# 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?

fresh_42
Mentor

## 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.

Mr Davis 97