Ratio distance similarity transformations: $|\varphi (a)- \varphi (b)| = k|a-b|$

Question

I am struggling with the following group geometry question. I am given that: A simililarity transformation is a non-constant map $\varphi : \mathbb{R^2} \to \mathbb{R^2}$ that leaves the ratios of distances invariant: for all $a,b,c,d \in \mathbb{R^2}$ with $a \neq b$ and $c \neq d$ we have $\frac{|\varphi(a)-\varphi(b)|}{|a-b|} = \frac{|\varphi (c)-\varphi(d)|}{|c-d|}$.

I have to use the isometries on $\mathbb{C}$, and prove the following:

1) The similarity transformation multiplies all distances by the same positive factor $k$, so basically $|\varphi (a)- \varphi (b)| = k|a-b|$.

2) The set of all similarity transformations $Sim(\mathbb{R}^2)$ is a subgroup of the permutation group $S(\mathbb{R}^2)$.

I am failing to understand how I can use the isometries on $\mathbb{C}$ for this, if someone can perhaps explain simply its relation to rotations, translations, etc in $\mathbb{R}^2$, that would clarify a lot I think.

Answer 1

For part 1), we start by picking any $c\neq d\in \mathbb{R}^2$. Apply $\varphi$ and set $k=\frac{|\varphi (c)-\varphi(d)|}{|c-d|}$. By definition we have $\frac{|\varphi(a)-\varphi(b)|}{|a-b|} = \frac{|\varphi (c)-\varphi(d)|}{|c-d|}$ for any $a$, $b\in \mathbb{R}^2$. This gives $$\frac{|\varphi(a)-\varphi(b)|}{|a-b|} = k$$ $$|\varphi(a)-\varphi(b)|=k|a-b|$$ For part 2), we need to show that these transformations are bijective (so that they are actually in $S(\mathbb{R})$ to begin with) and that their subgroup is closed under composition. Closure follows from 1) by writing $$\frac{|\varphi_2(\varphi_1(a))-\varphi_2(\varphi_1(b))|}{|a-b|}=\frac{k_2|\varphi_1(a)-\varphi_1(b)|}{|a-b|}=\frac{k_2k_1|a-b|}{|a-b|}=k_2k_1$$ which implies that the composition of two similarity transformations is again a similarity transformation, and the subgroup is closed.

That the transformations are injective is clear: $\varphi(x) = \varphi(y)$, $x\neq y$ implies $k=0$, but then $\varphi(a)=\varphi(x)$ for all $a\in \mathbb{R}^2$, and $\varphi$ is constant, which we don't allow.

For surjectivity, we can modify Qiaochu Yuan's answer here. I believe all that needs to be changed is the first line – we are mapping the sphere of radius $r$ to the sphere of radius $rk$, and it isn't an isometry. The rest still works. With that result, we know that the sphere of radius $r$ is mapped bijectively to the sphere of radius $rk$ centered at $\varphi(0)$. To show that a point $x$ is in the image of $\varphi$, we just take the image of the sphere of radius $r=\frac{|x-\varphi(0)|}{k}$.