$g^n = 1\Rightarrow g = 1$ in a group of size coprime to $n$

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 the 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 $\,\Rightarrow\, g^{\Large\color{#0a0}n}=1\,$ hence $\ \left[\, g^{\Large \color{#c00}m} = 1\:\!\right]^{\Large\frac{\bf\color{#c00}{1}}{\Large \color{#c00}m\vphantom{|_.}}\!\bmod \color{#0a0}n}\!\! \Rightarrow g^{\large\color{#c00}1}\! = 1\,$ (elaborated in prior link)

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${\bf Alternatively} \ \ o(g)\mid\color{#c00}m\ $ by $\,g^{\Large \color{#c00}m} = 1\, $ and Order Theorem,$\ \ o(g) := $ order of $\,g$

$\qquad\qquad\qquad\quad\, o(g)\mid \color{#0a0a}n = |G|\ $ by Lagrange's Theorem

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

${\bf Generally}\!: \ \ \ \ \ \ o(g)\mid m,n\!\iff\! o(g)\mid \gcd(m,n)\ $ by gcd Definition / Universal Property.

Key Idea $ $ The simple basic property that periods of an element are closed under subtraction so closed under gcd is ubiquitous in number theory and algebra - usually first encountered in simpler additive form for denominators of fractions, e.g. see here and here.