Fix an element $h\in G$. Define $\varphi_h:G\to G$ by $\varphi_h(g)=hg$. Is that an isomorphism from $G$ to itself?

Question

Fix an element $h$ $\in$ $G$. Define $\varphi_h:G\rightarrow G$ by $\varphi_h(g)=hg$

Is the following an isomorphism from $G$ to itself. How can I prove this?

Any hints or help would be great.

Edit: Using my knowledge of group homomorphisms, I have added an attempt at proving the question above.

For each $h \in G$ let $\varphi_h : G \rightarrow G$ be the morphism $\varphi_h(g) = \varphi_h(g) + \varphi_h(h) − \varphi_h(g + h)$

Then $\varphi_0(g) = \varphi(0) + \varphi(g) − \varphi(g + 0) = 0$ for all $g \in G$.

the image of $\varphi_h$ is a single point for all $h \in G$, and since $\varphi_h(0) = \varphi_h(0)+ \varphi_h(h)-\varphi_h(0+h) = 0$, that point must be $0$.

It follows that $\varphi_h(g) = 0$ for all $g, h \in G$, therefore we always have $\varphi_h(g) + \varphi_h(h) = \varphi_h(g + h)$ and $\varphi_h$ is a group isomomorphism.

This may be non-sensical for group isomorphisms. Apologies if so.

Answer 1

Hint: Isomorphisms send the identity element in the domain to that of the codomain. Indeed, $e=ee$, so for an isomorphism $\psi: G\to G$, we have

$$\begin{align} e\psi(e)&=\psi(e)\\ &=\psi(ee)\\ &=\psi(e)\psi(e), \end{align}$$

so cancel $\psi(e)$ on the right. Hence $e=\psi(e)$.

But $\varphi_h(e)=he=h$ while $h\neq e$.