Archive for the ‘Groups and Fields’ Category

Definition 1. Let n be an integer. A group G is called nabelian if (xy)^n=x^ny^n for all x,y \in G.
In other words, G is n-abelian if the map f : G \to G defined by f(x)=x^n is a group homomorphism.

Definition 2. If G is both n-abelian and m-abelian, for some integers m,n, then G is also mn-abelian because then for x,y \in G we will have


So the set

\text{E}(G)=\{n \in \mathbb{Z}: \ \ (xy)^n=x^ny^n, \ \ \forall x,y \in G\},

i.e., the set of those integers n for which G is n-abelian, is a multiplicative subset of \mathbb{Z}. Clearly 0,1 \in \text{E}(G). The set \text{E}(G) is called the exponent semigroup of G.

Remark 1. Since (xy)^n=x(yx)^{n-1}y, we have (xy)^n=x^ny^n if and only if (yx)^{n-1}=x^{n-1}y^{n-1}. So a group G is n-abelian if and only if (yx)^{n-1}=x^{n-1}y^{n-1} or, equivalently, (xy)^{1-n}=x^{1-n}y^{1-n} for all x,y \in G. So a group G is n-abelian if and only if it is (1-n)-abelian. In other words, n \in \text{E}(G) if and only if 1-n \in \text{E}(G).

Example 1. Every abelian group is obviously n-abelian for all n. It is also clear that 2-abelian groups are abelian because xyxy=(xy)^2=x^2y^2=xxyy gives yx=xy. As the next two examples show, there exists a non-abelian n-abelian group for any n > 2.

Example 2. Let n \ge 3 be an odd integer and consider the Heisenberg group G:=H(\mathbb{Z}/n\mathbb{Z}), which is a non-abelian group (why?). We show that n, n+1 \in \text{E}(G), i.e. G is both n-abelian and (n+1)-abelian. To see that, let

g= \displaystyle \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \in G.

An easy induction shows that

\displaystyle g^m= \begin{pmatrix}1 & ma & mb + \frac{m(m-1)}{2}ac \\ 0 & 1 & mc \\ 0 & 0 & 1 \end{pmatrix}

for all integers m \ge 1. Thus g^n=I, the identity matrix, and so g^{n+1}=g, for all g \in G.
Hence I=(xy)^n=x^ny^n and xy=(xy)^{n+1}=x^{n+1}y^{n+1} for all x,y \in G proving that G is both n-abelian and (n+1)-abelian.

Remark 2.  Notice that, in Example 2, if n is even, then \displaystyle \frac{n(n-1)}{2} \ne 0 (as elements of \mathbb{Z}/n\mathbb{Z}) and so g^n is not always the identity element in this case. However, there’s a way to fix this, as the next example shows. But first, notice that H(\mathbb{Z}/n\mathbb{Z}) can also be viewed as the group of  triples

G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}

with multiplication defined by


In the next example, we modify the above multiplication such that we get g^n=1 for all g \in G.

Example 3. Let n \ge 3 be any integer and consider the set G:=\{(a,b,c): \ \ a,b,c \in \mathbb{Z}/n\mathbb{Z}\}. Define multiplication in G by


It’s easy to see that (G,*) is a non-abelian group. We show that n, n+1 \in \text{E}(G). Let g=(a,b,c) \in G. A quick induction shows that

g^m=(ma, mb+m(m-1)ac, mc)

for all integers m \ge 1. So g^n=(0,0,0)=1_G for all g \in G and thus 1_G=(xy)^n=x^ny^n for all x,y \in G proving that G is n-abelian.
Also, since g^{n+1}=g for all g \in G, we have xy=(xy)^{n+1}=x^{n+1}y^{n+1} and so G is (n+1)-abelian.

Problem 1. Let G be a group. Show that

i) if n, n+1 \in \text{E}(G) for some integer n, then x^n \in Z(G) for all x \in G

ii) if n, n+1,n+2 \in \text{E}(G) for some integer n, then G is abelian.

Solution. i) Let x,y \in G. We have


and so yx^n=x^ny.

ii) Let x,y \in G. By i), we have yx^n=x^ny and yx^{n+1}=x^{n+1}y. Thus


and so xy=yx. \ \Box

Problem 2. Let G be an n-abelian group and let x,y \in G. Show that

i) x^{n-1}y^n=y^nx^{n-1}

ii) (xyx^{-1}y^{-1})^{n(n-1)}=1

Solution. i) By Remark 1, (ab)^{n-1}=b^{n-1}a^{n-1} for all a,b \in G. Thus

\begin{aligned} x^{n-1}y^{n-1}=(yx)^{n-1}=((yxy^{-1})y)^{n-1}=y^{n-1}(yxy^{-1})^{n-1}=y^{n-1}yx^{n-1}y^{-1}=y^nx^{n-1}y^{-1} \end{aligned}

and the result follows.

ii) Again, using the Remark 1, we have

\begin{aligned} (xyx^{-1}y^{-1})^{n(n-1)}=((x(yx^{-1}y^{-1}))^{n-1})^n=((yx^{-1}y^{-1})^{n-1}x^{n-1})^n=(yx^{-(n-1)}y^{-1}x^{n-1})^n \\ =y^n(x^{-(n-1)}y^{-1}x^{n-1})^n=y^nx^{-(n-1)}y^{-n}x^{n-1}=1, \ \ \ \text{by i)}.  \ \Box \end{aligned}

Remark 3. Let G be an n-abelian group for some integer n \ge 2. An immediate result of Problem 2, ii), is that if either G is torsion-free or G is finite and \text{gcd}(n(n-1), |G|)=1, then G is abelian.


Problem.  Show that if \displaystyle F \subseteq \mathbb{R} is a field and x_1, \cdots , x_n are positive real numbers such that \sum_{i=1}^n x_i \in F and x_i^2 \in F for all i, then x_i \in F for all i.

Solution. The proof is by induction over n. There’s nothing to prove for n=1. For n\ge 2, since \sum_{i=1}^n x_i \in F, we have

\displaystyle \sum_{i=2}^n x_i \in F(x_1)=F + Fx_1

and thus, by the induction hypothesis, x_i \in F+ Fx_1 for all i \ge 2. So x_2=a+bx_1 for some a,b \in F.

If b=0, then x_2 \in F and so \sum_{i \ne 2}x_i \in F. Hence, by the induction hypothesis, x_i \in F for all i.

If a=0, then x_2=bx_1 (so b > 0 because x_1,x_2 > 0). Thus

\displaystyle \sum_{i=1}^n x_i = (b+1)x_1 + \sum_{i \ge 3} x_i \in F

and so, by the induction hypothesis, x_i \in F for all i.

If a, b \ne 0, then x_2^2=a^2+b^2x_1^2+2abx_1 gives x_1 \in F and so \sum_{i \ge 2} x_i \in F implying, again by the induction hypothesis,that x_i \in F for all i. \ \Box

Example. Show that \mathbb{Q}(\sqrt{2}+ \sqrt{5}+ \cdots +\sqrt{n^2+1})=\mathbb{Q}(\sqrt{2}, \sqrt{5}, \cdots , \sqrt{n^2+1}). for all integers n \ge 1.

Solution. Let F:=\mathbb{Q}(x_1 + \cdots + x_n), where x_i=\sqrt{i^2+1} for all i. Since \sum_{i=1}^n x_i \in F and x_i^2 \in F for all i, we have, by the above problem, x_i \in F for all i. Thus \mathbb{Q}(x_1, \cdots , x_n) \subseteq F and we are done because obviously F \subseteq \mathbb{Q}(x_1, \cdots , x_n).

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


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 2k \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}.

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!

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 denote by N(H) and C(H) the normalizer and the centralizer of H in G. If H = \langle a \rangle, then we 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 |H|=p^{n-1} because |H| \mid p^n. 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