Proof of Euler's Theorem with groups

Question

I read that in order to prove it using groups, we can see that the order of the group of units is equal to Euler's totient function: $$|U(n)|= \phi (n)$$ After this, however, the author goes on to say that this implies that $a^{\phi (n)}=1$ for all $a \in U(n)$. The only reason I can make this connection is that $U(n)$ might be cyclic, in which $a$ to the power of the order of the group is the identity. But what if $U(n)$ is not cyclic? How can we make this logical conclusion then?

Answer 1

This is a consequence of Lagrange's theorem.

For $g\in G$, where $G$ is a group, order of $g$ divides $|G|$. This is because the size of the subgroup generated by $g$ must divide $|G|$, and that size must equal the order of $g$.

Therefore, for some integer $k$,

$$ g^{|G|}=g^{o(g)*k}=1^k=1 $$

Apply this result to group of units $U(n)$ with group operation multiplication.

Answer 2

You don't need Lagrange's theorem when $G$ is a finite abelian group:

If $G$ is a finite abelian group of order $n$ and $a\in G$, then $a^n=1$.

Indeed, consider the map $x \mapsto ax$. This map is injective because $G$ is a group and so is surjective because $G$ is finite. Thus, the map permutes the elements of $G$. Therefore, if the elements of $G$ are elements $a_1, \dots, a_n$, then $a_1 \cdots a_n = (a a_1) \cdots (a a_n) = a^n a_1 \cdots a_n$ and so $a^n=1$.