How can $(\mathbb{R}, +)$ be turned into a group of symmetries?

Question

I'm exploring the claim that group theory is about symmetry.

Given a group $G$ that isn’t a set of structure preserving maps with composition, can you always interpret it as some kind of symmetry?

My answer to this is yes, because of Cayley's theorem. The procedure as I understand it is to look at G as purely a set, and consider all the structure preserving maps on it. Some subset of this set of transformations equipped with composition will have the same behaviour as the original group (it will be isomorphic). Hence every group can be represented as a subset of group of symmetries.

My question is what happens specifically when you try to turn $(\mathbb{R}, +)$ into a subset of some group of symmetries?

Answer 1

Turning my comments into an answer:

I suspect you're overthinking this, and not looking closely enough at the details of Cayley's theorem.

The right way to state Cayley's theorem is the following:

If $\mathcal{G}=(G;*)$ is a group, then the map $$\mathit{Cayley}: \mathcal{G}\rightarrow \mathit{Sym}_G: g\mapsto{} "h\mapsto g*h"$$ is an injective group homomorphism, and so in particular we get an isomorphism $\mathcal{G}\cong ran(\mathit{Cayley})$.

This is completely explicit about how $\mathcal{G}$ is isomorphic to a group of permutations of some set. And we can then use this to directly compute examples: if we look at $(\mathbb{R};+)$, we have that $G=\mathbb{R}$, $*=+$, and $\mathit{Cayley}(g)$ is the map $h\mapsto g+h$.

This is obscured by snappier-but-less-informative versions of Cayley's theorem like "Every group is isomorphic to a group of permutations." And the "explicit" form of Cayley's theorem above is no harder to prove than this less-useful form (in fact I don't know of a proof of the latter which doesn't in fact show the former).

Answer 2

How can $(\mathbb{R}, +)$ be turned into a group of symmetries?

The isometry group of the Euclidean space $\Bbb E^n$ is $$ {\rm Isom}(\Bbb R^n)=E(n)=O_n(\Bbb R)\ltimes \Bbb R^n. $$ For $n=1$ this is the group $C_2\ltimes \Bbb R$, with the subgroup $(\Bbb R,+)$ of translations.

The same question for $(\Bbb Z;+)$ was asked here:

$\mathbb{Z}$ is the symmetry group of what?

For the relation between symmetry group and isometry group see also

Isometry group vs. Symmetry group

Answer 3

Well, I'm not sure if this is an answer per se, but this is too long to fit in the comments....

I think for a lot of students learning group theory, there is a lot of confusion between the group itself, and the set that the group acts upon. For example, $S_n$ is a group that acts upon $\{0,1,2,\ldots, n-1\}$, but $\{0,1,2,\ldots, n-1\}$ is not a group here, it is the set upon which $S_n$ acts upon. Anyway, I suspect that is what is happening here.

As for this confusion, indeed, groups act upon themselves too. In $(\mathbb{R},+)$, the set $\mathbb{R}$ itself is a group with a law of composition specified by the usual addition. The group $S_n$ also acts upon itself as well. Indeed, it is possible to write out a multiplication table for $S_n$.

Anyway, let us consider the following subgroup $H$ of $S_n$: $H = \{\pi_j; j \in \{0,1,2,\ldots, n-1\}\}$, where $\pi_j(i) = (i +j) \pmod n$. Then $|H|=n$, and $H$ is clearly commutative. Also, $H$ is indeed a subgroup of $S_n$ but for $n \ge 3$, clearly $H$ is not the whole of $S_n$. But, $H$ also acts upon itself; $\pi_{j_1}\pi_{j_2} = \pi_{(j_1+j_2) \pmod n}$ for each $j_1,j_2 \in \{0,1,2,\ldots, n-1\}$. But then, couldn't we just represent $H$ as follows then: $H=(\mathbb{Z}/n\mathbb{Z},+)$.

Anyways, put informally, $(\mathbb{R},+)$ could be thought of as this subgroup of the permutation group $S_{\mathbb{R}}$ of $\mathbb{R}$, where $(\mathbb{R},+) = \{\pi_x; x \in \mathbb{R}\}$, where $\pi_x$ is the element in $S_{\mathbb{R}}$ where $\pi_x(y)=x+y$ for each $y \in \mathbb{R}$.