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I wanted to find all non-abelian groups of order 8 up to isomorphism, but I ran into a contradiction and I can't find the error in my reasoning. My reasoning was as follows:

Since the groups we're after are non-abelian 2-groups, their center is neither trivial nor the whole group. So let's do a case distinction on the order of the center being either 2 or 4.

If the center $Z/(G)$ has order 4, then the quotient group $G/Z(G)$ has order 2 and is therefore cyclic. We now invoke the following theorem to reach a contradiction on G being non-abelian.

So the center has order 2 and the quotient group $G/Z(G)$ has order 4. The Klein four-group is the only non-cyclic group of order 4 (which is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$), so it holds that $G/Z(G) \cong G/\mathbb{Z}_2 \cong \mathbb{Z}_2 \times \mathbb{Z}_2$. But now all my elements have order 2, right? Which would mean the group is abelian again.

The actual solutions are the dihedral and and the dicyclic group and even though I understand that proof, I simply cannot find the error in my proof, mainly because I'm new to group theory. So any help is deeply appreciated.

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    $\begingroup$ You are saying that the quotient has all elements order 2 and so does the "denominator," therefore so does $G$. If you try to prove this statement on its own, you will see where it goes wrong. $\endgroup$ – Randall May 27 at 23:57
  • $\begingroup$ But since all elements be written in the form $ab$, where $a \in \mathbb{Z}_2$ and $b \in \mathbb{Z}_2 \times \mathbb{Z}_2$, if thought that if I multiply an element with itself, I get with the commutativity of the center that $abab = aabb = ee = e$. $\endgroup$ – LionCoder May 28 at 0:11
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    $\begingroup$ But $b$ is not an element of $\mathbb{Z}_2\times\mathbb{Z}_2$. It’s an element of $G$ that maps to an element of $\mathbb{Z}_2\times\mathbb{Z}_2$. That means that $b^2$ lies in the kernel of the map... but it need not be trivial. It could be equal that $b^2=a$. $\endgroup$ – Arturo Magidin May 28 at 0:13
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    $\begingroup$ @LionCoder I think you're assuming that the quotient splits (the product is direct and not semi- or something weaker), which it need not. This is the same as Arturo's comment. $\endgroup$ – Randall May 28 at 0:37
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    $\begingroup$ No one has said it, but your work up to that point was right on. $\endgroup$ – Randall May 28 at 0:48

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