I Does associativity imply bijectivity in group operations?

[valenumr]

Are you/we distinguishing between a "group operator" and a "group action"?

If I understand right, a group action is like a group operating on a set, but I'm still learning (think groups 101). Anyhow, my initial question seems silly in hindsight, or maybe I'm missing something. I just assumed that the operator was inherently bijective, but I never read anywhere that it was, and I'm not sure how to drive the truth of that question base on the definition of a group.

 

[valenumr]

If I understand right, a group action is like a group operating on a set, but I'm still learning (think groups 101). Anyhow, my initial question seems silly in hindsight, or maybe I'm missing something. I just assumed that the operator was inherently bijective, but I never read anywhere that it was, and I'm not sure how to drive the truth of that question base on the definition of a group.

And to clarify a little, it seems obvious for finite groups, but may be a little more messy for infinite groups. I expect aG (or Ga) to recover all elements of the group for all a in G. I haven't had time to revisit the concept with infinite groups just yet.

 

[fresh_42]

If I understand right, a group action is like a group operating on a set, but I'm still learning (think groups 101). Anyhow, my initial question seems silly in hindsight, or maybe I'm missing something. I just assumed that the operator was inherently bijective, but I never read anywhere that it was, and I'm not sure how to drive the truth of that question base on the definition of a group.

Consider a group homomorphism $\varphi \, : \, G \longrightarrow \operatorname{Sym}(X)$ which means $\varphi(g\cdot h)(x)=\varphi (g)(\varphi (h)(x)).$ $G$ is the group here, and $X$ a set, $\operatorname{Sym}(X)$ the symmetry group of $X,$ i.e. the group that shuffles the elements of $X$.

Now you can say $G$ acts on $X$, $G$ acts on $X$ via $\varphi$, $G$ operates on $X$, $G$ operates on $X$ via $\varphi$, or $(G,\varphi )$ is a representation of $G$ on $X$. These are all the same, and people often write $g.x$ instead of $\varphi (g)(x).$ The homomorphism property is then written as an operation property $(g\cdot h).x=g.(h.x).$

The term representation is often reserved for linear representations. This is when $X=V$ is a vector space and $G$ operates via regular linear functions, i.e. $\varphi \, : \,G\longrightarrow \operatorname{GL}(V).$ Then we call $V$ the representation space. $G$ still acts or operates on $V,$ only by certain functions (isomorphisms of $V$) and not all bijections of $X=V.$

 

[valenumr]

Consider a group homomorphism $\varphi \, : \, G \longrightarrow \operatorname{Sym}(X)$ which means $\varphi(g\cdot h)(x)=\varphi (g)(\varphi (h)(x)).$ $G$ is the group here, and $X$ a set, $\operatorname{Sym}(X)$ the symmetry group of $X,$ i.e. the group that shuffles the elements of $X$.

Now you can say $G$ acts on $X$, $G$ acts on $X$ via $\varphi$, $G$ operates on $X$, $G$ operates on $X$ via $\varphi$, or $(G,\varphi )$ is a representation of $G$ on $X$. These are all the same, and people often write $g.x$ instead of $\varphi (g)(x).$ The homomorphism property is then written as an operation property $(g\cdot h).x=g.(h.x).$

The term representation is often reserved for linear representations. This is when $X=V$ is a vector space and $G$ operates via regular linear functions, i.e. $\varphi \, : \,G\longrightarrow \operatorname{GL}(V).$ Then we call $V$ the representation space. $G$ still acts or operates on $V,$ only by certain functions (isomorphisms of $V$) and not all bijections of $X=V.$

Honestly, you are a bit ahead of me. I'm not quite clear on the difference between homomorphism and isomorphism, though they look very similar to me at the moment.

 

[valenumr]

Consider a group homomorphism $\varphi \, : \, G \longrightarrow \operatorname{Sym}(X)$ which means $\varphi(g\cdot h)(x)=\varphi (g)(\varphi (h)(x)).$ $G$ is the group here, and $X$ a set, $\operatorname{Sym}(X)$ the symmetry group of $X,$ i.e. the group that shuffles the elements of $X$.

Now you can say $G$ acts on $X$, $G$ acts on $X$ via $\varphi$, $G$ operates on $X$, $G$ operates on $X$ via $\varphi$, or $(G,\varphi )$ is a representation of $G$ on $X$. These are all the same, and people often write $g.x$ instead of $\varphi (g)(x).$ The homomorphism property is then written as an operation property $(g\cdot h).x=g.(h.x).$

The term representation is often reserved for linear representations. This is when $X=V$ is a vector space and $G$ operates via regular linear functions, i.e. $\varphi \, : \,G\longrightarrow \operatorname{GL}(V).$ Then we call $V$ the representation space. $G$ still acts or operates on $V,$ only by certain functions (isomorphisms of $V$) and not all bijections of $X=V.$

By the way, when you say symmetry group, do you me the maximal (I think) order group for X?

 

[fresh_42]

By the way, when you say symmetry group, do you me the maximal (I think) order group for X?

I mean the group of all bijective functions $X \longrightarrow X$.

 

[valenumr]

I mean the group of all bijective functions $X \longrightarrow X$.

Ok, I think that's in line with what I thought, basically Sn from my material.

 

[valenumr]

Ok, I think that's in line with what I thought, basically Sn from my material.

Eh, let me walk that back a little. I mean, I'm still working on things where group operators are essentially scalar operations. I'm not sure what to expect yet when it goes beyond that.

 

[valenumr]

Are you/we distinguishing between a "group operator" and a "group action"?

I thought about this for a while, because, as I've mentioned, pretty much all of the material examples I've encountered so far have been based on integers, with a sprinkling of reals, and mostly finite groups. Think like an advanced ninth grader or typical eleventh grader could probably easily handle the material.

In any case, my consideration would be a nice Persian rug I happen to own. It does have multiple symmetries, say left / right reflection, front back reflection, and top / bottom reflection.

So when I say "group operator", I'm thinking of the concept of one element of the group composed with another (single) element of the group. When you say "group action", which I expect is more appropriate, it's like rotating the carpet some multiple of pi / 2 on the floor, or maybe flipping it over. It affects the whole rug, but it is just applying the operator to every element.

I'm still trying to think about non-unitary stuff as well that might shrink my rug, for example. But I don't what to get ahead of the material.

In any case, my other comment about working with scalars is mostly a reflection. I suspect that whole "abstract" part of abstract algebra doesn't really care about the group elements or group operator, so long as the rules apply.