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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, $G(2)=\langle g^2 \mid g\in G\rangle$.

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    $\begingroup$ There is no "classification of finite groups". Certainly you just forgot the word "simple" there. $\endgroup$ – Johannes Hahn Jan 11 '15 at 20:02
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You have noted (accurately) that $G(p) =G$ unless $G$ has a cyclic Sylow $p$-subgroup. However, it is also clear that when $G$ has a cyclic Sylow $p$-subgroup and $G(p) \neq G,$ the group $G$ has a normal $p$-complement. For, otherwise, we have by Burnside's transfer theorem, there is a $p$-regular element $x \in N_{G}(P) \backslash C_{G}(P)$. Now we have $P = [P,x] \times C_{P}(x).$ Since $P$ is cyclic and $P \neq C_{P}(x),$ we must have $P = [P,x]$ and $C_{P}(x) = 1.$ But then $P \leq G^{\prime},$ a contradiction, since $[G:G(p)] =p$ and $G(p)$ is clearly a normal subgroup when it is a subgroup.

On the other hand, if $G$ has a normal $p$-complement and cyclic non-trivial Sylow $p$-subgroup, then clearly $G$ has a normal subgroup of index $p$, and that that subgroup is indeed $G(p)$.

The conclusion (when $|G|$ has order divisible by $p,$ is that $G(p)$ is a subgroup of $G$ if and only if one of the following cases occur:

$G(p) = G$, or, $G$ has a normal $p$-complement and a cyclic Sylow $p$-subgroup.

Notice that when $p =2$, every finite group with a cyclic Sylow $2$-subgroup is known to have a normal $2$-complement, so the condition given for $p$ odd to guarantee that $[G:G(p)] = p$ is somewhat analogous to what you observed when $p =2$. It is indeed sufficient that $G$ should have a cyclic Sylow $p$-subgroup when $p$ is the smallest prime divisor of $|G|$ to ensure that $[G:G(p)] = p$. It is not, however, necessary.

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  • $\begingroup$ Thanks! A good answer, and using the normal p-complement I was able to solve my 2nd question (when $G(p)$ is a non-trivial subgroup, it is the subgroup generated by $p$'th powers). $\endgroup$ – Ofir Gorodetsky Jan 14 '15 at 20:58
  • $\begingroup$ Yes, that is certainly true. $\endgroup$ – Geoff Robinson Jan 14 '15 at 21:15

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