Discussion
Let G be a group of order n and \(H = \{a \in G \mid a = a^{-1}\}\), then which of the following is true ?
Answer: A
The true statement is if the cardinality of H is odd then n is odd.
H consists of the identity together with the elements of order 2. The other elements of G pair off with their inverses, \(a \ne a^{-1}\), so
\[ n = |H| + 2 \times (\text{number of pairs}) \]
Hence \(|H|\) and n have the same parity. If \(|H|\) is odd, n is odd; and if n is even, \(|H|\) is even, which also shows the statement 'if \(|H|\) is even then n is odd' is false.
The other two fail on small examples. In \(S_{3}\) (order 6), H is the identity and the three transpositions, so \(|H| = 4\), which does not divide 6. In the dihedral group of order 8, H has the identity and five elements of order 2, so \(|H| = 6\), not a power of two.
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