# Error in determining non-abelian groups of order 8.

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.

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 $$\sigma^2G_1 = G_1$$.
• 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! Commented Oct 19, 2023 at 22:27