Conjugate Elements in a Group

Definition: Let $G$ be a group and let $g_1, g_2 \in G$. Then $g_1$ is said to be a Conjugate of $g_2$ if there exists an $a \in G$ such that $g_1 = ag_2a^{-1}$.

The following proposition tells us that when $G$ is an abelian group then elements in $G$ are conjugate with one another if and only if they're the same element.

Proposition 1:Let $G$ be a group. If $G$ is abelian and $g_1, g_2 \in G$ then $g_1$ is a conjugate of $g_2$ if and only if $g_1 = g_2$.

• Proof: Let $G$ be an abelian group and let $g_1, g_2 \in G$.

• $\Rightarrow$ Suppose that $g_1$ is a conjugate of $g_2$. Then there exists an $a \in G$ such that $g_1 = ag_2a^{-1}$, i.e., $g_1a = ag_2$. Since $G$ is abelian, we have that $g_1a = g_2a$. Multiplying on the right by $a^{-1}$ gives us $g_1 = g_2$.

• $\Leftarrow$ Suppose that $g_1 = g_2$. Let $e \in G$ be the identity element. Then $g_1 = eg_2e^{-1}$. So $g_1$ is a conjugate of $g_2$. $\blacksquare$

Proposition 2: Let $G$ be a group and let $g_1, g_2 \in G$. If $g_1$ is a conjugate of $g_2$ then the orders of $g_1$ and $g_2$ are the same.

• Proof: Suppose that $g_1$ is a conjugate of $g_2$. Then there exists an $a \in G$ such that $g_1 = ag_2a^{-1}$. Let $i_a : G \to G$ be the inner automorphism defined for all $g \in G$ by $i_a(g) = aga^{-1}$. Since $i_a$ is bijective, we have that $i_a(g_2) = ag_2a^{-1} = g_1$.

• If $g_2$ has order $n < \infty$ then $n$ is the smallest positive integer such that $g_2^n = e$. Thus $g_1^n =[i_a(g_2)]^n = i_a(g_2^n) = i_a(e) = e$. So $\mathrm{ord}(g_1) \leq n$. If $\mathrm{ord}(g_1) = m < n$ then $g_1^m = e$, so:

(1)

\begin{align} \quad i_a(g_2^m) = [i_a(g_2)]^m = [g_1]^m = e \end{align}

• But since $i_a$ is a bijection, $i_a(g_2^m) = e$ if and only if $g_2^m = e$. Since $m < n$, this contradicts $\mathrm{ord}(g_2) = n$. Thus the assumption that $\mathrm{ord}(g_1) < \mathrm{ord}(g_2)$ is false. Hence $\mathrm{ord}(g_1) = \mathrm{ord}(g_2)$.

• A similar argument shows that if $g_2$ has infinite order then so does $g_1$. $\blacksquare$