I'm struggling with another problem from Rotman, namely:
If a group G has exactly two conjugacy classes, and at least one element of finite order, prove that it is isomorphic to the cyclic group of order 2.
Here's what I've got.
A. Let a be an element of order n. Then any other non-identity element is a conjugate of a, and hence has order dividing n.
B. By a very similar argument, all non-identity elements have the same order.
C. Obviously, that order must be a prime.
D. If that prime equals 2, the result is easy to prove as follows:
For any non-identity a and b, ab has order 2 (or 1). So abab = 1. So bab = a.
We just proved that all conjugates of a equal a. Since every non-identity element of G is by assumption a conjugate of a, we're done.
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So what am I missing?
2007-12-14
02:59:21
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1 answers
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asked by
Curt Monash
7