Internal Direct Product

Question

Prove that $\mathcal D_{4}$ cannot be expressed as an internal direct product of two proper subgroups.

What I have so far:

We know that the order of $\mathcal D_{4}$ is $8$ since, in general, the order of $\mathcal D_{n}$ is equal to $2n$. So, if there exists an $H$ and a $K$ such that $\mathcal D_{4} = H \times K$, then either the order of $H$ must be $4$, and the order of $K$ must be $2$, or the order of $K$ must be $4$ and $H$ must be $2$.

Without loss of generality, take $K$ to be of order $2$, and $H$ to be of order $4$. Then $K$ must be isomorphic to $\mathbb Z_{2}$ and $H$ must be isomorphic to either $\mathbb Z_{4}$ or $\mathbb Z_{2}$ $\oplus$ $\mathbb Z_{2}$.

What next?

Well, you're almost there. You've shown that your subgroups $H$ and $K$ must be abelian, so what can you say about their direct product? — James, Apr 13 '15 at 20:11
What next ? See here for the solution. It is a duplicate, I think. — Dietrich Burde, Apr 13 '15 at 20:15

score 3 · Accepted Answer · answered Apr 13 '15 at 20:12

3

You are exactly right so far. You need but one more line to finish up:

The direct sum of abelian groups is abelian. Is $D_4$ abelian?

answered Apr 13 '15 at 20:12

Kaj Hansen

33,511

Well, since you ask this question, I am guessing that $D_{4}$ is not abelian. Is this the case for all dihedral groups? – EmaLee Apr 13 '15 at 20:14
Indeed it is the case that $D_n$ is not abelian for any $n>2$. The important thing is that you can convince yourself why. I'm willing to bet that your textbook has somewhere defined the dihedral group as a group generated by a rotation $R$ and a flip $F$, where the elements $R$ and $F$ obey some properties? – Kaj Hansen Apr 13 '15 at 20:16

Internal Direct Product

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