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How can I prove following problem in abstract algebra?

Let $G$ is a finite non-abelian group. show that there exist elements $a,g,h\in G$ such that $g\neq h, h=aga^{-1}$ and $gh=hg$.

Please help me. Thanks in advance.

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marked as duplicate by T. Bongers, some1.new4u, Shuchang, Branimir Ćaćić, Erick Wong Dec 1 '13 at 7:09

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What have you tried? –  Daniel Rust Nov 10 '13 at 15:26
Can g or h be trivial? –  Chris K Nov 10 '13 at 15:30
@ChrisK No, because if you apply $h=aga^{-1}$ you get $g=h$. –  Jack M Nov 10 '13 at 15:31
@JackM, true (g and h are otherwise g and h not unique)... stupid question, perhaps? –  Chris K Nov 10 '13 at 15:33
Do you know about source of problem? –  Babak Miraftab Nov 10 '13 at 17:12

1 Answer 1

Hint: I think using Conjugacy class equation: $$\displaystyle \left|{G}\right| = \left|{Z \left({G}\right)}\right| + \sum_{x_j\notin Z(G)} \left[{G : C_G \left({x_j}\right)}\right]$$ is effective here.

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Nice hint. :)+1 –  Sami Ben Romdhane Nov 10 '13 at 19:36

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