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I know that $S_3$ is the semidirect product of $\bigl\langle(1\ \ 2\ \ 3)\bigr\rangle \rtimes\bigl\langle(1\ \ 2)\bigr\rangle$, and I'm not sure where exactly the direct product property fails. Is it only because $\bigl\langle(1\ \ 2)\bigr\rangle$ is not normal in $S_3$?

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    $\begingroup$ Exactly. Good. +1 $\endgroup$ – DonAntonio Feb 4 '18 at 12:39
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That is enough, yes. If $G=H_1\times H_2$, then $\{e_{H_1}\}\times H_2$ and $H_1\times\{e_{H_2}\}$ are normal subgroups of $G$.

Besides, the direct product of two abelian groups is again abelian.

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    $\begingroup$ I think the last sentence really nails the question in the title: All groups of order $<6$ are abelian, hence if it were the direct product of smaller groups, it would have to be abelian. The same argiument applies to all nonabelian groups such that all proper divisors primes or $<6$, i.e., non-abelian groups of orders $8,9,10,14,15,21,22,25,\ldots$ (not for all these orders do nonabelian groups exist). $\endgroup$ – Hagen von Eitzen Feb 4 '18 at 12:51

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