# $p$-subgroup of normalizer of sylow $p$-subgroup of $G$ is a subgroup of sylow $p$-subgroup of $G$ [closed]

May I get a hint on how to prove the following statement:

Let $$G$$ be a group.

Let $$P=Syl_p(G)$$ be some sylow $$p$$-group of $$G$$.

Let $$H=N_G(P)$$.

Then forall $$P'$$ that is $$p$$-subgroup of $$H$$, $$P'\leq P$$.

The book I am using suggests I should look into natural homomorphism $$\phi:H\rightarrow H/P$$, but I have no idea how to proceed...

Your help would be greatly appreciated!

By the setting, $$P$$ has the maximal $$p$$-power order possible in $$G$$, hence in $$H$$ as well. Therefore, $$H/P$$ has order prime-to-$$p$$. If we think of $$f\colon P'\rightarrow H\rightarrow H/P$$, for $$g\in P'$$, $$f(g)$$ is of prime-to-$$p$$ order but also it has $$p$$-power order (since $$P'$$ is a $$p$$-subgroup). Hence $$f(g)=1$$ and we see $$g\in P$$.

• May I know why does $f(g)$ have $p$-power order? $g$ in $P'$ have $p$-power order, but how does that say about $f(g)$ in $H/P$? Why does homomorphism preserve order in this case? Apr 4, 2023 at 8:31
• If $g^q=1$ then $f(g)^q=1$, so the order of $f(g)$ is a divisor of $q$. Apr 4, 2023 at 11:22
• oooooh, thanks! Apr 4, 2023 at 16:16

Here's an alternative perspective on this problem, using the second isomorphism theorem.

Since $$P' \leq N_G(P)$$, $$P'P$$ is a subgroup of $$G$$, and $$P$$ is a normal subgroup of $$P'P$$. (Justify both these statements carefully!)

By the second isomorphism theorem, $$\frac{|P'P|}{|P|} = \frac{|P'|}{|P' \cap P|}.$$

From here, the steps are:

• Using the fact that $$P'$$ is a $$p$$-group, argue that the right-hand side of the equation is a power of $$p$$.
• Using the fact that $$P$$ is a Sylow $$p$$-group, deduce that $$|P'P| = |P|$$.
• Conclude that $$P' \leq P$$.

By the way, I didn't produce this from thin air. This idea is used in that part of the proof of Sylow's theorems where you show that the number of $$p$$-subgroups is congruent to $$1$$ mod $$p$$. There, the result you need is even more general: If $$Q$$ is a $$p$$-subgroup of $$G$$ (not necessarily contained within $$N_G(P)$$), then $$Q \cap N_G(P) = Q \cap P$$. Do have a go at proving this!

• Thanks for the response, but may I know what is the meaning of $AB$ for group A, B? I never saw this notation of directly juxtaposing two groups before xD Apr 4, 2023 at 8:39
• $AB$ is the subset $\{ ab : a\in A, b \in B \}$. In general, this is only a subset - it's not necessarily a subgroup. I'm asking you to show that if $B$ is a subgroup and $A \leq N_G(B)$, then $AB$ is in fact a subgroup; furthermore, $B$ is a normal subgroup of $AB$. Anyway, if you haven't seen this notation before, then I'm guessing you haven't come across the second isomorphism theorem either? Apr 4, 2023 at 8:49
• Ah yes, after reviewing my note I find this notation used in isomorphism theorems as well xD yeah i would have to admit I did not really understand the motivation behind 2nd and 3rd isomorphism theorem (and hence did not grasp it). Maybe I should relearn them when I got time, and will come back to this answer after that!! Apr 4, 2023 at 16:17
• @xade93 That's fair enough, take your time! Apr 4, 2023 at 16:22