I am attempting to determine all non-abelian groups of order $8$. So say $|G|=8$. Then as $G$ is a $2$-group, it has a cyclic tower where each quotient is the cyclic group of order $2$, say $G=G_0\ge G_1\ge G_2\ge G_3 = \{e\}$. Now $G_1$ has index $2$, so has order $4$ and can either be cyclic or the Klein $4$-group.

If it is cyclic, then call its generator $\tau$ and consider $G/G_1\cong Z/2Z$. Let $\sigma G_1$ be the generating coset. We observe that clearly $\sigma\notin G_1$ and conclude that $\sigma,\tau$ are generators of $G$. Now as $G_1$ is also normal (2 is the smallest prime dividing $|G|$ and $G_1$ has index 2) we know $\sigma$ acts on $G_1$ by conjugation. So there is some homomorphism $\langle \sigma\rangle\to \text{Aut}(G_1)$. I claim this homomorphism can't be trivial, as then $\sigma,\tau$ commute and $G$ is abelian. So we must have that $\sigma\tau\sigma = \tau^3$, which is the only choice of automorphism we have left. Now we have $\sigma^2=\tau^4=e$ and $\sigma\tau = \tau^3\sigma$ so I thought this gives $D_8$.

Now I know my goal is to produce $D_8$ and the quaternions, so I expected the Klein $4$-group to give me that. But upon looking at the quaternions, I realized it does not have a subgroup isomorphic to $K_4$. So something in my above proof has gone wrong. Can someone help me spot where? Thank you.


1 Answer 1


We also have $ij = j^3i$ in the quaternions. On the other hand, the dihedral group has a subgroup isomorphic to the Klein 4-group. Additionally, I don't see a proof that we must have $\sigma^2 = e$. All I can see from your writing is that $\sigma^2G_1 = G_1$.

So I think you have concluded prematurely, and have your two cases swapped around.

  • $\begingroup$ I see now, thank you! We cannot have $\sigma^2 = \tau$ or $\sigma^2 = \tau^3$ as otherwise $\sigma$ has order 8 and the group is cyclic. If $\sigma^2 = e$ then my proof shows it is $D_8$ and if $\sigma^2 = \tau^2$, we can get the quaternions! $\endgroup$
    – beeclu
    Commented Oct 19, 2023 at 22:27

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