Abelian group of finite order This problem was submitted to me as part of Abstract Algebra Homework.

Let $(G,\cdot)$ be an abelian group of finite order.
Show that if $o(G)=n$ then $a^n=e$ for all $a\in G $.
Hint : Show that if $G=\{e,a_1,a_2,\cdots,a_{n-1} \}$ then $G=\{a,a\cdot a_1, a \cdot a_2 , \cdots , a \cdot a_{n-1} \}$
with $o(\cdot)$ is the cardinality and $e$ is the identity element.

My first idea is that it reminded me of Cauchy's theorem for Abelian groups of finite order. However I did not manage to link it to it.
Edit 1: It was actually Lagrange theorem I was thinking of.
A colleague proposed :

*

*to prove that $\exists k \in [[1,n]], a^k =e$

*to prove the hint when $a \neq e$ (it is trivial otherwise), $\forall k < n, a^k \neq e$, "since otherwise the powers of $a$ form a smaller set of numbers than $G$ so you have to have $a^n =e$.

Edit 2: The proof must not use Lagrange's theorem.

Use the hint: let $b=ea_1a_2\dotsm a_{n-1}$. Then
$$
b=ae\cdot aa_1\cdot \dotsm aa_{n-1}=a^nb
$$
(because the group is abelian). Hence $a^n=e$ by cancelling $b$.
Why can we write $G=\{a,aa_1,\dots,aa_{n-1}\}$? Consider the map $\mu_a\colon G\to G$, $\mu_a(x)=ax$. This map is injective, hence also surjective.
