Preimage of cyclic subgroup under projection is Abelian

Question

Let $G$ be a finite non-Abelian group, let $Z(G)$ denote its centre, let $\pi: G \to G/Z(G)$ be the canonical projection, and let $H \vartriangleleft G/Z(G)$ be a cyclic normal subgroup of the quotient group.

Is it true that $\pi^{-1}(H)$ is an Abelian normal subgroup of $G$?

Suppose $s,t \in \pi^{-1}(H)$. Then $\pi(t),\pi(s) \in H$, so that $\pi(st) = \pi(s) \pi(t) \in H$, and hence $st \in \pi^{-1}(H)$. Similarly $\pi(s^{-1}) = \pi(s)^{-1} \in H$, so $s^{-1} \in \pi^{-1}(H)$. Hence $\pi^{-1}(H)$ is a subgroup of $G$.

Normality: Suppose $s \in \pi^{-1}(H)$ and $t \in G$: Then $\pi(t^{-1}st) = \pi(t)^{-1}\pi(s)\pi(t) \in H$, as $H< G/Z(G)$ is normal. Hence, $t^{-1}st \in \pi^{-1}(H)$.

I'm having some trouble proving Abelianity. Note that the above demonstrations only used the fact that $\pi$ is a homomorphism, but I think here we are really going to need some special structure of the projection.

My idea was to use the fact that, as $H$ is cyclic, it is generated by some $h \in H$, and so, if $s,t \in \pi^{-1}(H)$, then $\pi(s) = h^x$ and $\pi(t) = h^y$ for some $x$ and $y$, and then somehow use the fact that $h^{x+y} = h^{y+x}$ to show that $\pi^{-1}(H)$ is Abelian. But I seem to be getting lost in the group theoretic finesse. I think we may need to use the fact that we are quotienting by $Z(G)$ - in particular, by an Abelian group.

I would much appreciate it if someone could point me in the right direction.

Many thanks.

Answer 1

let $\langle h Z(G) \rangle=\overline H \vartriangleleft G/Z(G)$.

If $ \ g_1,g_2 \in \pi^{-1}(\overline H) \ $ then $\pi(g_k)=h^{x_k}Z(G)$ , ($k=1,2$).

Also $\pi (g_k)=g_kZ(G)$

Hence we have,

$g_1=h^{x_1}z_1$ And $g_2=h^{x_2}z_2$

From here it follows, $g_1g_2=g_2g_1$

Answer 2

You can prove another claim: If $G/Z(G)$ is cyclic, then in fact $G$ is abelian.

Hint:

There are several ways to prove this. Note that $G/Z(G)$ is isomorphic to the group of innermorphisms on $G$. (Why?) So then $\text{Inn}(G)$ is cyclic with some generator $\gamma_a$; this means that for any $g \in G$ there is $n \in \mathbb Z$ such that $\gamma_g = \gamma_a^n$. Then evaluating $\gamma_g(a)$ tells us that $a$ commutes with every $g \in G$. (Why?) Hence $\gamma_g = \gamma_a^n$ acts trivially on $G$, which implies that $g$ also commutes with every element of $G$. But $g$ was arbitrary, so we conclude that $G$ is abelian.

(Fill the the details!)