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I was asked to have a look at a problem:

There is no a finite non-trivial group $G$ that all its non-trivial elements can be commuted with exactly half elements of group .

For the first step, I saw I could not prove it directly so, I assumed we have such a group $G$, finite and satisfying above property. The property led $|G|$ to have an even order because $∀ (e≠)x\in G$, $|C_G(x)|=\frac{|G|}{2}$. Am I on the right way? Any hints will be appreciated. Thanks.

Mikasa
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    You don't need your reasoning to conclude $|G|$ is even, you can just observe that if $|G|$ is odd, then "exactly half the elements" doesn't even make sense... – Ben Millwood Jun 03 '12 at 09:58

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You're on the right track. Remember that $[G: C_G(x)]$ is the size of the conjugacy class of $x \in G$. Use the fact that $G$ is a disjoint union of conjugacy classes to complete your solution.

Mikko Korhonen
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  • :Yes. Thanks. Thanks. As, you hinted, $\Sigma_x[G:C_G(x)]$ taken on all non-trivial $x\in G$ is even and it leads $|G|$ to be odd number. I got it. :-) – Mikasa Jun 03 '12 at 10:18
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    @BabakSorouh: Yes, that's correct. You can generalize this problem by doing the same thing with "all non-trivial elements commute with exactly $\frac{|G|}{n}$ elements of the group" for any $n \geq 2$. – Mikko Korhonen Jun 03 '12 at 10:36
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    Yes, m.k. points out the key property which is quite general. You could say it this way too: "if $G$ is a finite group with trivial center, there are at least $3$ conjugacy classes of different cardinality (counting the class of the identity)". That statement is no longer true if we allow a non-trivial center, as the non-Abelian groups of order $8$ illustrate (for example). – Geoff Robinson Jun 03 '12 at 10:43