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!


2 Answers 2


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 $g\in P$.

  • $\begingroup$ 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? $\endgroup$
    – xade93
    Apr 4, 2023 at 8:31
  • $\begingroup$ If $g^q=1$ then $f(g)^q=1$, so the order of $f(g)$ is a divisor of $q$. $\endgroup$
    – Ayaka
    Apr 4, 2023 at 11:22
  • $\begingroup$ oooooh, thanks! $\endgroup$
    – xade93
    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!

  • $\begingroup$ 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 $\endgroup$
    – xade93
    Apr 4, 2023 at 8:39
  • $\begingroup$ $AB$ 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? $\endgroup$
    – Kenny Wong
    Apr 4, 2023 at 8:49
  • $\begingroup$ 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!! $\endgroup$
    – xade93
    Apr 4, 2023 at 16:17
  • $\begingroup$ @xade93 That's fair enough, take your time! $\endgroup$
    – Kenny Wong
    Apr 4, 2023 at 16:22

Not the answer you're looking for? Browse other questions tagged .