element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$

Question

Hello. I thank for your answer, in advance.

Let $G$ be a finite group and $G$ has an element of order $n$ such that $\pi(n)=\pi(G)$ where $\pi(n)$ denote the set of prime divisors of $n$ and $\pi(G)$ denote the set of prime divisor of $|G|$. What can be said about the structure of the group? I know in nilpotent group there exist such element.

I generalize my question. What is the relation between the $|\pi(G)|$ and the maximum $|\pi(n)|$, where $n$ range in all order elements of a finite group $G$?

Answer 1

There is a pathological example that pretty much demonstrates that the existence of such an element gives no significant information about the group:

Example 1. Let $H$ be any finite group, and let $\pi(H)=\{p_1,\dots, p_k\}$. Now let $C$ be a cyclic group of order $p_1\cdot p_2\cdots p_k$ with generator $c$. Then $\pi(H\times C)=\pi(H)$ and $H\times C$ has an element $g=(1,c)$ of order $n=p_1\cdot p_2 \cdots p_k$, i.e $\pi(n)=\pi(H)$.

@Someone has suggested a second example which allows one to construct perfect groups with the required property (and thereby deals with my earlier remark "My hunch is that if you prescribe that $G$ is perfect, i.e. $G=G'$, then there will never exist an element of the kind you seek... But even then I'm not sure"):

Example 2. Let $S$ be any simple group and let $\pi(S)=\{p_1,\dots, p_k\}$. Now let $G=\underbrace{S\times \cdots \times S}_k$ and observe that $\pi(G)=\pi(S)$. Now let $s_i$ be an element of order $p_i$ in $S$ and observe that $g=(s_1,\dots, s_k)\in G$ has order $\pi(S)$.

Both examples are also relevant to your generalized question.

Answer 2

In the case of solvable groups, this may not say much about the structure of the group. For example, if $G$ is a finite $\{p,q\}$-group, where $p,q$ are distinct primes (hence $G$ is solvable by Burnside's $p^{a}q^{b}$-theorem), then it is quite unusual (though certianly not impossible) for $G$ not to have an element of order $pq.$ If $G$ contains an elementary Abelian $p$-group of order $p^{2}$ and an elementary Abelian $q$-group of order $q^{2}$, then $G$ will contain an element of order $pq.$ If, for example, $G$ contains no elementary Abelian $q$-subgroup of order $q^{2},$ then the Sylow $q$-subgroups of $G$ are cyclic or generalized quaternion