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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 $G \cong H \times K$. –  user38268 Sep 23 '11 at 11:56

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up vote 7 down vote accepted

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 ? –  Tav Sep 23 '11 at 12:14
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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. –  Zhen Lin Sep 23 '11 at 12:42
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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. –  Jack Schmidt Sep 23 '11 at 14:26
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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). –  Jack Schmidt Sep 23 '11 at 16:44
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(I think @ Jack does not actually notify me. @Jack (no space) works.) –  Jack Schmidt Sep 23 '11 at 16:45

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