Given a group $G$ with neutral element $e$ and a subgroup $H \leq G$ as well as the equivalence relation $g_1 \sim g_2 \Leftrightarrow g_2^{-1} g_1 \in H$ (equivalence classes $[g]$). G be finite.
Prove that for all $g \in G: g^{|G|} = e$
This homework consisted of several prior tasks: Existence of bijective function between classes, Lagrange and existence of $k \in \mathbb{N}: g^k = e$, I was able to show those but am stuck on the final one.
So far I only found that if k divides |G|, then obviously $g^{n}=(g^k)^{\frac{n}{k} \in \mathbb{N}}=e$.
If you have Lagrange's theorem then you're most of the way home: you know that for any subgroup $H$ of $G$, $|H|$ divides $|G|$; if $k$ is the order of the subgroup $H=\langle g\rangle$, then what can you say about $g^k$?