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 $G$?



element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$
There is a pathological example that pretty much demonstrates that the existence of such an element gives no significant information about the group:



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> **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 $\pi(n)=\pi(H)$.
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@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"*):



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> **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 $\pi(S)$.
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Both examples are also relevant to your generalized question. 


