Let $G$ be a group of order $n$ and $m$ be relatively prime to $n$; given that $g^m=e$, show that $g$ is an identity element.

Question

Let $G$ be a group of order $n$ and $m$ be relatively prime to $n$; given that $g^m=e$, show that $g$ is an identity element.

(In other words, I will have to show its order is one.)

I think there are different approaches to solve this question, it could be solved by using one of the Lagrange's theorems results starting off with $g^n=e$ (and I know how to prove this part) , and $\gcd(m,n)=1$ but I seem to be stuck here. I can't seem to connect the information correctly to lead me to the answer required.

Kindly would anyone explain me the steps clearly ?

Answer 1

The fact that $m$ and $n$ are relatively prime means that there exist integers $r$ and $s$ such that $1 = mr + ns$. Therefore: $$\begin{aligned} g &= g^1 \\ &= g^{mr + ns} \\ &= g^{mr}g^{ns} \\ &= (g^m)^r (g^n)^s \\ &= 1^r 1^s \\ &= 1 \end{aligned}$$ Line 5 follows from line 4 because $g^m = 1$ by assumption and $g^n = 1$ by Lagrange's theorem.

Answer 2

Lagrange's Theorem $\Rightarrow g^n=1\,$ thus $\ g^m = 1^m\Rightarrow\, g = 1,\,$ by raising to power $\,\frac{1}m \bmod n\,$

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${\bf Or}\ \ \ o(g)\mid\color{c00} m\ $ by $\,g^m = 1\, $ and the Order Theorem

$\ \ \ \ \ \ \ \ o(g)\mid \color{#}n = |G|\ $ by Lagrange's Theorem

$\ \Rightarrow \ o(g) = 1\ $ by $\,\color{c00}m,\color{0a0}n\,$ coprime, by hypothesis.

Remark $ $ Generally $\,o(g)\mid m,n\iff o(g)\mid \gcd(m,n)\ $ by the gcd Universal Property.