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!


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