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

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 $G$?

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I'd like to suggest a rewording to state and focus the question a little better: For a finite group $G$, let $P(G)$ be the product of the prime divisors of $|G|$. It is known [or "I have proved"] that if $G$ is a nilpotent group, then there is an element $g$ for which $P(G)$ divides the order of $g$. Are there any non-nilpotent groups with this property? If so, what other properties do they share with the nilpotent groups (but not all groups)? (This suggestion, of course, is only appropriate if nilpotent groups are the only examples you know of.) –  Barry Cipra Apr 28 '13 at 15:46
You can easily constuct non-nilpotent examples by taking a direct product of an arbitrary group with a cyclic group of the required order. But I would be surprised if there were any nonabelian simple groups with this property. –  Derek Holt Apr 28 '13 at 16:46

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.

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Perfectness doesn't help: Just take the direct product of $k$ copies of $G$ where $k$ is the number of prime divisors of $G$. –  Someone Apr 30 '13 at 9:44
Good point, shall edit. –  Nick Gill Apr 30 '13 at 10:35
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
Also, there is a result of Lucido that if $p$, $q$ and $r$ are distinct prime divisors of $|G|$ (with $G$ solvable), then $G$ contains an element of at least one of the orders $pq$, $pr$, $qr$. –  Tobias Kildetoft Jul 25 '13 at 6:44