In a group $G$, if $a \in G$ is the only element such that its order is 2, prove that $G = C(a)$

Question

(i) As $C(a) \leq G $, $x \in C(a) \implies x \in G$. Hence $C(a) \subset G$.

(ii) How to prove $G \subset C(a)$?

I tried to start with $a^2 = e$ and manipulate this to get $ax = xa$, but it doesn't work.

Answer 1

Hint: What is the order of $x^{-1}ax$?

Answer 2

Fun fact: in general, if $p$ is the smallest prime dividing $|G|$ and $N \unlhd G$ with $|N|=p$, then $N \subseteq Z(G)$.

Proof note that $G=N_G(N)$ and $G/C_G(N)$ can be embedded in Aut$(N) \cong C_{p-1}$ ($G$ acts by conjugation on $N$ with kernel $C_G(N)$). Since $p$ is minimal this implies $G/C_G(N)=\overline{1}$, whence $G=C_G(N)$, being equivalent to $N \subseteq Z(G)$.

Note that the uniqueness of your element $a$, implies $\{1,a\}$ is a normal subgroup of $G$.