# Show that given conditions, element is in center of group G

## Homework Statement

Let $G$ be a finite group and $m$ a positive integer which is relatively prime to $|G|$. If $b\in G$ and $a^mb=ba^m$ for all $a\in G$, show that $b$ is in the center of $G$.

## The Attempt at a Solution

Let $|G| = n$ and $b\in G$. Note that by Bézout 's identity $nx + my = 1$ for some $x,y\in\mathbb{Z}$. Also, note that $a^m = ba^mb^{-1}$. So

\begin{align*} a^m &= ba^mb^{-1}\\ (a^m)^y&= (ba^mb^{-1})^y\\ a^{my} &= ba^{my}b^{-1}\\ a^{1-nx} &= ba^{1-nx}b^{-1}\\ a(a^n)^{-x}&= ba(a^n)^{-x}b^{-1}\\ a(e)^{-x}&= ba(e)^{-x}b^{-1}\\ a&= bab^{-1}\\ \end{align*}

Since $a$ is an arbitrary element of $G$, we see that $b\in Z(G)$ ☐

Is this the correct argument?

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## Homework Statement

Let $G$ be a finite group and $m$ a positive integer which is relatively prime to $|G|$. If $b\in G$ and $a^mb=ba^m$ for all $a\in G$, show that $b$ is in the center of $G$.

## The Attempt at a Solution

Let $|G| = n$ and $b\in G$. Note that by Bézout 's identity $nx + my = 1$ for some $x,y\in\mathbb{Z}$. Also, note that $a^m = ba^mb^{-1}$. So

\begin{align*} a^m &= ba^mb^{-1}\\ (a^m)^y&= (ba^mb^{-1})^y\\ a^{my} &= ba^{my}b^{-1}\\ a^{1-nx} &= ba^{1-nx}b^{-1}\\ a(a^n)^{-x}&= ba(a^n)^{-x}b^{-1}\\ a(e)^{-x}&= ba(e)^{-x}b^{-1}\\ a&= bab^{-1}\\ \end{align*}

Since $a$ is an arbitrary element of $G$, we see that $b\in Z(G)$ ☐

Is this the correct argument?
Yes. Because $m$ and $n$ are coprime, $a$ generates the same elements as $a^m$ so they can be interchanged.