Let $(G,\odot)$ be a finite group with identity element $e$. Show that for each $g \in G$, there is $k \in \Bbb N^+$ such that $g^k= e$

Question

This is Exercise $7.$ from Section 7: Groups and Homomorphisms, Chapter 1: Foundation, textbook Analysis I by Herbert Amann and Joachim Escher.

Exercise $7$: Let $(G,\odot)$ be a finite group with identity element $e$. Show that for each $g \in G$, there is $k \in \Bbb N^+$ such that $g^k := \underbrace{g \odot \cdots \odot g}_{k \text{ times}} = e$.

My attempt:

Assume the contrary that $g^k \neq e$ for all $k > 0$. Then $g^n \neq g^{n+p}$ for all $p>0$. If not, $g^n = g^{n+p}$ for some $p>0$ and thus $(g^{-1})^n \odot g^n = (g^{-1})^n \odot g^{n+p}$. It follows that $e = g^p$, which is a contradiction. So the set $\{g^k \mid k \in \Bbb N\}$ is infinite. This contradicts the fact that $G$ is finite.

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At the end of this exercise, the authors give a hint Use Exercise $1.$ where

Exercise $1$: Let $N$ be a subgroup of a finite group $(G,\odot)$. Show that $|G| = |N| \cdot |G/N|$ where $G/N$ is the set of left cosets of $G$ modulo $N$.

My questions:

1. Does my attempt look fine or contain gaps/flaws?

2. How to utilize Exercise $1$ to do Exercise $7$ as suggested by the authors?

Thank you for your help!

Answer 1

Your proof is fine. But here are some alternate proofs you may like:

Proof-I: Let, $g(\ne e)\in G$ since $G$ is a group so $g^n\in G$ for all $n\in\Bbb N$. Now since $G$ is finite so for any two distinct $m,n$ we have, $g^m=g^n\implies g^{m-n}=e$ set $t=m-n$ we have $g^t=e$. So, such $n>0$ exists.

Proof-II: Every finite cyclic group $(G,\circ)\cong_f \Bbb (Z_n,+)$ where $|G|=n$ and $f$ is the isomorphism. Since there always exists $m>0$ s.t. $mx=0,\forall x\in \Bbb Z_n$ so due to isomorphism $g^m=e$ where $f(g)=x$, for all $g\in G$.

Proof-III:[Using Excr. 1] Let, $g\in G\setminus N$ i.e. $g\notin N$ and suppose that $g^n\ne e$ for all $n>0$ then obviously $g^n\ne g^m$ for all $n\ne m$. So then the order/cardinality of the quotient space will be infinite hence, $|G/N||N|\ne |G|$ contradicting the theorem 1 for finite groups.

Hope this works.