Here is my question that I'm currently working on:

Show that the Quaternion group is indecomposable.

I have already shown that $S_3$ is indecomposable, so here is my approach with the Quaternion group (we haven't had much exposure to this group, but here's my shot at the problem):


We know that we can express the Quaternion group as the following:

$$Q = \{\pm 1,\pm i,\pm j,\pm k\}.$$

Also, $Q$ has order $8$. Suppose that $Q \cong H$ x $K$, where $H$ and $K$ are two nontrivial groups. Then, we have that |$Q$| = $8$ = |$H$| $\cdot$ |$K$|, where we have two possibilities. Either |$H$| = $2$ and |$K$| = $4$, or |$H$| = $4$ and |$K$| = $2$...

This is how far I got. Is this correct so far? If so, how would one progress further?


You're on the right track. Note that $H$ and $K$ are necessarily abelian (since all groups of order $2$ and $4$ are abelian), hence so is $H\times K$. But $Q_8$ isn't abelian.

  • 1
    $\begingroup$ Ah now I see. Seems I forgot to mention that $H$ and $K$ are necessarily abelian up on my proof, but you clarified it with your explanation. $\endgroup$ – Kyogre May 9 '17 at 1:15

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