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Let $G$ be an (affine) algebraic group. Fix a $g \in G$. Prove that the left multipication map $\varphi: x \mapsto gx$ is an isomorphisms of (affine) variety $G$.

Could you show me explicitly why this map is a morphism and, moreover, an isomorphism?

Note: Here are some definitions I use. Let $X$ be an (affine) variety. A regular funtions on $X$ (or a morphism $X\to \mathbb{A}^1$) is a restriction of a polynomial map on $X$. A morphism of varieties $X\to \mathbb{A}^n$ is a $n-$tuple of regular functions on X. A morphisms of varieties $X\to Y\subset\mathbb{A}^n$ is a morphism $X\to \mathbb{A}^n$ with image in $Y$. A morphism $X\to Y$ is an isomorphism iff its associative $k-$algebra homomorphism $k[Y]\to k[X]$ is an isomorphism.

Thank you very much.

Huy Hùng Lê
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    Since $G$ is an algebraic group, multiplication $\mu : G \times G \to G$ is a morphism. Left multiplication by an element $g$ is just the restriction of $\mu$ to the variety ${g} \times G$. Its inverse is multiplication by $g^{-1}$. – Exit path Nov 13 '17 at 22:29
  • What about this map: $\phi: x\mapsto g^{-1}xg$? Could you apply in the same way above? – Huy Hùng Lê Nov 14 '17 at 15:40
  • You can use a similar argument. In this case it's a composition $G \to G \times G \times G \to G \times G \to G$, where the first map sends $x$ to $(g^{-1}, x, g)$, the second to $(g^{-1}x, g)$ and the last to $g^{-1}xg$. – Exit path Nov 14 '17 at 21:32

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