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$H$ and $K$ are subgroups of a group $G$ with $H$ doesn't equal to $K$ and $|H| = |K| = 11$. Prove that $H \cap K = \{e\}$

Have no idea about this question. Dont know if there is some theorem about it.

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Hint: Lagrange's Theorem. – user21436 Apr 15 '12 at 16:19
Can we generalize this post for a prime $p$ so that this shall be a good original for subsequent posts? – user21436 Apr 15 '12 at 16:31
@KannappanSampath: That seems like a good idea. Would it be considered impolite for you to just go ahead and change it? – Tara B Apr 16 '12 at 7:57
@TaraB It would be impolite. May be, I would make a meta post in the near future. Thank you for dropping by to tell me what you feel. I'd ping you when I write a question at the meta. Regards, – user21436 Apr 16 '12 at 10:57

Hint: $H\cap K$ is a subgroup of $H$. What can you say about the subgroups of a group of prime order?

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I see. Then this question is just so easy. – Shannon Apr 15 '12 at 16:24

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