$G$ contains a normal $p$-subgroup

Question

Let $G$ be a non-abelian finite group with center $Z>1$.

I want to show that if $G/Z$ is solvable then $G$ contains a normal $p$-subgroup for some prime $p$ with $p\mid |G:Z|$.

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Since $G/Z$ is solvable, we have that $1=N_0\leq N_1\leq \dots \leq N_k=G/Z$ is a normal series of $G/Z$ ( $N_i\trianglelefteq G/Z$ ) where $N_{i+1}/N_i$ is abelian.

Could you maybe give me a hint how we could show that $G$ contains a normal $p$-subgroup?

Do we maybe use the correspondence theorem?

Answer 1

(With the comments of Derek Holt, this is correction in the earlier wrong answer.)

Since $G/Z$ is solvable, consider the smallest (non-identity) term, say $G_n/Z$, in the derived series of $G/Z$; it should be characteristic in $G/Z$, and it is also abelian. Thus, $G_n/Z$ is direct of Sylow subgroups, each of which is normal (characteristic) in $G_n/Z$, and since $G_n/Z$ is characteristic in $G/Z$, every Sylow subgroup of $G_n/Z$ is normal in $G/Z$. Correspondence theorem implies that the pull-back of Sylow subgroups of $G_n/Z$ are normal in $G$.

• Suppose no prime divisor of $|G_n/Z|$ divides $|Z|$

Let $H/Z$ be a Sylow-$p$ subgroup of $G_n/Z$. As seen above, $H/Z\trianglelefteq G/Z$, hence $H\trianglelefteq G$. Let $|H/Z|=p^k$. Then $|H|=|Z|.|H/Z|=|Z|.p^k$. Since $p\nmid |Z|$, it follows that $H$ contains a Sylow-$p$ subgroup of order $p^k$, say $P$. Then $H=PZ$ with $P\cap Z=1$ (orders being relatively prime), hence $H=P\times Z$ (since $Z$ is central). Then $P$ is normal Sylow $p$-subgroup of $H$, it should be characteristic in $H$. But $H\trianglelefteq G$, hence $P\trianglelefteq G$.

• Suppose some prime divisor of $|G_n/Z|$ also divides $|Z|$, say $q$.

In this case, simply consider Sylow-$q$ subgroup of $Z$. Clearly it is normal in $G$.

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I guess, the argument concludes something more: for every prime divisor $p$ of $|G\colon Z|$, $G$ contains a characteristic $p$-subgroup.