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I have to prove that $D_4$ cannot be the internal direct product of two of its proper subgroups.Please suggest.

Since the order of the group is $8$. Internal direct is possible if there exists two normal subgroups $H$ and $K$ of $D_4$ such that $D_4 = H \times K$.

Then, by Lagranges Theorem we can have $|H| = 2$ and $|K| = 4$ or vice a versa. I can see that both $H$ and $K$ are abelian groups. How to proceed further in this ??

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    The following theorem may help: Let $H,K$ be subgroups of $G$. Then if $H \cap K$ is the trivial subgroup, $HK =G$ and $H,K$ are normal in $G$, then $G \cong H \times K$.2011-09-23

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A hint: the direct product of abelian groups is abelian.

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    @ mt_ : Thank you for the hint. I think you mean that H x K is abelian but D4 is not. But I am not able to figure out how the direct product of two abelian groups is always abelian. Can I find its proof in some book ?2011-09-23
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    @Tav: Let $(h, k)$ and $(h', k')$ be two elements in $H \times K$. All you need to do is show that $(h, k) \cdot (h', k') = (h', k') \cdot (h, k)$, assuming $H$ and $K$ are abelian groups.2011-09-23
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    @ Zhen Lin : I am confused about this as the elements of the form (h, k )belong to the external direct product. Here H X K is an internal direct product ?2011-09-23
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    @Tav: Elements of H commute with each other, elements of K commute with each other, so all that is left is to show elements of H commute with elements of K. hkH = Hhk = Hk = kH = khH (since H is normal, gH=Hg, and since h in H, hH=Hh). Similarly, hkK = khK, so hk(kh)^-1 is in both H and K, but their intersection is the identity, so hk = kh. This is basically the chinese remainder theorem.2011-09-23
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    @ Jack Schmidt: I have a small doubt regarding this...is it not always true for an Internal direct product that elements from different subgroups commute. So in that case, it will always be the case that whether H and K are abelian does not matter. Because, hk(h^-1)(k^-1) is in H as well as K (using the fact that both H and K are normal ). Therefore, by the definotion of Internal Direct Product, hk(h^-1)(k^-1)= {e} and hence, hk=kh.2011-09-23
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    @ Jack Schmidt: Is the internal direct product always abelian, irrespective of the fact whether H and K are abelian or not ?2011-09-23
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    @tav: It is always true that H commutes with K. It is not always true that H commutes with H; that requires H to be abelian. Your proof is right, but it does not handle h1 * h2 = h2 * h1 (which need not be true if H is not abelian).2011-09-23
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    (I think @ Jack does not actually notify me. @Jack (no space) works.)2011-09-23
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    @JackSchmidt: Thanks for your help. I appreciate it.2011-09-23