IFF Criterion for y to be a Conjugate of x^n

# IFF Criterion for y to be a Conjugate of x^n

Proposition 1: Let $G$ be a group and let $x, y \in G$, $n \in \mathbb{Z}$. Then $y$ is a conjugate of $x^n$ if and only if $y$ is the $n^{\mathrm{th}}$ power of a conjugate of $x$. |

**Proof:**$\Rightarrow$ Let $y$ be a conjugate of $x^n$. Then there exists an $a \in G$ such that:

\begin{align} \quad y = ax^na^{-1} \end{align}

- Let $z$ be a conjugate of $x$. Then there exists a $b \in G$ such that $z = bxb^{-1}$. So $x = b^{-1}zb$. Substituting this into the above equation yields:

\begin{align} \quad y &= a(b^{-1}zb)^na^{-1} \\ &=a\underbrace{(b^{-1}zb)(b^{-1}zb)...(b^{-1}zb)}_{n \: \mathrm{times}}a^{-1} \\ &= ab^{-1}z^nba^{-1} \\ &= (ab^{-1})z^n(ab^{-1})^{-1} \end{align}

- So $y$ is a conjugate of $z^n$, i.e., $y$ is the $n^{\mathrm{th}}$ power of a conjugate $z$ of $x$.

- $\Leftarrow$ Suppose that $y$ is the $n^{\mathrm{th}}$ power of a conjugate of $x$. Then $y = z^n$ where $z$ is a conjugate of $x$. Since $z$ is a conjugate of $x$ there exists an $a \in G$ such that $z = axa^{-1}$. So:

\begin{align} \quad y = z^n &= (axa^{-1})^n \\ &= \underbrace{(axa^{-1})(axa^{-1})...(axa^{-1})}_{n\: \mathrm{times}} \\ &= ax^na^{-1} \end{align}

- So $y$ is a conjugate of $x^n$. $\blacksquare$