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Two elements $a$ and $b$ of a group $G$ are conjugate if there is an element $g$ in the group such that $b = g^{–1}ag$.

I was going through the Wikipedia page on conjugacy classes and was curious about this:

What can we say about a group $G$ in which all its elements are conjugate? Does it have any special properties? Can there exist such a non-trivial group $G$?

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    To make it more interesting, consider replacing "all elements are conjugate" with "all non-identity elements are conjugate". Then, the cyclic group of order $2$ would be such a group. – Geoffrey Trang May 03 '20 at 23:39
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    @saltandpepper: If the group is abelian, then every element is conjugate only to itself. – Arturo Magidin May 03 '20 at 23:47
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    A group in which every element is conjugate is trivial, since $e$ can only be conjugate to itself. A group with exactly two conjugacy classes, one for the identity and one for every other element, is either cyclic of order $2$, or must be infinite, and the constructions are difficult, and the question is then a duplicate. – Arturo Magidin May 03 '20 at 23:47

2 Answers2

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No, it cannot. That's because the identity element is conjugate with itself and only with itself.

José Carlos Santos
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  • This would no longer be a valid answer after my suggestion above to replace "all elements are conjugate" with "all non-identity elements are conjugate" is followed. – Geoffrey Trang May 03 '20 at 23:42
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That can't happen unless the group is trivial. The class equation always consists of at least one $1$, corresponding to the class of the identity.

If we change it as @GeoffreyTrang suggests, then we get only one conjugacy class of size one, and hence the center is trivial.

  • As with the other answer, this would no longer be a valid answer after my suggestion above to replace "all elements are conjugate" with "all non-identity elements are conjugate" is followed. – Geoffrey Trang May 03 '20 at 23:42
  • Ok @GeoffreyTrang. –  May 03 '20 at 23:46