# Does associativity imply bijectivity in group operations?

• I
• valenumr
In summary: In general, associativity does not imply bijectivity. It is possible to have a subgroup of S3 that only operates on 2 elements, which is isomorphic to S2. There is no specific concept or term for a group where the operator acts on all elements in a bijective way. If the operator is both injective and surjective, it is bijective. The relationship between the group operator on the group and the size of the generator set vs the order of the group is still being studied. The smallest order of a group must be n or greater. It is possible for a group with order |Gn| < n to
valenumr
TL;DR Summary
Are group operators always bijective?
Quick question: do the group axioms imply that the group operator is bijective? More in general, does associativity imply bijectivity in general?

I can think about a subgroup of S3 that only operates on 2 elements, but it is really isomorphic to S2.

But is there some concept or term for a group where the operator acts on all elements in a bijective way?

If it's both injective and surjective, then it's bijective, right?

sysprog said:
If it's both injective and surjective, then it's bijective, right?
Well, yeah. I'm asking if that is always true for a group operator. I guess my S3 subgroup operator still maps as bijective? Anyhow, I'm just thinking about the relationship between the group operator on the group and the size of the generator (? terminology) set vs the order of the group.

What got me thinking about it is the question of the smallest order of a group. I think it must be n (like Cn must be |Cn| = n) or greater.

Oof. Maybe I can state that better, or perhaps not. If you take one transpose group of S3, it is still a proper subgroup, equivalent to S2. But in say, S3 space, it is of a smaller order than 3, while in S2 space, it is equal to order 2. I don't know if that means anything at all, but I am wondering if there is some meaningful distinction for groups where |G| >= n vs the contrary.

sysprog
Have you looked at the SU(3) non-Abelian group? (it's part of why the existence of the top quark was predicted)

sysprog said:
If it's both injective and surjective, then it's bijective, right?
So back to the original question. It is by more or less definition? That's my impression.

sysprog said:
Have you looked at the SU(3) non-Abelian group? (it's part of why the existence of the top quark was predicted)
Hehe. I'm not that far into group theory, but working toward lie algebra.

So what I am thinking, is it possible to make a statement that any group with order |Gn| < n is necissarily a subgroup of some G'?

But again I'm about 1/6 along studying group theory.

valenumr said:
So what I am thinking, is it possible to make a statement that any group with order |Gn| < n is necissarily a subgroup of some G'?

But again I'm about 1/6 along studying group theory.
*Normal subgroup of G'.

valenumr said:
So what I am thinking, is it possible to make a statement that any group with order |Gn| < n is necissarily a subgroup of some G'?

But again I'm about 1/6 along studying group theory.
Kind of feeling silly on this... Not sure but, is it true that a group can be a subgroup of some other group. That wasn't what I was aiming for.

What do you mean by group operator here?

WWGD, martinbn and sysprog
PeroK said:
What do you mean by group operator here?
Yeah ##-## I was hoping that @valenumr would notice that 'non-Abelian' means (in the context) non-commutative . . .

There's a difference between group (binary) operation (elementary concept) and an operator on a group (more advanced concept).

sysprog
PeroK said:
There's a difference between group (binary) operation (elementary concept) and an operator on a group (more advanced concept).
Quite so, @PeroK ##-## I was trying to hint at the concept regarding whether the result of applying an operation to two elements of a group does or does not or may or may not depend on the order of the elements (thanks Niels Abel).

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sysprog said:
Quite so ##-## I was trying to hint at the more elementary concept regarding whether the result of applying an operation to two elements of a group does or does not or may or may not depend on the order of the elements (thanks Niels Abel).
Technically, it makes little sense to ask whether ##+:G \times G \rightarrow G## is a bijection. Unless we have the trvial group of a single element, then for all ##a \in G## we have ##a + (-a) = 0##.

sysprog
valenumr said:
So what I am thinking, is it possible to make a statement that any group with order |Gn| < n is necissarily a subgroup of some G'?

But again I'm about 1/6 along studying group theory.

It sounds to me like this is the main question. Is your goal here that |G'|=n? If so then no, because if ##G\subset G'## is a subgroup, you must have the order of G divides the order of G'. So is |G| =n-1 for example, you can't do it unless n=2.

If you don't care about the size of G' it's trivial, ##G'= G\times \mathbb{Z}_2##.

It's not clear to me if this is the overall question of the thread.

valenumr and sysprog
I think that I'm not alone in sometimes wishing that I could sometimes give more than one like per post.

PeroK said:
Technically, it makes little sense to ask whether ##+:G \times G \rightarrow G## is a bijection. Unless we have the trvial group of a single element, then for all ##a \in G## we have ##a + (-a) = 0##.
PS The left and right group operations are, of course, bijective:
$$L_a: G \rightarrow G, \ s.t. \ L_a(g) = ag$$$$R_a: G \rightarrow G, \ s.t. \ R_a(g) = ga$$

sysprog
A group operation is a homomorphism ##G \stackrel{\varphi }{\longrightarrow} \operatorname{Sym}(X)## where ##G## is the group, ##X## a set it operates on, and ##\operatorname{Sym}(X)## the permutation group ##S_{|X|}## of the elements of ##X.##

Hence it is a different concept than the structure of ##G##. That means that ##G## can operate on many different ##X## leading to many different operations ##\varphi ,## or its operations are not of inetrest. Of course, you can always define an operation ##G \longrightarrow \operatorname{Sym}(G)## on itself by conjugation ##x\stackrel{\varphi }{\longmapsto} (g\longmapsto xgx^{-1})## or it is naturally given as in the case of matrix groups (operating on vector spaces) or permutation groups (operating on ##\{1,2,\ldots,n\}),## but this is not automatically in the center of consideration.

Your answer has therefore to be 'no' because we can have different sets ##X## which a group can operate on, but different sets ##X## lead to different groups ##\operatorname{Sym}(X)## and they cannot all be isomorphic to ##G##. To name the trivial counterexample look at ##\varphi \equiv 1##. We can always define ##g.x :=x## as operation where every element of ##G## is mapped to the identity permutation of ##X.##

sysprog
PeroK said:
What do you mean by group operator here?
I mean the composition operator defined for the group.

sysprog said:
Yeah ##-## I was hoping that @valenumr would notice that 'non-Abelian' means (in the context) non-commutative . . .
Yes,I know what it means.

valenumr said:
I mean the composition operator defined for the group.
That is multiplication? Then you have ##G \times G \longrightarrow G.## How could this be an isomorphism?

Office_Shredder said:
It sounds to me like this is the main question. Is your goal here that |G'|=n? If so then no, because if ##G\subset G'## is a subgroup, you must have the order of G divides the order of G'. So is |G| =n-1 for example, you can't do it unless n=2.

If you don't care about the size of G' it's trivial, ##G'= G\times \mathbb{Z}_2##.

It's not clear to me if this is the overall question of the thread.
Thanks, this is mostly where I am trying to figure things out (working on quotient groups), and I think I just confused myself.

fresh_42 said:
That is multiplication? Then you have ##G \times G \longrightarrow G.## How could this be an isomorphism?
I think we are disconnected. If the group is just (Z, +), I mean '+' is the group operater.

valenumr said:
I think we are disconnected. If the group is just (Z, +), I mean '+' is the group operater.
Yes, but it is a binary operator. It has to variables, one on the left and one on the right. If you fix one of them you get the mappings @PeroK mentioned in post #18. Say we fix ##a\in G## and consider ##L_a(g)=ag##. Then we have a bijection between ##G## and ##aG##. But this isn't neither a homomorphism nor is ##aG## a group.

Edit: corrected, as @martinbn pointed out in post 27.

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fresh_42 said:
Yes, but it is a binary operator. It has to variables, one on the left and one on the right. If you fix one of them you get the mappings @PeroK mentioned in post #18. Say we fix ##a\in G## and consider ##L_a(g)=ag##. Then we have a bijection between ##G## and ##aG##. But this is neither a homomorphism nor is ##aG## a group.
I see. I think my original question is a bit silly. I was trying to see if '+', for example (the group operator) was always bijective just by definition of a group. I was working on a problem and got a weird answer, but I made a mistake.

fresh_42 said:
A group operation is a homomorphism ##G \stackrel{\varphi }{\longrightarrow} \operatorname{Sym}(X)## where ##G## is the group, ##X## a set it operates on, and ##\operatorname{Sym}(X)## the permutation group ##S_{|X|}## of the elements of ##X.##
Isn't the usual terminology "group action" rather than "group operation"?
fresh_42 said:
Yes, but it is a binary operator. It has to variables, one on the left and one on the right. If you fix one of them you get the mappings @PeroK mentioned in post #18. Say we fix ##a\in G## and consider ##L_a(g)=ag##. Then we have a bijection between ##G## and ##aG##. But this is neither a homomorphism nor is ##aG## a group.
Well, ##aG## is equal to ##G##, since ##a\in G##.

fresh_42
valenumr said:
I see. I think my original question is a bit silly. I was trying to see if '+', for example (the group operator) was always bijective just by definition of a group. I was working on a problem and got a weird answer, but I made a mistake.
This should have been obvious. It cannot be the case if the group has more than one element, because ##a+(-a)=0## or with multiplicative notation ##aa^{-1}=1## for all ##a##. Also for finite groups ##G\times G## has more elements than ##G##.

martinbn said:
Isn't the usual terminology "group action" rather than "group operation"?
Group action, group operation, and group representation are all the same.

The latter is often associated with linear operations on a vector space. Action is less ambiguous than operation (see the discussion here), but if it is used correctly, i.e. operates on, then it is the same.
martinbn said:
Well, ##aG## is equal to ##G##, since ##a\in G##.
Touché.

valenumr said:
I mean the composition operator defined for the group.
Are you/we distinguishing between a "group operator" and a "group action"?

Stephen Tashi said:
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 said:
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.

valenumr said:
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.##

fresh_42 said:
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.

fresh_42 said:
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?

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