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Let $G$ be a group, $H$ and $K$ normal subgroups of $G$ such that $H$ and $K$ are simple, $G=HK$, and $H\cap K = \langle e \rangle$. Show that either

  1. $H\cong K\cong \mathbb{Z}_p$ for $p$ a prime, or
  2. The only normal subgroups of $G$ are $\langle e\rangle$, $H$, $K$, and $G$.
Alex Kruckman
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Lila
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  • I really need help with this problem, have been stuggling with it for a long time. Thanks. – Lila Mar 15 '13 at 03:10
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    By "$H\cap K$ is the generator $e$," do you mean $H\cap K=\langle e\rangle$? (The first is nonsense, the second isn't. Also, almost always the letter $e$ is reserved for the identity element of a group, and it seems this is not the case here.) Where in the world does $N$ come from in your last sentence? – anon Mar 15 '13 at 03:24
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    What is N? And you should really be more careful with the way you write: it's very hard to understand what you say...for example, it *seems to be that $,H\cap K={e},$ , which means $,H,K,$ have trivial intersection... – DonAntonio Mar 15 '13 at 03:25
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    I've edited your question for readability - can you verify that you meant what I've written? – Alex Kruckman Mar 15 '13 at 03:29

1 Answers1

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Hints:

Assuming Alex's effort is the correct answer: since $\,H\,,\,K\lhd G\,\,,\,\,H\cap K=1$ then their product is isomorphic with their direct product:

$$G=HK\cong H\times K$$

Thus any normal subgroup of $\,H\,$ or $\,K\,$ is normal in $\,G\,$ , and if $\,N\lhd G\,$ , then $\,N\cap K\lhd K\;,\;\;N\cap K\lhd K\,$ .

Complete now the argument.

DonAntonio
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  • Hi, yes the way you wrote the intersection of H and K is correct. Sorry for the confusion. The last part, I checked the question is correct. The N just suddently appear at the end, it says, if N is normal to G, then N = e (inside the diamond sign), H, K, or G. – Lila Mar 15 '13 at 04:01
  • Stop saying "e (inside the diamond sign)"...:) . In group theory we use to denote $,\langle x\rangle,$ the cyclic group generated by $,x,$...this is what is meant here. – DonAntonio Mar 15 '13 at 04:03
  • Hi Antonio, thanks for the hint, I know about the direct product part. But how you relate it to H is isomorphic to K and Zp? I thought if H and K are simple groups, then their order is either equal to G or identity? I assume identity is 1? I am new to abstract algebra. – Lila Mar 15 '13 at 04:06
  • Hahahha....I know the diamond sign is silly, but I was afraid I said it wrong. Ok then. I won't say it again =) – Lila Mar 15 '13 at 04:07
  • Well, one possibility is $,H=K=\Bbb Z_p,$ . Remember that any cyclic group of order a prime is trivially a simple group. the hint tries to help you to deal with the case when $,H,K,$ are NOT the cyclic group of order $,p,$...BTW, now that I read the editing of your question, there is one more possibility: that $,H,K,$ are cyclic group of order different primes. Check this...Did you take this question from some book? – DonAntonio Mar 15 '13 at 04:12
  • Aside from the direct product, the book provides hint that elements in H commutes with elements in K. So is it means that HK is a subgroup of G? but I don't know if this can help to solve the Zp part. – Lila Mar 15 '13 at 04:18
  • The book doesn't says anything about different primes. Hmmm...do you think that is what the book says? – Lila Mar 15 '13 at 04:21
  • I think the book says different primes, because the intersection of H and K is the cyclic group generated by identity. Don't you agree? If H=K=Zp, then the intersection won't be that. – Lila Mar 15 '13 at 04:34
  • Oh, the can be different, though isomorphic, cyclic groups, @Lila...but I think I know what they mean here. I shall come back later with more since I must go now. – DonAntonio Mar 15 '13 at 04:42