Computation of 2nd homology using Hopf's formula Let $G$ be a group and $G'$ be a group obtained from $G$ by adding a one generator $x$ and relations $gx=xg$. 
That is, $G'=G\times\mathbb{Z}$ and $H_2(G')=H_2(G)$. 
Problem. Prove $H_2(G')=H_2(G)$ using Hopf's formula.
Suppose that we fix a presentation of $G$, say $G=F/R$. 
Then, $G'=F'/R'$, where $F'=F\ast \mathbb{Z}\langle x\rangle$ and $R'$ is a subgroup generated by $R$ and $[y,x]$ for all $y\in F$. 
Hopf's formula shows $H_2(G)=\frac{R\cap [F,F]}{[F,R]}$ and $H_2(G')=\frac{R'\cap [F',F']}{[F',R']}$.
Since $[y,x]\in [F',F']$ for all $y\in F$, $R'\cap [F',F']$ contains $R\cap [F,F]$ and all words of the form $[y,x]$. However, I could not see that why $[y,x]$'s lie in $[F',R']$.
Any help?

It follows from the Kunneth Theorem that $$H_2(G\times {\mathbb Z})=(H_2(G)\otimes H_0({\mathbb Z}))\oplus (H_1(G)\otimes H_1({\mathbb Z}))$$
$$=H_2(G)\oplus H_1(G)$$
So your statement is only true if $H_1(G)=0$.
