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

[Mr Davis 97]

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 $G$.

Homework Equations

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?

 

[fresh_42]

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 $G$.

Homework Equations

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.