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A group where every two elements different than 1 are conjugate has order 1 or 2.

I have to show this result:

If G is a finite group with exactly two conjugacy classes, then G is isomorphic to $\mathbb{Z}_2$

I have tried to show that $|G|=2$ but without success. I'm not supposed to use Sylow's theorems on this one. I have tried to make $g$ act upon itself through conjugation, but that does not lead me to any conclusion upon its order.

Can someone help me?

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Marra
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    When $G$ is finite, the order of a conjugacy class divides the order of $G$. There is a conjugacy class of order $|G|-1$, so..? See also this question – Mikko Korhonen May 01 '12 at 22:41
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    Why is there a conjugacy class of order $|G|-1$? Is it because one must be the conjugacy class of ${ 1}$ and the other one the class that includes all the other elements of $G$ (and since they partition G, the sum of the order of both is the order of G)? – Marra May 01 '12 at 22:44
  • Basically yes. One of them has to be ${1}$. Since conjugacy classes partition $G$ (ie. are disjoint and cover all of $G$), the other one must be $G\backslash {1}$. – Mikko Korhonen May 01 '12 at 23:02
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    Duplicate: http://math.stackexchange.com/q/137858/22405 – Brett Frankel May 02 '12 at 00:27

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