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Let $G$ be a finite group and $p$ divide $|G|$. Suppose $P$ and $Q$ are two subgroups s.t. $P\subset N_G(Q)$. Prove that $PQ$ is a $p-$subgroup of $G$.

I imagine that I have to use second isomorphism theorem, but I don't know how to do.

mesel
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Henri
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2 Answers2

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You do not need second isomorphism theorem:

Recall that the equality always true $$|HK|=\dfrac{|H||K|}{|H\cap K|}.$$

In general $HK$ need not to be a subgroup. Just notice that $HK$ is a group if $K$ normalizes $H$, that is, when $K\leq N_G(H)$.

mesel
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  • @Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact. – mesel Jul 30 '18 at 11:36
  • @Surb: you can also check from this: https://math.stackexchange.com/questions/168942/order-of-a-product-of-subgroups-prove-that-ohk-fracohokoh-cap-k – mesel Jul 30 '18 at 11:38
  • Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link) – Surb Jul 30 '18 at 11:39
  • Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$. – mesel Jul 30 '18 at 11:41
  • great. Thank you :) (and good answer too) (+1) – Surb Jul 30 '18 at 11:50
  • @Surb: you are welcme. What does "Je ne suis pas une pipe." realy mean? – mesel Jul 30 '18 at 11:54
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    It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-) – Surb Jul 30 '18 at 11:59
  • @Surb I see right now :) Thanks. – mesel Jul 30 '18 at 13:03
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Hint

Your idea by using 2nd isomorphism theorem is the right idea. You have that $P$ is a subgroup of $N_G(Q)$. Since $Q\lhd N_G(Q)$ you have that $PQ\leq N_G(Q)$, and thus you can use 2nd isomorphism theorem in $N_G(Q)$.

Surb
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