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I know that it is true for any finite abelian group that they can be generated by the set of elements of maximal order, but is it true for any finite group?

If it is true, could you give me a hint of proof? If it is false, please give me a counter example.

Thank you!

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    $\begingroup$ No. Try the smallest nonabelian group. $\endgroup$ – Derek Holt Nov 14 '17 at 18:38
  • $\begingroup$ @DerekHolt Oh! You mean dihedral group D6? (since maximal order is 4, and r and r^3 have the maximal order, but only two of them cannot generate D6) $\endgroup$ – JacobsonRadical Nov 14 '17 at 18:49
  • $\begingroup$ I think @Derek means $S_3$ the permutation group of order $6$ on three letters. $\endgroup$ – Stephen Meskin Nov 14 '17 at 20:19
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$S_3$ the permutation group on $\{1, 2, 3\}$ of order $6$ is a counter example.

$(123)$ and $(132)$ are the two elements of maximal order $3$ in $S_3$ but they are inverses of one another and generate a cyclic group of order $3$.

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    $\begingroup$ Let's go a little further. Your initial comment was partially correct. $D_6$ the dihedral group of order $12$ has $2$ elements of maximal order $6$; they are inverses of each other and generate the same cyclic group, as in $S_3$ although this time of order $6$. More generally, in any dihedral group $D_n$ of order $2n$ the elements of maximal order are of order $n$ and they generate a single cyclic subgroup of $D_n$. The rest of the elements of $D_n$ all have order $2$. In particular, $S_3$ is an instance of $D_3$ $\endgroup$ – Stephen Meskin Nov 15 '17 at 4:29

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