A p-Sylow subgroup of a subgroup is a p-sylow subgroup of the group

Question

Is that true? If it is, then it is unprovable. I guess it isn't. The question is: how can I show it unequivocally? The definition my teacher gave us is:

$P$ is a $p$-Sylow subgroup of $G$ if $|P|=p^k$ with $p$ prime whereas $|G|=p^km$ and $(m,p)=1$.

How can I stick to this definition in my proof?

Answer 1

No. $$G=C_2\times C_2$$ with subgroup $C_2$.

Answer 2

Evidently no: take $G=\mathbf Z/p^n\mathbf Z$. As a $p$-group, it is the only p-Sylow subgroup of itself. Its subgroups are $H_k=p^k\mathbf Z/p^n\mathbf Z\simeq \mathbf Z/p^{n-k}\mathbf Z\enspace (k=0,\dots,n)\,\,$ and are their own p-Sylow subgroups.