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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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    See http://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class2011-10-22

2 Answers 2

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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

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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.)