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Prove that $D_4$ cannot be expressed as the internal direct product of two proper subgroups.

I know that it can't be an internal direct product of two rotational subgroups because no rotation combination gives a reflection, but for the other cases I can't seem to make any progress.

Oliver G
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1 Answers1

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The only possible orders for nontrivial proper subgroups are $2$ and $4$. So, if $D_4$ were a direct product of such subgroups, there are only two possibilities.

One possibility is that all of the factors of the direct product have order $2$, so $D_4$ is isomorphic to $C_2 \times C_2 \times C_2$, where $C_2$ denotes a cyclic group of order $2$. But $C_2 \times C_2 \times C_2$ is abelian, and $D_4$ is not, so this possibility is excluded.

The other possibility is that one of the factors has order $4$, hence the other must have order $2$. So $D_4$ is isomorphic to $H \times C_2$, where $H$ has order $4$. But there are only two isomorphism types for a group of order $4$, namely $C_4$ or $C_2 \times C_2$. Both of these are abelian, so $H$ is abelian, hence $H \times C_2$ is also abelian. So this possibility is also excluded.

  • What does $C_2$ represent? – Oliver G Aug 18 '16 at 17:37
  • @OliverG Cyclic group of order $2$, also known as $\mathbb Z_2$. –  Aug 18 '16 at 17:38
  • What do you mean by all of the factors have order $2$? Are you referencing factor groups or do you mean all the subgroups have order $2$? – Oliver G Aug 18 '16 at 17:49
  • @OliverG I mean, all of the factors in the direct product. If $D_4$ is a direct product of nontrivial proper subgroups, then it must have one of two forms: either $A \times B \times C$ where $A$, $B$, and $C$ all have order $2$, or $D \times E$, where $D$ has order $4$ and $E$ has order $2$ (or vice versa). –  Aug 18 '16 at 17:55
  • I missed a word in the problem statement, it should read: "as a product of two proper subgroups". Edited for clarity. – Oliver G Aug 18 '16 at 17:57
  • @OliverG OK, that's even easier. There's only one possibility to consider (the second one in my answer). –  Aug 18 '16 at 17:58