This is Exercise 11 from Section 7: Groups and Homomorphisms, Chapter 1: Foundation, textbook Analysis I by Herbert Amann and Joachim Escher.
The proof of $\Longrightarrow$ is quite easy. Please help me verify that of $\Longleftarrow$. Thank you so much for your help!
Here $\mathrm{S}_G$ is the set of all bijections from $G$ to $G$ and $\circ$ denotes the composition of functions.
My attempt:
To simplify notation, I denote $g_1 \odot g_2$ by $g_1g_2$.
- $G$ is a group $\implies L$ is a subgroup of $\mathrm{S}_G$
For all $Lg_1, Lg_2 \in L$, we have $Lg_1 \circ Lg_2 = L(g_1g_2) \in L$. So $L$ is closed under $\circ$. For all $Lg \in L$, $Lg \circ Lg^{-1} = L(gg^{-1}) = Le = \operatorname{id}_G$ where $e$ is the identity element of $G$. So every $Lg \in L$ has the inverse $(Lg)^{-1} = Lg^{-1} \in L$. Hence $L$ is a subgroup of $\mathrm{S}_G$.
- $L$ is a subgroup of $\mathrm{S}_G \implies G$ is a group
If $Lg_1=Lg_2$ then $g_1 = g_2$. If not, there exists $g_1,g_2\in G$ such that $g_1 \neq g_2$ and $Lg_1=Lg_2$. Then $gg_1=gg_2$ for all $g \in G$ and thus $Lg$ is not injective for all $g \in G$, which is a contradiction.
For all $g \in G$, there exists $h\in G$ such that $Lg \circ Lh = L(gh) = \operatorname{id}_G = Le$. Then $gh=e$ and thus for all $g \in G$, there exists $g^{-1} = h \in G$ such that $gg^{-1} =e$.
We have $(Lg_1 \circ Lg_2) \circ Lg_3 = Lg_1 \circ (Lg_2 \circ Lg_3)$ and thus $(g_1g_2)g_3 = g_1(g_2g_3)$. Hence the operation on $G$ is associative. To sum up, $G$ is a group.
It seems fine to me.
However, there is one bit that could be more simple. If $g\odot g_1=g\odot g_2$ for all $g\in G$, and $G$ has an identity element $e$ by hypothesis, then, in particular, $g_1=e\odot g_1=e\odot g_2=g_2$.