On direct product

Asked Nov 13 '13 at 12:19

Active Nov 14 '13 at 13:16

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Let $G$ be a finite group, $H$ and $K$ are two normal subgroups of $G$ with the property that $G=HK$ and $H\cap K=1$.

Prove that $$\left(\left|H\right|,\left|K\right|\right)=1\Leftrightarrow\forall A\leq G,A=\left(A\cap H\right)\left(A\cap K\right)$$

About Dummit & Foote, Abstract Algebra, page 171:

Each elementof $HK$ can be written uniquely as a product of $hk$, for some $h\in H$ and $k\in K$.
$HK\cong H\times K$.

Maybe it's usefull.

edited Nov 13 '13 at 13:12

asked Nov 13 '13 at 12:19

Truong

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What are your thoughts and efforts on the problem so far? – Cameron Buie Nov 13 '13 at 12:21
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Please don't dump questions here with no source, no indication of why they're of interest to you, and no sign of any effort on your part. – Gerry Myerson Nov 13 '13 at 12:22
Which direction of the proof do you have problems with? – j.p. Nov 13 '13 at 13:21
Unfortunately, both two. – Truong Nov 13 '13 at 13:23
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OK, here is a hint for $\Rightarrow$. Let $hk \in A$ with $h\in H$, $k \in K$. By raising to suitable powers show that $h,k \in A$. For the converse, choose a prime $p$ dividing both $|H|$ and $|K|$ and find a subgroup of order $p$ that is not in either subgroup. – Derek Holt Nov 13 '13 at 13:46
Can you give me more details? Please – Truong Nov 13 '13 at 14:08
Maybe this post is usefull. http://math.stackexchange.com/questions/211185/expression-for-a-subgroup-of-a-inner-direct-product-group?rq=1 – Truong Nov 13 '13 at 14:44
How can you raising to suitable powers of $h$ and $k$ ? – Truong Nov 13 '13 at 16:12
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Both $h$ and $k$ are powers of $hk$. Remember that it is a direct product, so $hk=kh$. – Derek Holt Nov 13 '13 at 17:14
@Alexander Gruber: I think my problem doesn't duplicate with that problem. – Truong Nov 14 '13 at 10:16

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