[Math] Prove that the center of G has order divisible by p

Question

Let $G$ be a non-trivial finite group and $p$ a prime number. If every subgroup $H\leq$ G has index divisible by $p$, prove that the center of $G$ has order divisible by $p$.

So I have that $[G:H]=pk$ for some integer $k$, and we need to prove that $|Z(G)|=pl$ for some integer $l$. Let $|G|=n$, I can prove the case if assume $G$ is an abelian group, then $|Z(G)|=n$, so the center has order divisible by $p$.

How should I approach if $G$ is not abelian?

Answer 1

Note that the class equation of a finite group $G$ is

$|G|=|Z(G)|+\sum_{i=1}^n |cl(a_i)|\implies |Z(G)|=|G|-\sum_{i=1}^n |cl(a_i)|$ where $a_i$ are the distinct class representatives.

Now $|cl(a_i)|=\dfrac{|G|}{|C(a_i)|}$ where $C(a_i)=\{x\in G:xa_i=a_ix\}$.

Note that $C(a_i)$ is a subgroup of $G$ for each $1\leqslant i\leqslant n.$

Since every subgroup has index divisible by $p$ , hence $p\mid |cl(a_i)|$ for all $1\leqslant i\leqslant n.$

Also for any subgroup $H$ of $G$ we have $|G|=|{\dfrac{G}{H}}||H|$ ,

$p \mid \dfrac{|G|}{|H|}\implies p\mid |G|$

Since the RHS is divisible by $p$ so is the LHS.