Why is left multiplication on a group bijective?

Question

I noticed the fact "left multiplication on a group is always bijective" is a very common argument (for instance to prove Sylow's theorem)

I don't see why left (and right) multiplication is bijective in general. All I can see is that left multiplication is an action of a group on itself, so let $G$ be a group: $$\cdot: G \times G \longrightarrow G$$ $$(g,g') \longmapsto g \cdot g'=gg'$$ Properties of group actions hold. I tried to prove this by contradiction but I'm not able to find any useful argument. Could anyone enlighten me?

Answer 1

Suppose $gx=gy$. Then $g^{-1} gx=g^{-1} gy \implies x=y$. That proves injectivity.

Suppose $g'\in G$. Then $g (g^{-1}g')=g'$ implies surjectivity.

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In addition, in terms of actions, left multiplication by an element of a group is transitive and faithful. These are somewhat similar notions.

Answer 2

The map $\ell_g$ defined by $\ell_g(h)=gh$ is a bijection $G\to G$. Indeed, note that the map $\ell_{g^{-1}}=\ell_g^{-1}$ because $$ (\ell_{g^{-1}}\circ\ell_g)(h)=g^{-1}gh=h$$ and similarly $$ (\ell_g\circ \ell_{g^{-1}})(h)=gg^{-1}h=h.$$ The analogous argument shows that the right multiplication operator is a bijection. Note that none of these operators is a group homomorphism except for example $\ell_e=\Bbb{1}_G.$