Odd-order and ambivalent implies trivial

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Suppose G is an Odd-order group (?) (i.e., a finite group of odd order) that is also an Ambivalent group (?): every element is conjugate to its inverse. Then, G is the trivial group.

Facts used

  1. Order of element divides order of group


Suppose G has odd order, is ambivalent, and is nontrivial. Then, there exists a non-identity element a in G. By fact (1), a has odd order, so a \ne a^{-1}.

Since G is ambivalent, there exists b \in G such that bab^{-1} = a^{-1}. Then, b^2ab^{-2} = a, so a and b^2 commute. Again by fact (1), b has odd order, so \langle b^2 \rangle = \langle b \rangle. Since a commutes with b^2, it must commute with all elements in \langle b^2 \rangle, and hence with b. Thus, bab^{-1} = a. This forces a = a^{-1}, a contradiction.