Question

Let \( G \) be a finite group containing a non-identity element which is conjugate to its inverse. Then, which one of the following is TRUE?

• A. The order of \( G \) is necessarily even
• B. The order of \( G \) is not necessarily even
• C. \( G \) is necessarily cyclic
• D. \( G \) is necessarily abelian but need not be cyclic

Hint:

If a finite group contains an element that is conjugate to its inverse, then the group’s order must be even, because such elements cannot be paired symmetrically without an even number of elements.

Answer 1

The Correct Option is A

Let \( g \) be an element of the group \( G \) such that \( g \neq e \) and \( g \) is conjugate to its inverse, i.e., there exists some \( h \in G \) such that \( hgh^{-1} = g^{-1} \). Such an element is called an involution. The key fact here is that if there is such an element in a finite group, the order of the group must be even. This follows from the fact that if \( g \) is conjugate to \( g^{-1} \), then the conjugacy class of \( g \) must contain an element distinct from the identity element. As conjugacy classes come in pairs of distinct elements and the identity is fixed, the presence of such a conjugate element implies that the number of elements in the group must be even. Thus, the correct answer is (A): the order of \( G \) is necessarily even.