Problem. Let G be a finite group with |G| > 1 and suppose that G satisfies the following condition:

(*)    for every subgroup H of G there exists a group homomorphism f: G \to H such that f(h)=h, \ \ \ \forall h \in H.

Prove that G is isomorphic to a direct product of cyclic groups of prime orders.

Solution. We first prove a claim.

Claim. Every subgroup of Gsatisfies (*).

Proof. Let K be a subgroup of G and let H be any subgroup of K. We’re given that there exists a group homomorphism f: G \to H such that f|_H is the identity map. Then clearly g:=f|_K: K \to H is a group homomorphism and g|_H is the identity map. So K satisfies (*) too.

Now, back to the solution. One way to solve the problem is by induction over |G|. There’s nothing to prove if |G|=2. Now consider two cases.

Case 1: G is simple, i.e. G has no non-trivial normal subgroup. Suppose that G has a non-trivial subgroup H. Since G satisfies (*), there exists a group homomorphism f: G \to H such that f|_H is the identity map. Then K:=\ker f is a normal subgroup of G and so either K=(1) or K=G. If K=(1), then G=H and if K=G, then H=(1), which are both false because H is non-trivial. So G has no non-trivial subgroup and hence it’s a cyclic group of prime order. That solves the problem in this case.

Case 2: G is not simple. So G has a non-trivial normal subgroup H. Since G satisfies (*), there exists a group homomorphism f: G \to H such that f|_H is the identity map. Let K:=\ker f. Note that since H is non-trivial, K is non-trivial too. Now, since f is onto, G/K \cong H and so |G|=|K| \cdot |H|. Let x \in K \cap H. Then 1=f(x)=x and so K \cap H=(1). Thus |KH|=|K| \cdot |H|=|G| and hence G=KH. Therefore, since K,H are normal and K \cap H=(1), we have G \cong K \times H. The solution is now complete because, by the above claim, both H,K satisfy (*) and hence, by our induction hypothesis, both are a direct product of cyclic groups of prime orders and hence G \cong K \times H is also a direct product of cyclic groups of prime orders. \Box

Problem. Let f: \mathbb{R} \to \mathbb{R} be a continuous function and let G be the set of all anti-derivatives of f. Let c \in \mathbb{R} and define the binary operation *: G \times G \to G by u *v=u+v(c), for all u,v \in G. Show that (G,*) is an abelian group isomorphic to (\mathbb{R},+).

Solution. If u,v \in G, then u-v is a constant function (because both u,v are anti-derivatives of f) and so

u-v=u(c)-v(c). \ \ \ \ \ \ \ \ \ \ \ \ (\dagger)

1) * is well-defined because if u \in G, then (u+v(c))'=u'=f and so u*v=u+v(c) \in G.

2) * is commutative because, by (\dagger),


3) * is associative because It’s easy to see that (u*v)*w=u*(v*w)=u+v(c)+w(c) for all u,v,w \in G.

Now fix an element u_0 \in G.

4) u_0-u_0(c)=1_G because if u \in G, then


and so, by 2), we also have (u_0-u_0(c))*u=u, proving that u_0-u_0(c) is the identity element of G. Note that 1_G(c)=0.

5) u^{-1}=1_G-u(c) because (1_G-u(c))*u=1_G-u(c)+u(c)=1_G and we’re done by 2).

So we have shown that (G,*) is an abelian group.

To show that (G,*) \cong (\mathbb{R},+), we define the map

\phi:(G,*) \to (\mathbb{R},+), \ \ \ \ \ \ \ \phi(u)=u(c).

6) \phi is a group homomorphism because


7) \phi is injective because \phi(u)=\phi(v) gives u(c)=v(c) and so, by (\dagger), we have u=v.

And finally

8) \phi is surjective because if r \in \mathbb{R} and u:=1_G+r \in G, then \phi(u)=u(c)=1_G(c)+r=r. \ \Box

Problem. Let F be a finite field and let K be the subfield of F generated by the set \{x^3: \ \ x \in F\}. Show that either K=F or |F|=4.

Solution. Suppose that K \ne F. We have |F|=p^n for some prime number p and some positive integer n. So |K|=p^d for some d \mid n. Let F^{\times}, K^{\times} be the multiplicative groups of F,K, respectively. Since x^3 \in K^{\times} for all x \in F^{\times} and K^{\times} \ne F^{\times}, we have |F^{\times}/K^{\times}|=3. So

\displaystyle 3=|F^{\times}/K^{\times}|=\frac{|F^{\times}|}{|K^{\times}|}=\frac{p^n-1}{p^d-1}=p^{n-d}+p^{n-2d} + \cdots + p^d+1

and thus p=2, n-d=d=1 because p \ge 2. So |F|=4 and the solution is complete. \Box

Exercise. Let F be the field of order 4, and let K be the subfield of F generated by \{x^3: \ x \in F\}. Show that |K|=2.

Problem. Let A,B be two non-zero elements of M_n(\mathbb{C}) such that AB-BA=c(A-B) for some 0 \ne c \in \mathbb{C}. Show that A,B have the same eigenvalues.

Solution. Let I \in M_n(\mathbb{C}) be the identity matrix. Let p(x),q(x) \in \mathbb{C}[x] be the characteristic polynomials of A,B, respectively. See that AB-BA=c(A-B) gives


and thus, taking the determinant, we get


Therefore r(x+c)=r(x), where \displaystyle r(x)=\frac{p(x)}{q(x)}, i.e. c \ne 0 is a period of the rational function r(x), which is non-sense unless r(x) is constant. So r(x) is a constant function and hence, since p(x),q(x) are both monic, p(x)=q(x) and the result follows. \Box

Exercise. Let f be a non-constant rational function over \mathbb{C}. Show that f cannot be periodic. Does that remain true if we replace \mathbb{C} with an arbitrary field?

In this post, we take a look at the rings R with 1 that satisfy the following property

\forall a \in R \setminus \{0\}, \ \exists x \in R: \ \ \ ax+xa=1. \ \ \ \ \ \ \ \ \ \ \ \ (*)

In other words, rings in which the equation ax+xa=1 has a solution for every 0 \ne a \in R.

Remark 1. It is clear that a commutative ring satisfies (*) if and only if it is a field of characteristic \ne 2 because in this case (*) becomes 2ax=1. It is also clear that any division ring R of characteristic \ne 2 satisfies (*) because if 0 \ne a \in R and x=2^{-1}a^{-1}, then ax+xa=1.

Remark 2. The equation ax+xa=1 could have more than one solution in a ring. For example, in the division ring of real quaterninos \mathbb{H}=\{\alpha+\beta i + \gamma j + \delta k, \ \ \alpha, \beta, \gamma, \delta \in \mathbb{R}\}, choose a=i+j and x=\beta i-\left(\frac{1}{2}+\beta\right)j, where \beta is any real number, and see that ax+xa=1.
However, if a=\alpha+\beta i + \gamma j + \delta k \in \mathbb{H} and \alpha \ne 0, then the equation ax+xa=1 has the unique solution x=2^{-1}a^{-1} (why ?).

We now show that any ring that satisfies (*) is a simple domain of characteristic \ne 2.

Proposition. Let R be a ring and suppose that R satisfies (*). Then R is a simple domain and its center is a field of chracateristic \ne 2.

Proof. Let I \ne (0) be an ideal of R and choose 0 \ne a \in I. There exists x \in R such that ax+xa=1 and so 1=ax+xa \in I implying that I=R. So R is a simple ring and thus its center is a field (see Remark 1 in this post!). For a=1, there exists x \in R such that 2x=ax+xa=1 and so 2 \ne 0 in R, i.e. the characteristic of the center of R is \ne 2.

It remains to show that R is a domain. We’ll do that by first proving two claims.

Claim 1. If a,x \in R and ax+xa =1, then a^2x=xa^2.

Proof. a^2x=a(ax)=a(1-xa)=a-axa=(1-ax)a=(xa)a=xa^2. \ \Box

Claim 2. If a \in R and a^2=0, then a=0.

Proof.  Suppose, to the contrary, that there exists 0 \ne a \in R such that a^2=0 and let x \in R be such that ax+xa=1. Then

a=a(ax+xa)=axa, \ \ \ (ax)^2=ax \ \ \ \ \ \ \ \ \ \ \ \ (1)

and, as a result, ax \ne 0. So there exists y \in R such that

axy+yax=1. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

But then, by (1) and Claim 1, axy=(ax)^2y=y(ax)^2=yax and so by (2), \ \ 2axy=1 and hence a=2a^2xy=0, contradiction! \Box

We are now ready to show that R is a domain. Suppose that ab=0 for some a,b \in R and a \ne 0. To show that R is a domain, we just need to prove that b=0. We have (ba)^2=b(ab)a=0 and so, by Claim 2,

ba=0. \ \ \ \ \ \ \ \ \ \ \ \ (3)

Also, since a \ne 0, there exists x \in R such that ax+xa=1 and so


by (3). Hence, by Claim 2, b=0 and that completes the proof. \Box

Remark 3. Since an artinian domain is a division ring, it follows from the above problem and Remark 1 that an artinian ring satisfies (*) if and only if it is a division ring of characteristic \ne 2.

Question. Is it true that a ring R satisfies (*) if and only if R is a division ring of characteristic \ne 2 ? In other words, is there an example of a  ring that satisfies (*) but the ring is not a division ring?

Problem 1. Let G be a group with the center Z(G). Given an odd integer m \ge 1, let

H:=\{g \in G: \ \ g^m=1\}.

Show that if \displaystyle \frac{G}{Z(G)} is abelian, then H is a normal subgroup of G.

Solution. It is clear that 1 \in H and g \in H implies that g^{-1} \in H. Also, if x \in G and g \in H, then (xgx^{-1})^m=xg^mx^{-1}=xx^{-1}=1 and so xgx^{-1} \in H. Thus the only thing we need to prove is that H is multiplicatively closed. So let g_1,g_2 \in H. We want to show that g_1g_2 \in H. Let


Claim 1: g \in Z(G).

Proof. Since \displaystyle \frac{G}{Z(G)} is abelian, g_1^{-1}Z(G) and g_2^{-1}Z(G) commute, i.e. g_1^{-1}g_2^{-1}Z(G)=g_2^{-1}g_1^{-1}Z(G), and thus gZ(G)= g_1g_2g_1^{-1}g_2^{-1}Z(G)=Z(G).

Claim 2: g_1^kg_2=g_2g_1^kg^k for all integers k \ge 1.

Proof. By induction over k. By Claim 1, g and g_2g_1 commute and so


and that proves the claim for k=1. Now suppose the claim is true for k. Then


and that completes the induction.

Claim 3: g \in H.

Proof. By Claim 2, g^k=g_1^kg_2g_1^{-k}g_2^{-1} for all integers k \ge 1 and so, in particular, for k=m we get


Claim 4: (g_1g_2)^k=g_1^kg_2^kg^{-k(k-1)/2} for all integers k \ge 1.

Proof. The proof is by induction over k. There’s nothing to prove for k=1. Suppose now that the claim is true for k. By Claim 2, g_2g_1^k=g_1^kg_2g^{-k} and thus



which completes the induction and the proof of the claim.

Now, if in Claim 4, we put k=m, then we get

(g_1g_2)^m=g_1^mg_2^mg^{-m(m-1)/2}=g^{-m(m-1)/2}, \ \ \ \ \ \ \ \ \ \ (*)

because g_1,g_2 \in H. But since m is odd, m divides m(m-1)/2 and hence g^{-m(m-1)/2}=1 because, by Claim 3, g \in H. Therefore, by (*), \ (g_1g_2)^m=1, i.e. g_1g_2 \in H, and that completes the solution. \Box

Problem 2. Let G be a non-abelian group of order p^3, where p is a prime number. Let Z(G) be the center of G and let H:=\{g \in G: \ \ g^p=1\}. Show that

i) |Z(G)|=p

ii) if p is odd, then H is a normal subgroup of G and |H| \ge p

iii) if p=2, then H is not always a subgroup.

Solution. i) Since G is a non-abelian p-group, 1 < |Z(G)| < p^3=|G| and So either |Z(G)|=p or |Z(G)|=p^2. If |Z(G)|=p^2, then \displaystyle \frac{G}{Z(G)} would be a group of order p, hence cyclic. But then G would be abelian, which is non-sense. So |Z(G)|=p.

ii) By i), \displaystyle \frac{G}{Z(G)} is a group of order p^2 hence abelian. So, by Problem 1, H is a normal subgroup of G. It is clear that |H| \ge p because, by i), Z(G) \subseteq H. You may also argue that |H| > 1 because, by Cauchy, G has an element of order p. Thus since |H| divides |G|=p^3, we must have |H| \ge p.

iii) Consider D_8, the dihedral group of order 8

D_8=\langle a,b: \ \ \ a^2=b^4=1, \ \ aba=b^{-1}\rangle.

Then a^2=(ab)^2=1 but (aab)^2=(a^2b)^2=b^2 \ne 1, i.e. in this case, H is not closed under multiplication hence not a subgroup. \Box

Exercise. Let R be a commutative ring with unity and let G be the Heisenberg group over R. Show that for any odd integer m \ge 1, the set \{g \in G: \ \ g^m=1\} is a normal subgroup of G.

Let G be a group and let H be a subgroup of H. Let g \in G. If H is normal, then gHg^{-1}=H. If H is not normal but gHg^{-1} \subseteq H and H is finite, then again gHg^{-1}=H because |gHg^{-1}|=|H|. But what if H is not normal, gHg^{-1} \subseteq H and H is not finite? Would that again imply gHg^{-1}=H ? Of course not, here is an example.

Example. Consider the group G:=\text{GL}(2,\mathbb{Q}), the group of 2 \times 2 invertible matrices with rational entries. For any given k \ge 1, consider the set

\displaystyle H_k:=\begin{pmatrix}1 & k\mathbb{Z} \\ 0 & 1\end{pmatrix}.

So H_k is the set of all 2 \times 2 matrices in the form \displaystyle \begin{pmatrix}1 & kn \\ 0 & 1\end{pmatrix} for some integer n. See that H_k is a subgroup of G. Now let g:=\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix} \in G and see that

\displaystyle gH_kg^{-1}=\begin{pmatrix}1 & 2k\mathbb{Z} \\ 0 & 1\end{pmatrix}.

So gH_kg^{-1} \subset H_k but gH_kg^{-1} \ne H. \ \Box

Definition. Let R be a ring. An element x \in R is said to be right quasi-regular if there exists y \in R such that x+y=xy. Similarly, x is said to be left quasi-regular if x+y=yx for some y \in R.

Remark 1. If R has 1, then x+y=xy is equivalent to (1-x)(1-y)=1. So, in this case, x is right quasi-regular if and only if 1-x has a right inverse. Similarly, x is left quasi-regular if and only if 1-x has a left inverse.

Remark 2. If R is a ring and x \in R is (right) quasi-regular, then x^2 is not necessarily (right) quasi-regular. For example, choose R=\mathbb{Z}_3 and x=-1. The converse however is always true, as the following problem shows.

Problem. Let R be a ring and let x \in R. Show that if x^n is (right) quasi-regular for some integer n \ge 2, then x is (right) quasi-regular too.

Solution. So there exists y \in R such that



I:=\{xr-r: \ \ \  r \in R\}.

See that I is a right ideal of R. To solve the problem, we only need to show that x \in I. In I we put r=x^{k-1}y, where k \ge 1 is any integer, to get x^ky-x^{k-1}y \in I (note that for k=1, the term x^{k-1}y simply means y not 1 \cdot y, which is not necessarily meaningful since R may not have 1). Also, if in I we put r=x^{k-1}, where k \ge 2 is any integer, we get x^k-x^{k-1} \in I. Thus

\displaystyle \begin{aligned}x^{n-1}=x^n-(x^n-x^{n-1})=x^ny-y-(x^n-x^{n-1})=\sum_{k=1}^n(x^ky-x^{k-1}y)-(x^n-x^{n-1}) \in I\end{aligned}

and hence

\displaystyle x=x^{n-1}-\sum_{k=2}^{n-1}(x^k-x^{k-1}) \in I. \ \Box

Problem. Let G be a group and suppose that |G| > 2. Show that |\text{Aut}(G)| > 1, i.e. G has a non-trivial group automorphism.

Solution. We consider three cases.

Case 1: G is non-abelian. In this case, let Z(G) be the center of G and choose g \in G \setminus Z(G). Then the map f: G \to G defined by f(x)=gxg^{-1} is clearly an automorphism of G and it is non-trivial because if f(x)=x for all x \in G, then gx=xg for all x \in G and so g \in Z(G), which is not the case.

Case 2: G is abelian and there exists a \in G such that a^2 \ne 1. In this case, the map f: G \to G defined by f(x)=x^{-1} is clearly an  automorphism of G and it is non-trivial because f(a) \ne a.

Case 3: G is abelian and x^2 = 1 for all x \in G. In this case, looking at G as an additive group, the condition x^2=1 becomes 2x=0  and so we may consider G as a vector space over the field F:=\mathbb{Z}_2.
Since |G| > 2, we have \dim_F G > 1. Let

B:=\{x_i: \ \ i \in I\}

be a basis for G. Since |B| \ge 2, we may choose two distinct elements x_i, x_j \in B. Now the map f: G \to G defined on B by f(x_i)=x_j, \ f(x_j)=f(x_i) and f(x_k)=x_k, \ \ k \in I \setminus \{i,j\}, and then  extended to the entire G lineraly, is clearly a non-trivial automorphism of G. \ \Box

Problem (Miklós Schweitzer Competition, 2019). Let R be a noncommutative finite ring with multiplicative identity element 1. Show that if the subring generated by I \cup \{1\} is R for each nonzero two-sided ideal I, then R is simple.

Solution. I show that, more generally, the result holds true for any noncommutative (left) artinian ring R. Let Z be the center of R. For any two-sided ideal I \ne (0) of R, the subring Z+I contains I \cup \{1\} and so

Z+I=R. \ \ \ \ \ \ \ \ \ (*)

As a result, non-zero ideals of R are noncommutative because R is noncommutative.
We now show that Z is a domain. Suppose, to the contrary, that ab=0 for some a,b \in Z \setminus \{0\}. Then, by (*), we have Z+Ra=R and so Zb=Rb implying that Rb \ne (0) is a commutative ideal of R, contradiction. So Z is a domain.
Now let J be the Jacobson radical of R. We show that J=(0). So suppose, to the contrary, that J \ne (0). Since R is artinian, J is nilpotent. Thus there exists the smallest integer n \ge 2 such that J^n=(0). Let x \in J^{n-1} \setminus \{0\}. We have Z+J=R, by (*), and so Rx=xR=Zx, i.e. Rx \ne (0) is a commutative ideal of R and that’s a contradiction. So J=(0) and thus, by the Artin-Wedderburn’s theorem, R is a finite direct product of some matrix rings over division rings

R \cong \prod_{i=1}^kM_{n_i}(D_i).

Let F_i be the center of D_i. Then Z \cong \prod_{i=1}^k F_i which is possible only if k=1 because, as we showed, Z is a domain. So R \cong M_n(D) for some integer n and some division ring D and hence R is simple. \Box

Remark. The above result is not true for commutative artinian rings. For example, let R:=\mathbb{Z}/n\mathbb{Z}, where n > 1 is not prime. Clearly R is not simple and the only subring of R that contains 1 is R itself. So the subring generated by I \cup \{1\} is R for any non-zero ideal I of R.