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Let $G$ be a finite group. How can we show that $|G/G^{'}|\leq |C_G(x)|$ for all elements $x\in G$?

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Is $G'$ the commutator subgroup of $G$? – zarathustra Aug 10 '14 at 8:15
Yes it is the commutator subgroup. – Z. Fo Aug 10 '14 at 9:06
up vote 2 down vote accepted

This follows from the fact that the order of a conjugacy class $$|Cl_ G(x)| \leq |G'|$$. Namely, define a map $f : Cl_G(x) \rightarrow G'$ by $f(g^{-1}xg)=[x,g]$. This map is clearly injective. Finally, $|Cl_G(x)|=[G:C_G(x)]$.

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This is a consequence of the fact that $G$ has $[G:G^{\prime}]$ linear characters (which are certainly irreducible)- I refer to complex characters here, and if $\lambda$ is a linear character of $G,$ we have $|\lambda(g)|^{2} = 1$ for all $g \in G.$ On the other hand, the orthogonality relations show that if $x \in G,$ we have $\sum_{\chi}|\chi(x)|^{2} = |C_{G}(x)|$ for each $x \in G,$ where $\chi$ runs over all complex irreducible characters of $G,$ so certainly forcing $|C_{G}(x)| \geq [G:G^{\prime}]$ for all such $x.$

If you prefer a direct group-theoretic group, note that for each $x \in G,$ there are $[G:C_{G}(x)]$ different possibilities for $x^{-1}g^{-1}xg$ as $g$ runs though $G,$ and these elements all lie in $G^{\prime},$ so $|G^{\prime}| \geq [G:C_{G}(x)].$

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Thank you. But I need a group theory proof. – Z. Fo Aug 10 '14 at 9:07
The sentence beginning "if you prefer a direct group-theoretic proof" gives a group theory proof which does not depend on what is written before. I have now put it as a separate paragraph (which I intended to do anyway). – Geoff Robinson Aug 10 '14 at 9:12

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