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In the problem posed in this question of mine we want to show that a particular group is both perfect and solvable, and therefore trivial, and this turns out to be useful in proving the result.

What other combinations of properties required of a group imply that it must be isomorphic to the trivial group?

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$G$ is trivial if any of the following hold.

  • $|G|$ is odd and every element is conjugate to its inverse

  • $G$ is cyclic and some group $H$ exists such that $H/Z(H)\cong G$ (equivalently, $G\cong \operatorname{Inn}(H)$)

  • $|G|=n^2$ and $G$ has an irreducible representation of dimension $n$

  • given any group $H$, there is precisely one group homomorphism $f:G\rightarrow H$.

  • given any group $H$, there is precisely one group homomorphism $f:H\rightarrow G$.

  • $G$ is solvable, not isomorphic to $S_3$, and all of its conjugacy classes have distinct sizes

  • $G$ is finitely generated, nilpotent, not $\mathbb{Z}_2$, and every automorphism is inner

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Nice, thanks. Under what conditions is an element conjugate to its inverse? If $g^{-1}xg=x^{-1}$ then $xg=gx^{-1}$, so $g$ commutes with $x$ up to taking inverses. Is there a formal way to describe this? – Alex Petzke Feb 5 '14 at 13:35
@AlexPetzke Hm, can you clarify your question? My inclination is to reply that your comment is precisely the formal description, but I'm sure that's not what you want. – Alexander Gruber Feb 5 '14 at 18:11
I suppose I was just hoping for something interesting that's at a higher level than just simple computation with elements, involving properties of $C_G(x)$ since this influences $\mathrm{cl}(x)$, or something along those lines. If there isn't anything like that I can certainly accept that too. I'm just trying to think and go a little deeper. – Alex Petzke Feb 6 '14 at 1:44

If the automorphism group og a group $G$ is trivial, then $G$ must be the trivial group or $\mathbf{Z}/2$. This is a nice qualifying-exam-type exercise. Although this includes two possibilities it is (hopefully) in the spirit of what you asked.

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$G$'s got to be finite too, I think. – Alexander Gruber Feb 4 '14 at 0:55
Thanks Alex Gruber. I realized as soon as I posted but you were too fast before my edit! – user44441 Feb 4 '14 at 0:56
I believe finiteness is not needed although the proof is bit harder. – user44441 Feb 4 '14 at 0:56
Finiteness is certainly not needed - although choice may be – zcn Feb 4 '14 at 1:49
You're right. I was thinking of something else (where a Tarski monster was a counterexample). – Alexander Gruber Feb 4 '14 at 4:05

Don't know if this is welcome:

If $G$ has exactly 1 element

If $G$ is finite and divisible

If $G$ is abelian, simple, and does not have prime order

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But the trivial group is not simple, so there are no groups at all that satisfy your third condition. – Derek Holt Feb 4 '14 at 9:02

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