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I'd like to prove group $G$ is simple if and only if for every pair $g,h \in G$ ($g \ne 1$), $h$ can always be written product of finite number conjugates of $g$ and $g^{-1}$.

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Have you shown implication in one of the directions? If $G$ has some non-trivial proper normal subgroup, can you find elements $g$ and $h$ not satisfying the given? –  Tobias Kildetoft Oct 14 '13 at 12:19
    
Your question would likely attract more positive attention if you said what you'd tried to do to work out the answer for yourself (or where you're stuck). –  TooTone Oct 14 '13 at 12:34
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1 Answer

Hint: look at the subgroup of $G$ generated by the conjugacy class of $g \in G$ and observe that this subgroup is normal.

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Nice, simple and short accurate hint. +1 –  DonAntonio Oct 14 '13 at 12:43
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