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I'm trying to understand the following proof that no group of order $96$ is simple:

A group of order 96 has either 1 or 3 subgroups of order 32. If there is only one such subgroup, it is normal and we are done. If not, let H and K be distinct subgroups of order 32. By Lemma 37.8, H ∩ K must have order 16, and is normal in both H and K, being of index 2. Thus N[H ∩ K] has order a multiple > 1 of 32 and a divisor of 96, so the order must be 96. Thus H ∩ K is normal in the whole group.

Lemma 37.8 talks about the formula $|H\cap K| = \frac{|H||K|}{|HK|}$. But how does he find $|H\cap K| = 16$? It's assuming that it knows $|HK|$ to put in the formula, but how? How can he know $|HK|$?

Also: not a duplicate of https://math.stackexchange.com/a/142839/166180 because I'm trying to understand a different proof

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Note that $|H\cap K|$ divides $32$ by Lagrange theorem as $H\cap K$ is a subgroup of $H$. If $|H\cap K |<16$ then $|HK|$ has more elements than $G$, which is impossible.

More clearly, $$|HK|=\dfrac{32*32}{|H\cap K|}\leq|G|=96$$ $$\dfrac{32}{3}\leq|H\cap K|$$

Thus, we must have $|H \cap K |=16$.

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  • $\begingroup$ Thanks. One more thing... How does the order of the normalizer of $H\cap K$ being the entire group proves that $H\cap K$ ir normal in $G$? $\endgroup$ – Guerlando OCs Jun 20 '17 at 0:35
  • $\begingroup$ $N_G(H\cap K)\geq H,K$. Thus, $|N_G(H\cap K)|\geq |HK|=64$. Can you conclude now ? $\endgroup$ – mesel Jun 20 '17 at 1:29
  • $\begingroup$ @GuerlandoOCs: Apparently, you did not conclude ? $\endgroup$ – mesel Jul 6 '17 at 11:57

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