Elements which have only a finite number of conjugates in a group is a characterstic group. Prove that in any group $G$, the subset of all elements which have only a finite number of conjugates in $G$ is a characteristic subgroup of $G$.
My approach: If some element has finite number of conjugates, then conjugates will start repeating but I was not able to proceed further
 A: Let $G$ be a group, and let $H$ be the set containing group elements with finite conjugacy class. 
To show that $H$ is a subgroup, consider the product $ab^{-1}$ of arbitrary elements of $H$. For any $g\in G$, we have the equality $$g(ab^{-1})g^{-1}=(gag^{-1})(gb^{-1}g^{-1}).$$ This implies that any conjugate of $ab^{-1}$ is just a product of conjugates of $a$ and $b$, of which there are only finitely many. Since there are only finitely many products, $ab^{-1}$ has a finite conjugacy class. Hence, $H$ is a subgroup.
Now, let $\varphi$ be some automorphism of $G$, and $a$ be an element of $H$. For any $g\in G$, there exists some $h\in G$ such that $\varphi(h) = g$. Consequently, it follows that we have the following equality for conjugation: $$g\varphi(a)g^{-1}=\varphi(h)\varphi(a)\varphi(h^{-1})=\varphi(hah^{-1}).$$ This implies that any conjugate of $\varphi(a)$ is the image of a conjugate of $a$, of which there are only finitely many. Hence, $\varphi(a)$ has finitely many conjugates, so $\varphi(a)\in H$. Therefore, $H$ is characteristic. 
A: Write $Cl_G(a)$ for the conjugacy class of $a \in G$, and $C_G(a)$ for the centralizer of $a$ in $G$. Observe that $|Cl_G(a)|=|G:C_G(a)|$.
Let the set $F=\{ a \in G : |Cl_G(a)| \lt \infty \}$. In general, if $H,K$ are subgroups of a group $G$  with $|G:H| \lt \infty$ and $|G:K| \lt \infty$, then $|G:H \cap K| \lt \infty$. Hence, if $a,b \in F$, then $C_G(a) \cap C_G(b)$ has finite index in $G$. Since $C_G(a) \cap C_G(b) \subseteq C_G(ab)$, it follows that $ab \in F$.
Finally, it is easy to show that $C_G(a)=C_G(a^{-1})$, whence if $a \in F$, then $a^{-1} \in F$. So $F$ is a subgroup.
If $\alpha \in Aut(G)$, then clearly (use that $\alpha$ is homomorphic and a bijection), $\alpha[C_G(a)]=C_G(\alpha(a))$, which shows that $F$ char $G$. 
A: Let $H=\{\text{elements of}\, G\, \text{having finite conjugates}\}$. Note that if $A=\{\text{conjugates of} \,g \, \text{in} G\}$ then $|A|=|G:C(g)|$. Now if $f$ is an automorphism on $G$ then $f(C(g))=C(f(g))$ i.e. if $B=\{ \text{conjugates of} \, f(g) \,\text{in} \,G\}$ then $|B|=|G:C(f(g))|=|A|$. This implies that $f(g)$ has finitely many conjugates in $G$.So $f(H)$ is a subset of $H$. Similar process show that $f^{-1}(H)$ is a subset of $H$. Hence $f(H)=H$ that is $H$ char $G$.
