3

Let $G$ be a non abelian group of order $p{^4}$,$p$ is a prime.Let $N$ be a normal subgroup of $G$ with |$N$|=$p$ and $G/N$ is abelian.Then prove that $N$ is a subgroup of $Z(G)$ and $Z(G)$/$N$ is a cyclic group.

So far what I get as following:|$Z(G)$|=$p$,$p{^2}$. Since $G$/$N$ is abelian $G{^1}$(Commutator subgroup of G) is contained in $N$.Again since $G$ is non abelian $G{^1}$=$N$.Now if |$Z(G)$|=$p{^2}$, |$G/Z(G)$|=$p{^2}$ so $G/Z(G)$ is abelian hence $G{^1}$ is contained in $Z(G)$.So we are done if |$Z(G)$|=$p{^2}$.I can not show $G{^1}$ =$Z(G)$ if |$Z(G)$|=$p$.Please help .thank you.

Panja
  • 107

1 Answers1

1

We know that in a nilpotent group every normal non-trivial subgroup contains a non-trivial central element, meaning:

$$(G\;\;\text{nilpotent and}\;\;1\neq N\lhd G)\implies N\cap Z(G)\neq 1$$

In our case, since $\;|N|=p\;$ (and thus cyclic, of course), we get that

$$\;1\neq N\cap Z(G)\le N\implies N\cap Z(G)=N\iff N\le Z(G)\;$$

and we're done ( even without dividing in the cases $\;|Z(G)|= p\;,\;\;|Z(G)|=p^2\;$), and in any case we have that $\;\left|Z(G)/N\right|=1\,,\,p\;$ , so the quotient is cyclic.

Timbuc
  • 34,191