Group operator bijectivity

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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?
 

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  • #2
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If it's both injective and surjective, then it's bijective, right? :oldconfused:
 
  • #3
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If it's both injective and surjective, then it's bijective, right? :oldconfused:
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.
 
  • #4
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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.
 
  • #5
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Have you looked at the SU(3) non-Abelian group? (it's part of why the existence of the top quark was predicted) :confused:
 
  • #6
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If it's both injective and surjective, then it's bijective, right? :oldconfused:
So back to the original question. It is by more or less definition? That's my impression.
 
  • #7
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Have you looked at the SU(3) non-Abelian group? (it's part of why the existence of the top quark was predicted) :confused:
Hehe. I'm not that far into group theory, but working toward lie algebra.
 
  • #8
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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.
 
  • #9
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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'.
 
  • #10
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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.
 
  • #11
PeroK
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What do you mean by group operator here?
 
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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 . . . :oops:
 
  • #13
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There's a difference between group (binary) operation (elementary concept) and an operator on a group (more advanced concept).
 
  • #14
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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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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##.
 
  • #16
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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.
 
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  • #17
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I think that I'm not alone in sometimes wishing that I could sometimes give more than one like per post.
 
  • #18
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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:
$$L_a: G \rightarrow G, \ s.t. \ L_a(g) = ag$$$$R_a: G \rightarrow G, \ s.t. \ R_a(g) = ga$$
 
  • #19
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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.##
 
  • #20
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What do you mean by group operator here?
I mean the composition operator defined for the group.
 
  • #21
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Yeah ##-## I was hoping that @valenumr would notice that 'non-Abelian' means (in the context) non-commutative . . . :oops:
Yes,I know what it means.
 
  • #22
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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?
 
  • #23
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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.
 
  • #24
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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.
 
  • #25
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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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