# Is the set a group if left translations form a group?

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 $$g_4 = g_2$$.

For convenience, I denote $$Lg$$ by $$L_g$$.
Let $$a,b\in G$$.
Then $$L_aL_b=L_c$$ for some $$c\in G$$.
Hence $$(L_aL_b)(e)=L_c(e)\implies a(be)=ce\implies ab=c.$$ Thus for every $$a,b\in G$$, $$L_aL_b=L_{ab}$$ By using the information you obtained, this shows that the operation on $$G$$ is associative.
Let $$a\in G$$. Then there exists $$b\in G$$ such that $$L_aL_b=\iota=L_bL_a.$$ So we have $$(L_aL_b)(e)=\iota(e)=(L_bL_a)(e)\implies L_{ab}(e)=e=L_{ba}(e)\implies (ab)e=e=(ba)e\implies ab=e=ba.$$