under what assumptions will $\phi_G$ be surjective

Question

Let $G$ be a group. For each $g\in G$, define $L_g:G\to G$ by $L_g(h)=gh$. Define $\phi_G : G\to Bi(G)$ by $\phi_G(g)=L_g$ where $Bi(G)$ denotes the bijection of $G\to G$. Now under what assumptions will $\phi_G$ be surjective.

My attempts : It's easy to show that $\phi_G$ is injective. Now we want $\phi_G$ be surjective. Then it will be bijective. So if G is finite , lets say $|G|=n$. Then $|Bi(G)|=n!$. And bijection needs $|G|=|Bi(G)|$, i.e $n=n!$. So $n=1$ or $n=2$. But what about if G is infinite. Any hints would be helpful.

Answer 1

If $G$ has at least three distinct elements $1,a,b$, then there exists a bijection $G\to G$ that swaps $a\leftrightarrow b$ while keeping everything else (including $1$) fixed. However, any left multiplication that fixes $1$ must be the identity.

Answer 2

Hagen von Eitzen has given an argument that works in general, but since you gave a nice cardinality-based argument in the case that $n$ is finite, I thought I should mention that your argument also works if $n$ is infinite!

The key point is that the set $\operatorname{Bij}(A)$ of bijections of a set $A$ has greater cardinality than $A$ for any set $A$ of cardinality at least 3. See this other math.SE question. So there cannot exist a bijection from $A$ to $\operatorname{Bij}(A)$.