helle fox,
So, there is a proof of the first Sylow theorem using the following lemma:
Let $H$ be a subgroup of $G$ and $S$ a $p$-Sylow of $G$, so there exist an element $g$ of $G$ such that $gSg^{-1}\cap H$ is a $p$-Sylow of $H$.
Here is the proof of the lemma: Consider the action of $G$ on the quotient set $(G/S)_{g}$ by multiplication, so the stabliliser of an element $gS$ of $(G/S)_{g}$ is $gSg^{-1}$.
Then we take the restriction of the action of $G$ on $(G/S)_{g}$, namely:$$H\times (G/S)_{g}\mapsto (G/S)_{g}$$ the stabliliser of an element gS of $(G/S)_{g}$ by this new action is $gSg^{-1}\cap H$, and they say this is a $p$-group of $G$.
My question is: Why $gSg^{-1}\cap H$ is a $p$-group of $H$, since $gSg^{-1}\cap H$ can be the $\{1\}$.
Notice: a p-group is a group of order $p^{\alpha}$, whereh $\alpha$ is an integer $\geq 1$.
Thanks for your help.