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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user1729
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Can g or h be trivial? – Christopher K Nov 10 '13 at 15:30
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2@ChrisK No, because if you apply $h=aga^{-1}$ you get $g=h$. – Jack M Nov 10 '13 at 15:31
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@JackM, true (g and h are otherwise g and h not unique)... stupid question, perhaps? – Christopher K Nov 10 '13 at 15:33
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Do you know about source of problem? – Bobby Nov 10 '13 at 17:12
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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.
Mikasa
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