Central extension of an Abelian group

Question

How to prove that in the group described by the presentation $$\langle a,b\,|\,a^{p^3}=b^{p^3}=[[a,b],a]=[[a,b],b]=1\rangle$$ the commutator $[a,b]$ has order $p^3$?

I cannot find a reasonable method to prove it, any ideas?

Answer 1

Denote the given group by $G$.

Notice that $$(C_{p^3}\times C_{p^3})\rtimes C_{p^3}=\langle x,a,b|x^{p^3}=a^{p^3}=b^{p^3}=1,x^a=x^b=1,a^b=ax\rangle$$ is generated by $a,b$ and satisfies the relations for $G$ so must be a quotient of $G$. Since $x=[a,b]$, this implies that the order of $[a,b]$ is at least $p^3$.

Now, $a^b=a[a,b]$ and $[a,b]^b=[a,b]$, so $a=a^{b^{p^3}}=a[a,b]^{p^3}$. Hence $[a,b]^{p^3}=1$ so $[a,b]$ must have order $p^3$.

In fact, this shows that all the relations defining $(C_{p^3}\times C_{p^3})\rtimes C_{p^3}$ are satisfied by $G$ so $G\cong (C_{p^3}\times C_{p^3})\rtimes C_{p^3}$.