I need to prove the following without using Lagrange's Theorem:
Show that for an abelian group $G$, $\forall \; a \in G:$ $a^{o(G)}=e$ .
This is a generalization of the Euler-Phi Theorem.
So I looked at the proof I have for the Euler-Phi theorem, which shows that $a^{\phi (m)} = e (mod \; m)$ by showing that $(\mathbb{Z} / m \mathbb{Z})^x = \{[b_1], ..., [b_{\phi (m)}]\} = \{[a \cdot b_1], ..., [a\cdot b_{\phi (m)}]\}$, using the $gcd$ operator. I'm stuck on what set I would use here to prove this in a similar way for $a^{o(G)}$.
The proof of the Euler-Phi theorem works by exhibiting an isomorphism in order to calculate $a^{\phi(n)}$. This proof can be done by considering the isomorphism $\varphi_a : G \rightarrow G$ by sending $b$ to $ab$. In particlar, note that $$\prod_{b \in G} b = \prod_{b \in G} ab$$ and see if you can conclude the proof. This last step is where the hypothesis that $G$ is abelian is essential.