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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?

Shaun
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Davide Radaelli
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    Note: "symmetric group" means something specific in group theory. – Shaun Mar 17 '22 at 18:35
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    One way to do it is to see that it is isomorphic to the group of translations $t_a(x)=a+x$ of the real line under composition. – Randall Mar 17 '22 at 18:35
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    I'm not sure I understand the question. What exactly is the problem with applying Cayley's theorem to $\mathbb{R}$? – freakish Mar 17 '22 at 18:37
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    You have a bit of contradiction between 'look at $G$ purely as a set' and 'consider structure preserving maps' — there is no structure (to speak of) to be preserved on a set. As for what happens when you do that — it goes through perfectly fine. $Sym(\mathbb{R})$ is the set of all 1:1 functions $f(\cdot):\mathbb{R}\mapsto\mathbb{R}$ with composition as the group operation, and $(\mathbb{R},+)$ is a subgroup of this. – Steven Stadnicki Mar 17 '22 at 18:38
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    Continuing Randall's thought, the group of translations is in particular a subset of all bijections $\Bbb R \to \Bbb R$, which is a symmetric group of uncountably many objects. – Travis Willse Mar 17 '22 at 18:38
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    (And as a note, $(\mathbb{R},+)$ does preserve some structure — it preserves the metric of $\mathbb{R}$. It also preserves the total order relation on $\mathbb{R}$, and in fact the group of symmetries of $\mathbb{R}$ that preserve both of these things is isomorphic to $(\mathbb{R},+)$; this is a nice exercise.) – Steven Stadnicki Mar 17 '22 at 18:40
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    @StevenStadnicki In my experience "1:1" usually means injective, not bijective. – Noah Schweber Mar 17 '22 at 18:45
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    I suspect you're overthinking this. Suppose $\mathcal{G}$ is a group with underlying set $G$ (so $G$ is what we get from $\mathcal{G}$ by "forgetting" the group operation; e.g. if $\mathcal{G}=(\mathbb{R};+)$, then $G=\mathbb{R}$). Cayley's theorem tells us that $\mathcal{G}$ is isomorphic to a subgroup of the group $Sym_G$ of all self-bijections of $G$ under composition. For example, in the $\mathcal{G}=(\mathbb{R};+)$-case, a real number $r$ corresponds to the function $$F_r:\mathbb{R}\rightarrow\mathbb{R}: s\mapsto r+s.$$ (Note that there's nothing special about $(\mathbb{R}; +)$ here.) – Noah Schweber Mar 17 '22 at 18:47
  • If the above doesn't address your question, I think you need to be clearer: what exactly are you looking for that isn't part of that corresopndence? – Noah Schweber Mar 17 '22 at 18:51
  • I understand now. @NoahSchweber you were right I was overthinking it. I now see how $(\mathbb{R}, +)$ is isomorphic to the group of translations on the real line (equipped with composition), which is an example of the general case where groups are isomorphic to a subset of their symmetric group. – Davide Radaelli Mar 17 '22 at 19:20
  • One follow up question I'm still not sure about is: do we have a guaranteed technique for finding this subset of the symmetric group that's isomorphic? – Davide Radaelli Mar 17 '22 at 19:21
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    @DavideRadaelli Yes, Cayley's theorem tells you exactly where to look! Given any group $\mathcal{G}$ with underlying set $G$, for $g\in G$ let $$F_g:G\rightarrow G:h\mapsto gh.$$ Then we have $F_{g^{-1}}=(F_g)^{-1}$, $F_g=F_h$ iff $g=h$ and $F_g\circ F_h=F_{gh}$ - each of these points is worth checking, and it's a good exercise to see what goes wrong if we look instead at $G_g: G\rightarrow G: h\mapsto h*g$ - which means that ${F_g:g\in G}$ is a subgroup of $Sym_G$ isomorphic to $\mathcal{G}$ via the map $g\mapsto F_g$. – Noah Schweber Mar 17 '22 at 19:36
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    Re: intuition I think my post here might be helpful, incidentally; while it's written with a focus on the finite case, the ideas work fine for the general case (just replace "$S_n$" with "$Sym_G$"). Everything in Cayley's theorem is as explicit as one could hope. – Noah Schweber Mar 17 '22 at 19:39

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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

Dietrich Burde
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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).

Noah Schweber
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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 $y \in \mathbb{R}$.

Mike
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  • "Lets' also forget for a moment that $\mathbb{R}$ is uncountable." Why? How does cardinality enter the picture at all? – Noah Schweber Mar 17 '22 at 19:56
  • It probably does not, and I took that statement out. I am a discrete mathematician, and I am unfamiliar with how things work w uncountable sets. – Mike Mar 17 '22 at 19:58
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    If you imagine first writing down/looking at/picturing in your head the full set of symmetries of the set and only after that start looking for the relevant subgroup, then cardinality matters because in the uncountable case this first step is extremely scary and off-putting and you might risk never reaching the second step at all. (All this not withstanding that one can of course always jump directly to the subgroup regardless of cardinality). So forgetting that $\mathbb{R}$ is uncountable is good advise from a purely psychological standpoint, I'd say. – Vincent Mar 18 '22 at 13:19
  • Yes, it was psychological as much as anything else. It is much easier to imagine a permutation on a finite set, or even a permutation on the set of integers. I think many of us find it harder to imagine a permutation on an uncountable set such as $\mathbb{R}$. – Mike Mar 18 '22 at 14:01