Let $G$ be a set with an operation $\odot$ and identity element (i.e. there exist $e \in G$ such that $e \odot g = g \odot e = g$ for all $g \in G$). For $g \in G$, define the left translation $Lg: G \rightarrow G, h \mapsto g \odot h$. Suppose $Lg$ is bijective for each $g \in G$. Prove that if the left translations form a group with respect to function composition, then $G$ is a group.
I can't even prove that $\odot$ is associative.
What I have so far:
- $\odot$ is associative if and only if $Lg_1 \circ Lg_2 = L(g_1 \odot g_2)$ for all $g_1, g_2 \in G$. This is a trivial result, since $(Lg_1 \circ Lg_2)(h) = g_1 \odot (g_2 \odot h)$ and $L(g_1 \odot g_2)(h) = (g_1 \odot g_2) \odot h$ for all $g_1, g_2, h \in G$.
- Let $g_1, g_2 \in G$. Since the left translations form a group, there exists $g_3 \in G$ such that $Lg_1 \circ Lg_2 = Lg_3$. Since $Lg_1$ is bijective, there exists $g_4$ such that $g_3 = g_1 \odot g_4$. I don't know how to show $g_4 = g_2$.