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Prove or contradict: A finite group with exactly $2$ conjugacy classes always isomorphic to $\mathbb{Z}_2$.

At first I was trying to work with familiar groups to contradict it(permutations, cyclic, dihedral and general linear) but I could not find any counter example.

So I was trying to prove it.Let $G$ be a group that has exactly $2$ conjugacy classes, obviously one of those classes is $e$, and the other class is the rest of the elements.

So I'll define $f:G \rightarrow \mathbb{Z}_2$ as follows - $f(e) = 0$, and $f(g) = 1$ for every other $g \in G$. It is easy to see that $f$ is homomorphism, and also that $\ker (f) = e$, and by the first isomorphism theorem, we get that $G \cong \mathbb{Z}_2$.

Is it correct?

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    $\begingroup$ I don't think it's at all obvious that $f$ is a homomorphism. Can you expand on that? $\endgroup$ – Christopher Sep 5 '18 at 16:09
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    $\begingroup$ $Z_2$ is the only finite group with exactly two conjugacy classes, but there are other infinite examples. $\endgroup$ – Derek Holt Sep 5 '18 at 16:35
  • $\begingroup$ @Christopher Rethinking about it, I guess that it isn't homomorphism. How else can I prove it? $\endgroup$ – ChikChak Sep 5 '18 at 17:22
  • $\begingroup$ @Derek Holt do you mean there are groups of inifinte order with two conjuacy classes, or that there are inifinte finite such groups? $\endgroup$ – ChikChak Sep 5 '18 at 17:23
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    $\begingroup$ "Infinite finite group"??? See here $\endgroup$ – Derek Holt Sep 5 '18 at 17:51
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Hint: use the Orbit-Stabiliser Theorem, with $G$ acting on itself by conjugation. (So the orbits are the conjugacy classes of $G$).

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    $\begingroup$ So we got $2$ classes, one is $e$, and the other one is $G \setminus \{e\}$. Since conjugacy classes are subgroups, their order must divide $|G|$, and we get from it that $|G|-1$ divides $|G|$, so $|G|=2$(?) $\endgroup$ – ChikChak Sep 5 '18 at 17:45
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    $\begingroup$ This assumes that $G$ is finite and, as I said in my previous comment, there are infinite groups with exactly two conjugacy classes. $\endgroup$ – Derek Holt Sep 5 '18 at 17:50
  • $\begingroup$ @Derek Holt I saw your example now, very interesting. Thank you both very much(I indeed forgot that the group is assumed to be finite). $\endgroup$ – ChikChak Sep 5 '18 at 18:01
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    $\begingroup$ @ChikChak: Is a conjugacy class really a subgroup? Is $a$ always conjugate to $a^{-1}$? $\endgroup$ – Robert Lewis Sep 5 '18 at 18:22
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    $\begingroup$ @Robert Lewis No, not always. So maybe a different approach - $|G|$ should be equal to the sum of the sizes of the orbits( which are the conjugacy classes in our case). We got $2$ classes and therefore $2$ orbits. One orbit of size $1$, so the other must be of size $|G|-1$. And orbits are indeed subgroups, so $|G|=2$. Is it correct now? $\endgroup$ – ChikChak Sep 5 '18 at 20:09

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