Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$H$ 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.

share|cite|improve this question
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?

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.