I am trying to make my way through Grove's Groups and Characters but am having some trouble with the following seemingly benign exercise:
If a group $G$ has a normal $p$-complement $K$ show that $K$ is unique, and hence $K$ char $G$. Show that $K = \{x \in G : p \nmid |x|\}$
I am not really sure where to go with this. A promising route seems to be via the second isomorphism theorem and comparing divisors of the order of the groups, but this hasn't led anywhere. A hint or solution would be much very much appreciated.
I can't quite see how the Second Isomorphism Theorem would help, but that doesn't mean it wouldn't. The key idea is that for any homomorphism $\phi : G \to H$, we have that $|\phi(g)|$ divides $|g|$. Stop reading now, if you want to think about that for a bit.
Consider $g \in G$, and the homomorphism $\pi: G \to G/K$. We know that:
$\bullet$ The order of $\pi(g)$ divides the order of $g$. Thus, if $p$ divides $\pi(g)$, $p$ must also divide $|g|$.
$\bullet$ The order of $G/K$ is a power of $p$, hence all non-identity elements of $G/K$ must have orders divisible by $p$.
If $p$ doesn't divide $|g|$, does $\pi$ have any choice about where to map $g$?