Thread: Finite Groups with exactly two conjugacy classes

Dummit and Foote:Abstract Algebra section 4.3 exercise 13 states:

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Find all finite groups which have exactly two conjugacy classes.
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I am unsure about how to approach this problem.

Can anyone please help?

Peter

well, if G is abelian, every element has just one conjugate, itself. this means that the only abelian groups which fit the bill have just 2 elements.

on the other hand, if G is not abelian, then we have at least |Z(G)| conjugacy classes. since we are assuming Z(G) ≠ G, this means the center has to be trivial.

now we can only have one other conjugacy class, which by the class equation has size the index of the normalizer [G:N(a)] for some non-central (which in this case means non-identity) element a.

so we have [G:N(a)] = |G| - 1, or |N(a)|(|G| - 1) = |G|, or better still: |N(a)| = |G|/(|G| - 1). |G|/(|G| - 1) has to be a positive integer, and |G| = 2 is not allowed (we covered that in the abelian case). how is this possible?