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?

1. The order of G is necessarily even
2. The order of G is not necessarily even
3. G is necessarily cyclic
4. G is necessarily abelian but need not be cyclic

Answer: The order of G is necessarily even

Solution :-

We have a Result that say: If G is a group of odd order then G has no non-identity element which is conjugate to its inverse.

Given that G has a non-identity element that is conjugate to its inverse. So, the order of group G can never be odd.
Hence, Option 1. is right and Option 2. is wrong.

Let us consider two groups
(1) $\left(G_1,\cdot \right)=\{-1,1\}$ and
(2) $G_2=Q_8=\{-1,1,i,-i,j,-j,k,-k\}$

Here, in $G_1$ all non-identity elements of this group is conjugate to its inverse and order of this group is even and also this group is abelian and cyclic.

Now, in $G_2$ all non-identity elements of this group is conjugate to its inverse and also order of this group is even but this group is neither abelian nor cyclic.

So, we can observe that a group G, which have a non-identity element which is conjugate to its inverse then that group G is not necessarily cyclic as well as not necessarily abelian.
Hence, Option 3. and 4. are also wrong.

So, the correct answer is option-1, $\to$ The order of G is necessarily even