$G$ has a normal Sylow $2$-subgroup, then every subgroup of $G$ has a normal Sylow $2$-subgroup.

Reading M.Isaacs book Finite Group Theory (FGT) I found the following statement.

"If a group $G$ has a normal Sylow $2$-subgroup, it is easy to see that every subgroup of $G$ has a normal Sylow $2$-subgroup".

My question. How one can sufficiently easy see this?

I try to be more concrete. Is it really important that we deal with $2$-groups, or it doesn't matter?

I can prove this for arbitrary $p$ (not necessary $2$) using the following fact (excercise 2.B.6) from FGT. I give it here.

A group $G$ has a normal Sylow $p$-subgroup iff every subgroup of the form $<x,y>$ has a normal Sylow $p$-subgroup, where $x,y$ are conjugate elements of $G$ having $p$-power order.

But to to prove the last statement I have used Baer's theorem, which is not very easy itself. So it seems to me that the author meant that there is more simple proof of the fact about normal Sylow $2$-subgroups.

Let $p$ be a prime dividing the order of a finite group $G$ and let $H$ be a subgroup. Assume that $G$ has only one Sylow $p$-subgroup $S$, that is $n_p(G)=1$ and $S \unlhd G$.
If $p$ divides $|H|$ then pick a Sylow $p$-subgroup in $H$, say $P$. Since $P$ is a $p$-subgroup of $G$ it must be contained in some Sylow $p$-subgroup of $G$, hence $P \subseteq S$, so $P \subseteq H \cap S$.
On the other hand, reasoning within $H$, $S \cap H$ is a normal $p$-subgroup of $H$ and is hence contained in some Sylow $p$-subgroup of $H$, which is a conjugate of $P$. But $S \cap H$ is normal, so it must be contained in $P$. We conclude: $P=S \cap H$, is the only Sylow $p$-subgroup of $H$.
Note: in general $n_p(H) \leq n_p(G)$. And if $H \unlhd G$, even $n_p(H) \mid n_p(G)$.
• Thank you very much! I really complicated the solution. Although this doesn't concern the question can you give a hint how I can prove that if $H \unlhd G$, then $n_p(H) \mid n_p(G)$? – Mikhail Goltvanitsa Jan 3 '18 at 14:00
• Mikhail, thanks and thanks for the upvote, yes it is basically the Frattini argument. Now try yourself: if $H \unlhd G$, then $n_p(G/H) \mid n_p(G)$. – Nicky Hekster Jan 3 '18 at 14:34
• Nicky,I try to explain, how I solve your problem. Firstly I prove that every Sylow $p$-subgroup $A$ of $\bar{G} = G/H$ is an image of some Sylow $p$-subgroup $S$ of $G$. Let $S$ be a Sylow p-subgroup of $G$, then $n_p(\bar{G}) =|\bar{G}:N_{\bar{G}}(\bar{S})|$. Also $\overline{N_G(S)}\subseteq N_{\bar{G}}(\bar{S})$ and $n_p(G) = |G:N_G(S)|$. So, $n_{p}(G) = n_p(\bar{G})\cdot|H:H\cap N_G(S)|\cdot\frac{N_{\bar{G}}\bar{S}}{\overline{N_G(S)}}$. What do you think about it?) – Mikhail Goltvanitsa Jan 3 '18 at 16:04