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Let H, K be two subgroups of a finite group G such that $|H|>\sqrt{G}$ and $|K|>\sqrt{G}$. Show that $|H \cap K|>1$.

Attempt of solution:

It can't be $0$ since $H \cap K$ is also a subgroup, so it must have at least the trivial element.

If H or K is normal in G, then HK is another subgroup, and it's order is $|H||K|/|H \cap K|$. That number is greater than $|G|$ when $|H \cap K|=1$ which is a contradiction. But I don't know how to argue if H and K are not normal.

santm
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    The equality $|HK|=\frac{|H|\cdot |K|}{|H\cap K|}$ always holds. If $H$ and $K$ are both not normal, $HK$ might not be a subgroup, but it is a subset of $G$. Look here for a proof of its cardinality: https://math.stackexchange.com/questions/2436538/why-is-hk-frachkh-cap-k – Mark Jun 30 '23 at 00:05
  • Aw, true. Thank you! – santm Jun 30 '23 at 00:13
  • Instead of $\sqrt{G}$ you mean $\sqrt{|G|}$. – Dietrich Burde Jun 30 '23 at 19:49

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