X_i)$ and $G_i$ is an abelian subgroup of ${\rm{Sym}}(X_i).$ Thus, by the induction hypothesis

$|G_i| \leq \alpha(n_i),$

for all $i.$ Now, define $\varphi : G \longrightarrow \bigoplus_{i=1}^k G_i$ by $\varphi(g)=(g_1, g_2, \ldots , g_k)$ for all $g \in G.$ It is obvious that $\varphi$ is one-to-one and so

$|G| \leq |\bigoplus_{i=1}^k G_i|=\prod_{i=1}^k |G_i| \leq \prod_{i=1}^k \alpha(n_i) \leq \alpha(\sum_{i=1}^k n_i)=\alpha(n). \ \Box$

Remark. The map $\alpha: \mathbb{N} \longrightarrow [1, \infty)$ defined by $\alpha(p)=3^{p/3},$ for all $p \in \mathbb{N},$ satisfies both conditions in the above Problem. So if $|X|=n$ and if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq 3^{n/3}.$

Degree of an irreducible factor of composition of two polynomials

Posted: November 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Let $F$ be a field and suppose that $f(x),g(x), p(x)$ are three polynomials in $F[x].$ Prove that if both $f(x)$ and $p(x)$ are irreducible and $p(x) \mid f(g(x)),$ then $\deg f(x) \mid \deg p(x).$

Solution. Let $\mathfrak{m}$ and $\mathfrak{n}$ be the ideals of $F[x]$ generated by $f(x)$ and $p(x),$ respectively. Let $E=F[x]/\mathfrak{m}$ and $L = F[x]/\mathfrak{n}.$ Since both $f(x)$ and $p(x)$ are irreducible, $E$ and $L$ are field extensions of $F.$ Now, define the map $\varphi : E \longrightarrow L$ by $\varphi(h(x) + \mathfrak{m})=h(g(x)) + \mathfrak{n},$ for all $h(x) \in F[x].$ We first show that $\varphi$ is well-defined. To see this, suppose that $h(x) \in \mathfrak{m}.$ Then $h(x)=f(x)u(x)$ for some $u(x) \in F[x]$ and hence $h(g(x)) = f(g(x))u(g(x)) \in \mathfrak{n},$ because $p(x) \mid f(g(x)).$ So $\varphi$ is well-defined. Now $\varphi$ is clearly a ring homomorphism and, since $E$ is a field and $\ker \varphi$ is an ideal of $E,$ we must have $\ker \varphi = \{0\}.$ Therefore we may assume that $F \subseteq E \subseteq L$ and hence $\deg p(x) = [L : F] = [L: E][E:F]=(\deg f(x))[L : E]. \ \Box$

Throughout $G$ is a group with the center $Z(G)$ and the commutator subgroup $G'.$ The goal is to prove that if $G/Z(G)$ is finite, then $G'$ is finite too. We will also find an upper bound for $|G'|$ in terms of $|G/Z(G)|.$

Notation. For any $a,b \in G,$ we define $[a,b]=aba^{-1}b^{-1}$ and $a^b = bab^{-1}.$

Problem 1. Let $a,b,c \in G.$

1) $[a,b]=b^ab^{-1}, \ a^a=a, \ ba=a^bb$ and $[a,b]^c=[a^c,b^c].$

2) $c[a,b]=[a^c,b^c]c.$

3) If $[G:Z(G)]=n,$ then $[a,b]^{n+1}=[a,b^2][a^b,b]^{n-1}.$

Proof. 1) The first three identities are trivial and the fourth one is true because the map $f:G \longrightarrow G$ defined by $f(g)=cgc^{-1}=g^c$ is a group homomorphism.

2) By 1) we have $[a^c,b^c]c =[a,b]^cc=c[a,b].$

3) So $g^n \in Z(G)$ for all $g \in G,$ because $[G:Z(G)]=n.$ So $[a,b]^nb^{-1}=b^{-1}[a,b]^n$ with 1) give us

$[a,b]^{n+1}=[a,b][a,b]^n =b^ab^{-1}[a,b]^n = b^a[a,b]^n b^{-1}=b^a[a,b][a,b]^{n-1}b^{-1}$

$=b^ab^ab^{-1}[a,b]^{n-1}b^{-1}= (b^2)^ab^{-2}b[a,b]^{n-1}b^{-1}=[a,b^2](b[a,b]b^{-1})^{n-1}=[a,b^2]([a,b]^b)^{n-1}$

$=[a,b^2][a^b,b^b]^{n-1}=[a,b^2][a^b,b]^{n-1}. \ \Box$

Problem 2. Let $C=\{[a,b]: \ a,b \in G\}.$ If $[G:Z(G)]=n,$ then $|C| \leq n^2.$

Proof. Define the map $\varphi : C \longrightarrow G/Z(G) \times G/Z(G)$ by $\varphi([a,b])=(aZ(G),bZ(G)).$ If we prove that $\varphi$ is one-to-one, we are done because then $|C| \leq |G/Z(G) \times G/Z(G)|=n^2.$ So suppose that $\varphi([a,b])=\varphi([c,d]).$ Then $aZ(G)=cZ(G)$ and $bZ(G)=dZ(G)$ and hence $a^{-1}c \in Z(G)$ and $b^{-1}d \in Z(G).$ Therefore

$[a,b]=aba^{-1}b^{-1}=ab(a^{-1}c)c^{-1}b^{-1} =a(a^{-1}c)bc^{-1}b^{-1}=cbc^{-1}b^{-1} =cbc^{-1}(b^{-1}d)d^{-1}$

$=cb(b^{-1}d)c^{-1}d^{-1}=cdc^{-1}d^{-1}=[c,d]. \ \Box$

Schur’s theorem. If $[G:Z(G)]=n,$ then $\displaystyle |G'| \leq n^{2n^3}.$

Solution. Let $C=\{[a,b]: \ a,b \in G\}$ and $c \in G'.$ Then $c = c_1c_2 \ldots c_m,$ where $c_i \in C.$ We will choose the integer $m$ to be as small as possible, i.e. if $c = c_1'c_2' \ldots c_k',$ with $c_i' \in C,$ then $k \geq m.$ Now, we know from Problem 2, that the number of elements of $C$ is at most $n^2.$ So if we prove that each $c_i$ can occur at most $n$ times in $c = c_1c_2 \cdots c_m,$ then $m \leq n|C| \leq n^3$ and thus there will be at most $(n^2)^m \leq n^{2n^3}$ possible values for $c$ and the problem is solved. To prove that each $c_i$ can occur at most $n$ times in $c = c_1c_2 \ldots c_m,$ suppose, to the contrary, that, say $c_j,$ occurs $r \geq n+1$ times in the product. Then by part 2) of Problem 1, we can move each $c_j$ to the right hand side of the product to get $c = c_1'c_2' \ldots c_s' c_j^r,$ where $c_i' \in C$ and $r + s = m.$ But by part 3) of Problem 1, $c_j^{n+1}$ is a product of $n$ elements of $C$ and hence, since $c_j^r=c_j^{r-(n+1)}c_j^{n+1},$ we see that $c_j^r$ is a product of $r-1$ elements of $C.$ Therefore $c$ is a product of $s+r-1=m-1$ elements of $C,$ which contradicts the minimality of $m. \ \Box$

If $G$ is finitely generated, then the converse of Schur’s theorem is also true, i.e. if $G'$ is finite, then $G/Z(G)$ is finite too. It’s not hard, try to prove it!

Partitions of a group and normal subgroups

Posted: October 9, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Let $G$ be a group, $N$ a normal subgroup of $G$ and $P=\{g_iN: \ i \in I \}$ the set of cosets of $N.$ Then $P$ is a partition of $G$ and $g_iN g_jN = g_ig_j N \in P$ for all $i,j \in I.$ We’d like to consider the converse of this.

Problem. Let $G$ be a group and suppose that $P$ is a set partition of $G$ which satisfies the following condition:

$(*)$ for every $Q_1,Q_2 \in P,$ there exists $Q \in P$ such that $Q_1Q_2 \subseteq Q.$

Let $N$ be the element of $P$ which contains the identity element of $G.$ Prove that $N$ is a normal subgroup of $G$ and $P$ is the set of cosets of $N$ in $G.$

Solution. Notice that an obvious result of $(*)$ is that if $Q_1 \in P$ and $g, h \in G,$ then $gQh \subseteq Q_2$ for some $Q_2 \in P.$ Now, if $a \in N,$ then $a = a \cdot 1 \in N^2$ and so $N \subseteq N^2.$ We also have $N^2 \subseteq Q$ for some $Q \in P,$ by $(*).$ Thus $N \subseteq Q$ and so $N=Q.$ Therefore $N=N ^2,$ i.e. $N$ is multiplicatively closed. Let $a \in N.$ Then $Na^{-1} \subseteq Q$ for some $Q \in P$ and since $1 \in Na^{-1},$ we get $1 \in N \cap Q.$ Thus $Q=N$ and so $Na^{-1} \subseteq N.$ Hence $a^{-1} \in N$ and so $N$ is a subgroup. To prove that $N$ is normal, let $g \in G.$ Then, by $(*),$ there exists $Q \in P$ such that $gNg^{-1} \subseteq Q.$ But then $1 \in Q \cap N,$ because $1 \in gNg^{-1} \subseteq Q,$ and so $Q=N.$ Thus $gNg^{-1} \subseteq N,$ i.e.  $N$ is normal. Finally, let $g \in G$ and choose $Q, Q_1 \in P$ such that $g \in Q$ and $gN \subseteq Q_1.$ Then $g \in Q_1 \cap Q,$ because $g \in Ng \subseteq Q_1,$ and so $Q_1=Q.$ Hence $g N \subseteq Q.$ Let $Q_2 \in P$ be such that $N \subseteq g^{-1}Q \subseteq Q_2.$ Then $1 \in Q_2 \cap N,$ because $1 \in g^{-1}Q \subseteq Q_2,$ and so $Q_2=N.$ Hence $gN=Q. \ \Box$

As usual, for a subgroup $H$ of a group $G,$ we will denote by $N(H)$ and $C(H)$ the normalizer and the centralizer of $H$ in $G.$ If $H = \langle a \rangle,$ then we will write $N(a)$ and $C(a)$ for $N(H)$ and $C(H)$ respectively.

Problem. Let $p$ be a prime number and let $G$ be a group of order $p^n(p+1), \ n \geq 1.$ Prove that $G$ is not simple.

Solution. If $G$ is abelian, there is nothing to prove. So we suppose that $G$ is non-abelian and simple and we will get a contradiction. By Sylow theorem, the number of Sylow $p$-subgroups is $p+1$ and so $P=N(P)$ for every Sylow $p$-subgroup $P$ of $G.$ Now, we consider two cases.

Case 1 . $n=1.$ Let $P$ be a Sylow $p$-subgroup. Then $|P|=p$ and so $P \subseteq C(P).$ Suppose that $P \neq C(P)$ and choose $a \in C(P) \setminus P.$ Then $P \subseteq C(a).$ Let $Q=aPa^{-1}.$ We have $P \cap Q = \{1\}$ because $|P|=|Q|=p$ and $a \notin P=N(P).$ Thus $PQ \subseteq C(a)$ and so $|C(a)| \geq |PQ|=p^2$ which implies that $C(a)=G$ because $|C(a)| \mid |G|=p(p+1).$ Thus $\langle a \rangle$ is in the center of $G$ and hence it’s a normal subgroup of $G,$ contradicting our assumption that $G$ is simple. Therefore $N(P)=P=C(P)$ and we are now done by the Burnside’s normal complement theorem.

Case 2.   $n \geq 2.$ The idea for this case is similar to the one we used for case 2 in this problem. Let $P$ and $Q$ be two distinct Sylow $p$-subgroups of $G$ and put $H=P\cap Q.$ Then

$\displaystyle \frac{p^{2n}}{|H|} = |PQ| \leq |G|=p^n(p+1)$

which gives us $|H|=p^{n-1}$ because $|H| \mid |G|=p^n(p+1).$ As a result, $H$ is a non-trivial normal subgroup of both $P$ and $Q.$ Therefore $PQ \subseteq N(H)$ and so $|N(H)| \geq |PQ|=p^{n+1}.$ But $|N(H)| \mid |G|=p^n(p+1)$ and so $N(H)=G,$ i.e. $H$ is a normal subgroup of $G.$ This contradicts our assumption that $G$ is simple. $\Box$

Irreduciblity of a polynomial over a field extension

Posted: September 16, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Suppose that $p(x)$ is an irreducible polynomial over some field $F.$ Let $E/F$ be a finite field extension. The following problem investigates the irreducibility of $p(x)$ over $E.$

Problem. Suppose that $E/F$ is a finite field extension and $p(x) \in F[x].$ Prove that if $p(x)$ is irreducible over $F$ and $\gcd(\deg p(x), [E:F])=1,$ then $p(x)$ is irreducible over $E.$

Solution. We may assume, without loss of generality, that $p(x)$ is monic. Let $a$ be a root of $p(x)$ in some extension of $E$ and let $q(x)$ be the minimal polynomial of $a$ over $E.$ Clearly

$\deg q(x) \leq \deg p(x). \ \ \ \ \ \ \ (1)$

We also have

$[E(a):E)][E:F]=[E(a):F(a)][F(a):F]. \ \ \ \ \ \ (2)$

It now follows from $(2)$ and $\gcd(\deg p(x), [E:F]) = 1$ that $\deg p(x) \mid \deg q(x)$ and so by $(1)$

$\deg p(x)=\deg q(x). \ \ \ \ \ \ \ (3)$

Let $r(x)=q(x)-p(x) \in E[x].$ Then $\deg r(x) < \deg q(x)$ by $(3)$ and the fact that $p(x)$ and $q(x)$ are monic. We also have $r(a)=0$ and thus $r(x)=0$ by the minimality of $q(x).$ Hence $q(x)=p(x). \ \Box$

Note that it is possible for $p(x)$ to be irreducible over $E$ but $\gcd(\deg p(x), [E:F]) \neq 1.$ For example consider $p(x)=x^2+1, \ F = \mathbb{Q}$ and $E = \mathbb{Q}(\sqrt{2}).$

A finite non-abelian group in which every proper subgroup is abelian is not simple

Posted: June 8, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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The symmetric group $S_3$ is an example of a finite non-abelian group in which every proper subgroup is abelian. This group is not simple because its Sylow 3-subgroup is normal. In this post, we’ll show that this is the case for any finite (non-abelian) group all of whose proper subgroups are abelian.

Notation. Let $G$ be a group and let $H$ be a subgroup of $G.$ We will denote by $N(H)$ the normalizer of $H$ in $G.$ We will also define $\mathcal{C}(H)$ to be the union of conjugates of $H,$ i.e. $\mathcal{C}(H)=\bigcup_{g \in G} gHg^{-1}.$

Lemma 1. Let $G$ be a finite group and let $H$ be a subgroup of $G$ with this property that $gHg^{-1} \cap H = \{1\},$ for all $g \notin H.$ Then
1) the intersection of two distinct conjugates of $H$ is the identity element;
2) $|\mathcal{C}(H)|=|G|-[G:H] + 1$ and thus $|G \setminus \mathcal{C}(H)|=[G:H]-1.$

Proof. There is nothing to prove if $H=\{1\}.$ So suppose that $H \neq \{1\}.$ Clearly $H=N(H)$ because if $g \in N(H) \setminus H,$ then $H=gHg^{-1} \cap H = \{1\},$ which is a contradiction. Thus the number of conjugates of $H$ is $[G:N(H)]=[G:H].$
1) Suppose that $g_iHg_i^{-1}, \ i=1,2,$ are two distinct conjugates of $H$ and $g \in g_1Hg_1^{-1} \cap g_2Hg_2^{-1}.$ Then $g=g_1h_1g_1^{-1}=g_2h_2g_2^{-1},$ for some $h_1,h_2 \in H$ and hence $h_2 \in g_2^{-1}g_1 H g_1^{-1}g_2 \cap H.$ So, by hypothesis, either $h_2=1$ or $g_2^{-1}g_1 \in H.$ If $g_2^{-1}g_1 \in H,$ then $g_1Hg_1^{-1}=g_2Hg_2^{-1},$ which is a contradiction. Thus $h_2=1$ and so $g=1.$
2) Since every conjugate of $H$ has $|H|$ elements and, by 1), two distinct conjugates of $H$ have only one element in common , we have

$|\mathcal{C}(H)|=[G:H](|H|-1)+1=|G|-[G:H]+1. \ \Box$

Lemma 2. Let $G$ be a group and let $H_1$ and $H_2$ be two abelian subgroups of $G.$ Let $H$ be the subgroup generated by $H_1$ and $H_2.$ Then $H_1 \cap H_2$ is a normal subgroup of $H.$

Proof. Since $H_1$ and $H_2$ are abelian, $H_1 \cap H_2$ is a normal subgroup of both $H_1$ and $H_2.$ Thus $H_1 \cap H_2$ is also a normal subgroup of $H$ because an element of $H$ is a finite product of elements of $H_1$ and $H_2. \ \Box$

Problem. Let $G$ be a finite non-abelian group in which every proper subgroup is abelian. Prove that $G$ is not simple.

Solution. Suppose, to the contrary, that $G$ is simple and let $H$ be a maximal subgroup of $G.$ Clearly $N(H)=H$ because $H \subseteq N(H)$ and $H$ is a maximal subgroup of $G.$ Let $g \notin H.$ Then $g \notin N(H)$ and so $gHg^{-1} \neq H.$ Hence the subgroup generated by $H$ and $gHg^{-1}$ is $G$ because it contains $H$ strictly. Therefore $gHg^{-1} \cap H$ is a normal subgroup of $G,$ by Lemma 2. Hence $gHg^{-1} \cap H= \{1\}$ and so, by Lemma 1

$|G \setminus \mathcal{C}(H)|=[G:H]-1. \ \ \ \ \ \ \ \ (*)$

In particular $\mathcal{C}(H) \neq G$ and so we can choose $a \notin \mathcal{C}(H).$ Let $K$ be a maximal subgroup of $G$ which contains $a.$ Then again, exactly as we proved for $H,$ we have $gKg^{-1} \cap K = \{1\}$ and so , by Lemma 1

$|\mathcal{C}(K)|=|G|-[G:K]+1. \ \ \ \ \ \ \ \ \ (**)$

Now let $g_1,g_2 \in G.$ Note that $g_2Hg_2^{-1}$ is a maximal subgroup of $G$ because $H$ is a maximal subgroup of $G.$ Since $a \in K$ and $a \notin \mathcal{C}(H),$ we have $g_1Kg_1^{-1} \neq g_2Hg_2^{-1}$ and thus the subgroup generated by $g_1Kg_1^{-1}$ and $g_2Hg_2^{-1}$ is $G.$ Therefore $g_1Kg_1^{-1} \cap g_2Hg_2^{-1}$ is a normal subgroup of $G,$ by Lemma 2. Hence $g_1Kg_1^{-1} \cap g_2Hg_2^{-1}=\{1\}$ and so $\mathcal{C}(H) \cap \mathcal{C}(K)=\{1\}.$ Thus we have proved that

$\mathcal{C}(K) \setminus \{1\} \subseteq G \setminus \mathcal{C}(H).$

It now follows from $(*)$ and $(**)$ and the above inclusion that

$|G| \leq [G:H]+[G:K]-1 \leq |G|/2 + |G|/2 - 1 =|G|-1,$

which is non-sense. This contradiction shows that $G$ is not simple. $\Box$

In fact, the result still holds for finite non-cyclic groups because a simple abelian group is cyclic (of prime order).