Real element that is not a commutator Find a non-trivial finite group with no index-2 subgroups,
such that it has a real element that is not a commutator.
Ps. $x \in G$ is called real if $x$ is conjugate to $x^{-1}$. In particular, every involution is real.
 A: Let $N$ be elementary abelian of order $2^{2n}$ generated by $\{a_i,b_i : 1 \le i \le n\}$ for some $n > 0$.
Let $H = \langle t \rangle$ be cyclic of order $3$ and let $H$ act on $N$ by $a_i^t = b_i$, $b_i^t = a_ib_i$ for $1 \le i \le n$.
Let $G = N \rtimes H$ be the semidirect product with this action.
Now we can define a central extension $1 \to Z \to E \to G \to 1$, with $|Z| = 2^{n(n-1)/2}$ generated by $\{ z_{ij} : 1 \le i < j \le n \}$, such that, using the same symbols for generators of $E$ as for their images in $G$, we have  $t^3=1$,  $a_i^2=b_i^2=1$, $a_i^t=b_i$, $b_i^t=a_ib_i$  for all $i$, for each  $i,j$:
$$[a_i,a_j]=[b_i,b_j] = [a_i,b_j]=z_{ij},$$
and all other commutators of the generators of $N$ are trivial.
Since $Z \le [E,E]$, $E$ has no subgroup of index $2$.
The total number of commutators in $E$ is at most $|G|^2 = 9 \times 2^{4n}$, which is less that $|Z|$ for $n \ge 10$, so there must be elements of $Z$ that are not commutators.
Note that a real element of odd order must be a commutator, but you can have real elements of any even order that are not commutators.
