What are some properties that imply that a group must be the trivial group?

Question

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?

Answer 1

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.

Answer 2

$G$ 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

Answer 3

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