On page 94, the Corollary states:
If $H$ and $K$ are subgroups of $G$ and $H \leq N_G(K)$, then $HK$ is a subgroup of $G$. In particular, if $K \trianglelefteq G$ then $HK \leq G$ for any $H \leq G$.
I'm having a hard time understanding this proof. Let me follow the proof line by line and specify which items I do not get. I would appreciate some clarifications of these items.
Proof: We prove $HK = KH$.
[I'm not sure why exactly they prove this fact. I understand that there is a reference to Proposition 14 on the same page. But I don't understand how exactly Proposition 14 and Corollary 15 are related.]
Let $h \in H$, $k \in K$. By assumption, $hkh^{-1} \in K$,
[I don't see how this is the case.]
The rest of the corollary is OK to my current understanding.
$\endgroup$ 11 Answer
$\begingroup$Proof: We prove $HK = KH$.
You are trying to prove that $HK$ is a subgroup of $G$. Proposition 14 says that for any subgroups $H$ and $K$, $HK$ is a subgroup if and only if $HK=KH$. So if you can prove that $HK=KH$, Proposition 14 implies that $HK$ is a subgroup, which is what you are trying to prove.
Let $h \in H$, $k \in K$. By assumption, $hkh^{-1} \in K$,
This is exactly what it means to say $H\leq N_G(K)$: for every $h\in H$, $h\in N_G(K)$, i.e. for every $k\in K$, $hkh^{-1}\in K$.
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