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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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    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](http://math.stackexchange.com/questions/52350/finite-groups-with-exactly-n-conjugacy-classes-n-2-3)2012-05-01
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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)?2012-05-01
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    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 $G\backslash \{1\}$.2012-05-01
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    Duplicate: http://math.stackexchange.com/q/137858/224052012-05-02

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