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

Answer 1

The Correct Option is A

To solve this problem, we need to determine which statement must be true about a finite group \(G\) that contains a non-identity element conjugate to its inverse. The provided options suggest conclusions about the structure and properties of such a group.

First, let's understand the given conditions:

• The group \(G\) is finite.
• There exists a non-identity element \(g \in G\) such that \(g\) is conjugate to \(g^{-1}\). This means there is an element \(x \in G\) such that \(xgx^{-1} = g^{-1}\).

Now, let's analyze the implications:

1. The order of \(G\) is necessarily even:

If an element \(g\) is conjugate to its inverse \(g^{-1}\), it implies that the element pairs up with its inverse. In a finite group, if there are elements that pair up, and the identity element remaining unpaired, this leads to the conclusion that the group order must be even. To illustrate:

• Consider conjugation: \(xgx^{-1} = g^{-1}\). Conjugation is an automorphism of the group, meaning any operation which reorders the group's elements without adding or removing any is achievable in the group.
• Each element \(g \neq e\) paired with its inverse \(g^{-1}\) results in an even count of elements, contributing to the group order's evenness.

2. The order of \(G\) is not necessarily even:

This option would suggest that the pairing condition does not imply any particular order restriction, but as explained above, the pairing implies an even count.

3. \(G\) is necessarily cyclic:

This option suggests that the group can be generated by a single element. However, pairing an element with its inverse doesn't necessarily imply the cyclic nature of the entire group.

4. \(G\) is necessarily abelian but need not be cyclic:

An abelian group requires commutative property. The conjugation condition given (for a single element) does not extend to establish commutativity for all pairs in \(G\).

Conclusion:

The correct conclusion is that the group \(G\) must have an even order due to the necessity of element-inverse pairing for non-identity elements. Thus, the correct answer is:

• The order of \(G\) is necessarily even.