# Show that a group is isomorphic to $\Bbb R$

$\def\R{\mathbb R}$Let $$G = \mathrm{Aff}(\R) := \{ f\colon\R\to\R\mid f \colon x\mapsto ax+b, a \in \R^*, b \in \R \}$$ Then $(G,\circ)$ is a group where $\circ$ denotes function composition $(f \circ g)(x) = f(g(x))$. Now let $$N := \{ f\in G\mid f\colon x\mapsto x+d, d \in \R \}$$ and $$H := \{ g\in G \mid x \mapsto cx, c \in \R^*\}$$ I have shown that $G$ is a group and that $N$ and $H$ are subgroups of $G$. Now I have to show that $N$ is isomorphic to $\R$ and that $H$ is isomorphic to $\R^*$. After that I have to show that $G$ is the semidirect product of $N$ by $H$.

I am stuck at showing that they are isomorphic. Help is appreciated. Thanks

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The bijections are implicitly given when the sets $N,H$ were given by formulas: each $f\in N$ corresponds to its determining parameter $d\in\Bbb R$, and each $g\in H$ corresponds to its parameter $c\in\Bbb R^*$. The only thing to prove is that they preserve the given group operation, $+$ in case of $N$ and $\cdot$ in case of $H$. – Berci Nov 23 '12 at 12:07

At first try to show that $G\simeq N \rtimes H$. I think that's easy to show. This is not requested but I do believe it's a good exercise.

Now my hint. To do what you want demonstrate that: $\forall \tau_d \in N$ such that $\tau_d(x)= x+d$, $\varphi: N \rightarrow \mathbb{R}$ such that $\varphi(\tau_d)= d$ is an isomorphism. "Same" thing try to do for $H$.

In this sense $G\simeq \mathbb{R}\rtimes \mathbb{R}^{*}$.

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ok so i have shown that φ(τd) is an isomorphism for both N and H. Only thing left now is to show that G is a semidirect product of N and H, G≃N⋊H. – grendizer Nov 26 '12 at 11:24
If you want an idea consider: $\chi: N \rtimes H \rightarrow G$ such that $\chi((a,b))=\tau_{a,b}$, considering that $\tau$ such that $\tau(x)=ax+b$. – Ivan Nov 26 '12 at 23:22
thanks! got it all done. – grendizer Nov 29 '12 at 9:38
You're welcome! – Ivan Nov 29 '12 at 10:54

Well, let's look at the mapping $\theta\colon \mathbb{R} \to N$ defined like this: for every $d \in \mathbb{R}$ set $\theta(d) = f$, where $f\colon \mathbb{R} \to \mathbb{R},\ f(x) = x + d \quad \forall x \in \mathbb{R}$. In human words, $\theta$ maps each $d \in \mathbb{R}$ to the function that "shifts" all numbers by $d$.

Can you prove that $\theta$ is a homomorphism from $\mathbb{R}$ to $N$? That it is an isomorphism?

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