But we have read that only bijective mappings exist inversely. If we already assume sigma inverse, then there is nothing left to prove. So why...........?
You probably mixed sigma(inverse of g) with inverse of sigme(g). That both are the same thing is not immediately obvious, it has to be proved, it is not given from start!
Maybe my intuition is wrong but sounds like it could be possible to prove the axiom 2 from axiom 1; here's a start: in e.(g1.a1) = (e×g1).a1, the subexpression (g1.a1) produces some element a2, so let's rewrite the expression e.a2. On the other hand, (e×g1).a1 is equal to g1.a1. But we already rewrote that as a2. That means that e.a2 = a2. But maybe that's how far it goes... I'm stuck at trying to prove that the same goes for a1. (which is _any_ element whereas a2 is _some_ element)
Here's a rather silly example which shows why axiom 2 is independent from axiom 1. Let G be any nontrivial group and let A be any set containing more than 1 element. Fix an element x in A. Then define the "action" of G on A by g∙a = x for every g in G and for every a in A. This _does_ satisfy axiom 1 of a group action (in a rather silly way). For any g1 and g2 in G and any a in A, (g1*g2)∙a = x (by definition of the "action" since g1*g2 is an element of G and since every element of G sends every element of A to the fixed element x). Additionally, g1∙(g2∙a) = g1∙x = x (by the same reasoning). So (g1*g2)∙a = g1∙(g2∙a) for all g1 and g2 in G and for all a in A. But this "action" doesn't satisfy axiom 2, since A has more than one element. Say y is an element of A not equal to x. Then by definition of the "action", e∙y = x. Since this fake action satisfies axiom 1 but not axiom 2, you cannot prove axiom 2 from axiom 1.
OK, for all g in G, sigma_g is an injection; but is the set of all sigma_g's (some set of permutations of A) uniquely determined by G? Of course, as groups they will all be isomorphic, but can we tell in general how many different ones (as sets) there are?
I think you're asking: given group G and set A, could there be more than one way in which you could define a group action of G on A? The answer is yes. (Or, as you asked the question, no, the group action is not unique.) Consider A = complex numbers, G = the group of order 2, which I'll represent as ({0,1},+), i.e. addition modulo 2. Represent elements of A as (x+iy). Then 0.(x+iy) = x+iy 1.(x+iy) = -x+iy is a group action, and 0.(x+iy) = x+iy 1.(x+iy) = x-iy is also a group action. Thinking more on this: there are two other group actions in this system. There is always a trivial group action g.a=a for all g in G, a in A. And in this example we have in addition 0.(x+iy) = x+iy 1.(x+iy)= -x-iy So I have four group actions available (I assert without proof that these are the only group actions of cyclic group C2 on complex numbers Z) and these four group actions look very much like the Klein 4 group (i.e. are in some sense isomorphic to it). I bet there is some deep maths behind this, and maybe it will even be covered soon in this course.
Seriously this guy is way better than my algebra teacher. 2-sided inverse to prove the bijection is golden.
Yet another EXCELLENT video.
Very well explained. Thank you very much.
Thank you SO MUCH!!! This is very helpful.
These videos are great, thanks!
E X C E L L E N T V I D E O that helps out an idiot math major like me lol
But we have read that only bijective mappings exist inversely. If we already assume sigma inverse, then there is nothing left to prove. So why...........?
I also thought that there is something circular!
You probably mixed sigma(inverse of g) with inverse of sigme(g). That both are the same thing is not immediately obvious, it has to be proved, it is not given from start!
Maybe my intuition is wrong but sounds like it could be possible to prove the axiom 2 from axiom 1; here's a start: in e.(g1.a1) = (e×g1).a1, the subexpression (g1.a1) produces some element a2, so let's rewrite the expression e.a2. On the other hand, (e×g1).a1 is equal to g1.a1. But we already rewrote that as a2. That means that e.a2 = a2. But maybe that's how far it goes... I'm stuck at trying to prove that the same goes for a1. (which is _any_ element whereas a2 is _some_ element)
Hm, a quick web search says that you'll need the both axioms. Damn.
Here's a rather silly example which shows why axiom 2 is independent from axiom 1.
Let G be any nontrivial group and let A be any set containing more than 1 element. Fix an element x in A. Then define the "action" of G on A by g∙a = x for every g in G and for every a in A.
This _does_ satisfy axiom 1 of a group action (in a rather silly way).
For any g1 and g2 in G and any a in A,
(g1*g2)∙a = x (by definition of the "action" since g1*g2 is an element of G and since every element of G sends every element of A to the fixed element x).
Additionally, g1∙(g2∙a) = g1∙x = x (by the same reasoning).
So (g1*g2)∙a = g1∙(g2∙a) for all g1 and g2 in G and for all a in A.
But this "action" doesn't satisfy axiom 2, since A has more than one element. Say y is an element of A not equal to x. Then by definition of the "action", e∙y = x.
Since this fake action satisfies axiom 1 but not axiom 2, you cannot prove axiom 2 from axiom 1.
thankyou ssoooooo much!!
OK, for all g in G, sigma_g is an injection; but is the set of all sigma_g's (some set of permutations of A) uniquely determined by G? Of course, as groups they will all be isomorphic, but can we tell in general how many different ones (as sets) there are?
Can you clarify your question? I think I should be able to answer you, but I need to be sure of what you are asking.
I think you're asking: given group G and set A, could there be more than one way in which you could define a group action of G on A? The answer is yes. (Or, as you asked the question, no, the group action is not unique.) Consider A = complex numbers, G = the group of order 2, which I'll represent as ({0,1},+), i.e. addition modulo 2.
Represent elements of A as (x+iy). Then
0.(x+iy) = x+iy
1.(x+iy) = -x+iy
is a group action, and
0.(x+iy) = x+iy
1.(x+iy) = x-iy
is also a group action.
Thinking more on this: there are two other group actions in this system. There is always a trivial group action g.a=a for all g in G, a in A. And in this example we have in addition
0.(x+iy) = x+iy
1.(x+iy)= -x-iy
So I have four group actions available (I assert without proof that these are the only group actions of cyclic group C2 on complex numbers Z) and these four group actions look very much like the Klein 4 group (i.e. are in some sense isomorphic to it). I bet there is some deep maths behind this, and maybe it will even be covered soon in this course.
Somebody has roosters hehe.