For any finite group $G$ and $n$ a divisor of $|G|$, consider the following subset of elements of "co-order" dividing $n$:

$$G(n) = \{ g \in G \mid g^{|G|/n} = 1 \}$$

- By a classical theorem of Frobenius, $\frac{|G|}{n} \mid |G(n)|$.
- A conjecture of Frobenius (which was proved via the classification of finite simple groups) states that if there's equality in the above divisibility, $G(n)$ is a normal subgroup.
- $G(n)$ contains the identity and is closed to taking inverse and conjugation.

I'm interested in the following question:

For a prime $p$, when is $G(p)$ a subgroup of $G$?

- By Lagrange's Theorem, if $G(p)$ is a subgroup of $G$, then $|G(p)| \in \{ |G|, \frac{|G|}{p} \}$.
- When $G$ is abelian, $G(p)$ is a evidently a subgroup.
- When $G$ is a non-cyclic $p$-group, $G(p)=G$.
- In fact, as long as the $p$-Sylow subgroup of $G$ is not cyclic, $G(p)=G$.

The case $p=2$ is very elegant: composing the natural maps $G \to S_G \to \{\pm 1\}$, we obtain $G(2)$ as the kernel of this composition, hence it's a normal subgroup.

The case $p=3$ is false in general (as seen from the example $G=S_3$), and so I guess an additional condition should be imposed - perhaps $p$ should be the smallest prime dividing $|G|$.

An additional question is:

When $G(p)$ is a subgroup, does is have an alternative characterization?

For example, for $p=2$, if the 2-Sylow subgroup of $G$ is cyclic, $G(2)=\langle g^2 \mid g\in G\rangle$.