TITLE: Is a finite group generated by representatives of its conjugacy classes?
QUESTION [16 upvotes]: 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 :)

REPLY [8 votes]: This was asked, and answered, on MathOverflow some time ago: https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class

REPLY [5 votes]: 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.)