Question

Question: Let G be an infinite group and H the subset of all elements of G that have only a finite number of distinct conjugates in G. Prove that H is a subgroup of G.

Short answer

It is proved that $-$H is a subgroup of G.

Step-by-step solution

Conjugate

LetGbe group. If $a,b\in G$then, bis said to be a conjugate ofainGif there exists an element $c\in G$such that $b=c^{-1}ac$.

Clearly, we have, $e\in H$.

Suppose that $a,b\in H$for any $g\in G$, we have that $g^{-1}ab{g}^{-1}bg$the conjugates of the product are products of conjugates.

To prove H is a subgroup

Since a and b have finite number of conjugates, this implies that also does, and so $ab\in H$.

Suppose that $a\in H$and $g\in G$.

Then, $g^{-1}{a}^{-1}g={\left(g^{-1}ag\right)}^{-1}$the conjugate of an inverse is the inverse of a conjugate.

Since $a$has a finite number of conjugates, so does $a^{-1}$and therefore, $a^{-1}\in H$.

Since $e\in H$and $H$is closed under multiplication and inverses, it is a subgroup of $G$.

Thus, H is a subgroup of G.