Why can different symmetry groups be isomorphic?

Question

What I understand :

A group is characterized by how any two of its elements interact. It is the "structure" of the group that describes it. Thus two groups having the same structure are basically the same (isomorphic).

The additive group $\Bbb R$ can be regarded as the set of all sliding symmetries of the "number" line. Any number can be seen as the sliding action that takes $0$ to that number.

Similarly the multiplicative group $\Bbb R^+$ is the set of all stretching/squishing symmetries of the number line. A number here represents the action that takes $1$ to that number.

The problem :

It seems to me that the two groups mentioned above are very different in the sense that sliding and stretching/squishing are very different types of actions. But, the two groups are isomorphic. I would always see two isomorphic groups as being the same; however, after looking at $\Bbb R$ and $\Bbb R^+$ as groups of symmetries of the number line, I'm not so sure I would consider them as being the same.

P.S. I am not familiar with the concept of a "group action"

I'd suggest adjusting to ask a specific question. Either way, for what it's worth, these groups are isomorphic for the same reason that the same data can be plotted on a graph with a linear or logarithmic scale. — Travis Willse, Jun 10 '18 at 10:00
You should become comfortable with group actions, and then try to prove some form of a theorem that argues that two isomorphic groups act on the same things — Exit path, Jun 10 '18 at 11:13
@leibnewtz since many answers used this term I am currently looking into this concept (which is completely new for me) — Not Euler, Jun 10 '18 at 11:14
Well, {0,1} is a simple group. How can we think about it interacting with a square? Well, it could invert it along one of the diagonals. Or it could rotate it 180 degrees. Those are two diferent symmetries of the square, but do you think we should study two different {0, 1} groups separately? Why, they would have same multiplication tables, therefore everything same, so we just say it's the same thing. And we just abstract the square away, calling it an action. — Serge Seredenko, Jun 10 '18 at 14:29
An isomorphism is exactly what the definition says it is: a bijective homomorphism $f:G \to G'$. There's no reference in there to the flavor of the group multiplication, the labels of the underlying set, etc. — anomaly, Jun 10 '18 at 17:09

score 7 · Answer 1 · answered Jun 10 '18 at 10:18

Everything I talk about can be formally defined, but as your question seems to be one on intuition I will leave everything as intuition.

Two groups $G$ and $H$ are isomorphic if the have the same group structure - that is we can relabel elements of $G$ and the group operation to get $H$, in your example we replace '$0$' with '$1$' and $x$ with $e^x$ say to go from $\mathbb{R}$ to $\mathbb{R}^+$ (and obviously $+$ with $\times$).

But the structure you refer to in the question is the action of $G$. Two actions of a group need not be the same in structure (despite $G$ obviously being isomorphic to itself). For example $C_2\times C_2$ can be thought of as the subgroup of $S_4$ generated by $(1,2)$ and $(3,4)$ or the subgroup generated by $(1,2)(3,4)$ and $(1,3)(2,4)$. These actions are not equivalent (one is transitive, the other isn't). It so happens that the two actions you give are equivalent, by the same relabelling that gives the isomorphism.

score 6 · Accepted Answer · answered Jun 10 '18 at 19:58

Let me add a bit of spin to the existing answers: what's going on here is that you find group actions more natural things than groups "on their own." This is totally reasonable and personally I agree, but the point is that the notion of group "on its own" is specifically intended to abstract away from the specific action. Why do we want to do this? Well, one answer is that groups, rather than group actions, are also interesting in their own right; a less "pure" answer would be that - as often happens - studying the more "abstract" objects can yield data in the "concrete" case.

However, at the same time it's worth noting that there is also a notion of isomorphism of group actions: if I have an action $\alpha$ of a group$G$ on a set $X$ and an action $\beta$ of a group $H$ on a set $Y$, then $\alpha$ and $\beta$ are isomorphic if there are bijections $f:G\rightarrow H$ and $g:X\rightarrow Y$ which "make everything commute," that is, such that for all $c\in G$ and $x,z\in X$, we have $\alpha(c,x)=z\iff \beta(f(c),g(x))=g(z)$. With this notion the actions of $\mathbb{R}$ on $\mathbb{R}$ and $\mathbb{R}^+$ on $\mathbb{R}$ you describe are not isomorphic! This is a good exercise; as a hint, think about fixed points (can an action of the latter ever move $0$?).

That said, plenty of actions we may think of as not being the same are in fact isomorphic; e.g. the mutliplicative of $\mathbb{R}^+$ on $\mathbb{R}^+$ is isomorphic in the above sense to the additive action of $\mathbb{R}$ on $\mathbb{R}$. So there is still some potential non-intuitiveness here.

The point is that there are two distinctions you need to have in mind: equality versus isomorphism and groups versus group actions. Once you internalize these, I think the confusion will disappear.

Why does Robert Chamberlain say that the two actions I gave are equivalent? — Not Euler, Jun 10 '18 at 20:24
@HritRoy You'd have to ask him, but I suspect he interpreted your second action in the manner of my second-to-last paragraph: $\mathbb{R}^+$ acting on itself, not on $\mathbb{R}$. — Noah Schweber, Jun 10 '18 at 20:27

José Carlos Santos · Answer 3 · 2018-06-10T10:04:34.230

3

There is no contradiction. What you wrote is about the way each group acts naturally on $\mathbb R$. It has nothing to do with the way the elements of each group interact with each other. And this interaction is what determines that they are isomorphic.

edited Jun 10 '18 at 10:04

answered Jun 10 '18 at 09:58

José Carlos Santos

427,504

Sorry it was a typing mistake. I corrected it. – Not Euler Jun 10 '18 at 09:59
I've edited my answer. I hope that everything is clear now. – José Carlos Santos Jun 10 '18 at 10:04
I understand why they are isomorphic. What I don't understand is why groups are considered the same when they are isomorphic if they can determine two different set of symmetries of the same thing? – Not Euler Jun 10 '18 at 10:06
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(1) Groups being considered the same when they are isomorphic is just a definition of a sense of "same". (2) In this example, these aren't symmetries of the "same thing": (A) The underlying objects are different (one is $\Bbb R$, the other is $\Bbb R^+$), and (B) in one case we're looking at symmetries of the additive structure, in the other the multiplicative structure. – Travis Willse Jun 10 '18 at 10:12
@Travis can't we think of them as symmetries of an infinite line just like we think of $D_8$ as the group of symmetries of a square? – Not Euler Jun 10 '18 at 10:27
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@HritRoy Consider the group $(\mathbb{R},+)$. It acts on $\mathbb R$ through adding $\lambda$. And it also acts on $\mathbb R$ through multiplication by $e^\lambda$. Geometrically, the actions are different. So, by your argument, $(\mathbb{R},+)$ wouldn't be isomorphic to itself. Actually, no group other than the trivial group would be isomorphic to itself. – José Carlos Santos Jun 10 '18 at 10:31
Correct me if I'm wrong, but isn't the term "natural action" associated with the action of the permutation group $A(S)$ on the set $S$? Were you using the term loosely? – Not Euler Jun 10 '18 at 20:19
@HritRoy I was using it in this sense: if $(G,\cdot)$ is a group, a natural action of $G$ is the action of $G$ on itself such that the action of the element $g$ is defined by $h\mapsto g.h$. – José Carlos Santos Jun 10 '18 at 20:49
@JoséCarlosSantos 1. In case of a non Abelian group isn't there another action which should be called by the same name? That is, multiplying by $g$ on the right? 2. What about the action of $\Bbb R^+$ on $\Bbb R$? How would you call that natural? The way multiplication usually works? – Not Euler Jun 10 '18 at 20:52
@HritRoy 1. SInce the group is abelian, multiplying on the left or on the right is the same thing. 2. You are right about $\mathbb{R}^+$ and $\mathbb R$. But note that what I called natural action of $\mathbb{R}^+$ on itself extends naturally (no pun intended) to the action that you decribed in the original question. – José Carlos Santos Jun 10 '18 at 21:14
@JoséCarlosSantos Multiplying real numbers by positive numbers the way we usually do, yes? – Not Euler Jun 10 '18 at 21:17
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@HritRoy Yes, that is what I mean. – José Carlos Santos Jun 10 '18 at 21:26
@HritRoy If by symmetries you mean the isometries (distance-preserving maps from the line to itself), that's actually a third (in fact) nonabelian group, since this allows (like $D_8$ in the square case) reflections. If we require that the maps also preserve orientation of the line, then the resulting group of oriented isometries is $\Bbb R$. – Travis Willse Jun 11 '18 at 13:01
@HritRoy But notice that the usual isomorphism between $(\Bbb R, +)$ and $(\Bbb R^+, \cdot)$ and the usual distance function $d(x, y)$ then induce a (logarithmic) notion of distance of $(\Bbb R^+, \cdot)$, namely, $d'(a, b) := \log | a / b |$, and by construction this notion of distance is preserved by scaling ("stretching"), since $d'(\lambda a, \lambda b) = \log | (\lambda a) / (\lambda b) | = \log | a / b | = d'(a, b)$. In other words, with these distance functions, sliding on $\Bbb R$ and stretching on $\Bbb R^+$ are equivalent. – Travis Willse Jun 11 '18 at 13:05

JonathanZ · Answer 4 · 2018-06-10T16:32:47.270

As you said in a comment:

What I don't understand is why groups are considered the same when they are isomorphic if they can determine two different set of symmetries of the same thing?

This happens a lot. Here's a super-simple example that may be more instructive than the examples in your question:

Let's write $\mathbb{Z}_2$ as $\{I, A\}$, where $I$ is the identity and $A^2 = I$. We let $\mathbb{Z}_2$ act on $\mathbb{R}^2$ two different ways:

1) $A(x,y) = (-x,y)$

and

2) $A(x,y) = (x, -y)$

What the heck, why not add a third:

3) $A(x,y) = (y,x)$

When we look at groups as actions on spaces there are two components: things pertaining to the group itself, and properties of the space being acted on. $\mathbb{R}^2$ happens to have a very uniform structure so there are lots of ways $\mathbb{Z}_2$ can act on it. When we decide to do group theory we choose to "factor out" the properties of the group being acted on and do our best to ignore them, or at least get results that are independent of the. [BTW, you can look at representation theory to see what happens when we decide we want to care about the set we're acting on again.]

Another very simple analogy that might help is an answer to the question "What is 2?". One response is that 2 is the thing in common between two apples, two cars, two teachers, etc.. This is sometimes stated as "take two apples, two cars, two teachers and remove the apple-ness, the car-ness, the teacher-ness. Keep doing this for all pairs of things and what you have left is 2.". Similarly, groups were first noticed as actions on things, but we've tried to remove the dependence on things and ended up with modern group theory.

score 0 · Answer 5 · answered Jun 10 '18 at 12:23

From your intuitive description it seems that the following may be helpful. Suppose we have an arithmetic progression where there is a constant difference $\,d\,$ between consecutive terms. Another way to describe this is that given $\,d\,$ there is a mapping $\, x \mapsto x + d \,$ that takes each term to the next term. Similarly with geometric progressions where there is a constant quotient $\,q\,$ between any two consecutive terms. Thus given $\,q\,$ there is a mapping $\, x \mapsto x \, q \,$ that takes each term to the next term. What this gives us is a mapping between arithmetic progressions and geometric progressions and the two actions you describe behave similarly and are isomorphic. The isomorphism can be made explicit by an exponential/logarithmic map. In fact, John Napier invented logarithms in this way using parallel arithmetic and geometric progressions. Consult the MSE question 47927 "Motivation for Napier's Logarithms" for details.

score 0 · Answer 6 · answered Jun 10 '18 at 12:43

Groups are naturally thought of as "sets of symmetries".

One natural way they arise is by looking at how different, more or less geometric or algebraic objects are symmetric.

But one group can arise as the symmetry group of different objects: one example is the one you gave. The very same group ($(\mathbb{R},+)$ and $(\mathbb{R}_+^\times, \times)$ are isomorphic as you mentioned and so they are "essentially the same group") can arise both as the translations of the real line and the dilatations of the real line.

Now these two types of symmetries are geometrically different, but the way they behave is the same: that's why the groups are isomorphic.

There are other examples; from the most elementary ones (the one you described for instance, but also how $\mathfrak{S}_n$ can be seen as certain symmetries of a vector space; how $\mathbb{Z}/2\mathbb{Z}$ can be seen as switching things) to the more advanced ones (the fundamental group of a "nice" topological space that can be seen as the symmetries of its covering spaces).

The point of this is: even if two types of symmetries "look" very different "geometrically", they can behave very similarly, and that's what the isomorphism of two groups of such symmetries can reveal. Studying how different those types of symmetries may look geometrically for a given group is one of the ideas underlying representation theory.

The sentence "the group $G$ arises as a symmetry group of $X$" is the intuitive way to see that $G$ acts on $X$ in a manner that preserves its structure (whatever that means). Since you're not familiar with group actions, there's no need to go any further than that; but that's just to say that the concept of group action is one of the most natural ones: historically, groups appeared first as actions, before becoming the abstract groups that are now taught; and group actions are what first motivated the group axioms.

Why can different symmetry groups be isomorphic?

6 Answers6