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!