# Is a finite group generated by representatives of its conjugacy classes?

Suppose $G$ is a finite group with conjugacy classes $C_1,C_2,\dots,C_\ell$. Suppose we take one element from each conjugacy class: $g_i \in C_i$ for all $i=1,\dots,\ell$.

Is it true that $G = \langle g_1,g_2,\dots,g_\ell \rangle$ (i.e. $G$ is generated by these elements)?

If this is true, references? Hard to prove?

Thanks!!

Edit: Thanks again everyone! I guess I should have looked around more on overflow first :)

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–  David Speyer Oct 22 '11 at 2:07

Yes. Suppose not: then there will be some maximal subgroup $M\le G$ intersecting each conjugacy class. Then, because $G$ is the union of its conjugacy classes, $G$ is the union of conjugates of $M$. But this is impossible. (Can you see why? Try counting how many elements one can have in the union of $M$ and all its conjugates, noting there are at most $[G:M]$ such conjugates.)