Given a group G and a set J. J={phi: G>G : phi is an isomorphism}. Prove J is a group (NOT a subgroup!}
So out of four properties of a Group, I've been able to get 2.
1.) Identity:
Here we have the identity function. phi_e(g)=g such that every g put in comes out. Here we know the function is a bijection (defined in class). Since the elements are not changed, we know that it preservers the operation. Hence, phi_e is in J.
2.) Inverse:
Given any phi, since it is a bijection it lends itself to having an inverse. In addition to that, we know from class that an inverse isomorphism is an isomorphism. So, for every phi in J there is a phi^{1}.
I'm stuck on proving closure and associativity.
An odd group theory questions
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Re: An odd group theory questions
I assume the group operation is function composition, right?
So you want to show that the composition of two automorphisms of G is again an automorphism. You can show this directly by proving bijectivity and the homomorphism property, which both behave nicely under composition.
For associativity, isn't it true that function composition in general is associative?
So you want to show that the composition of two automorphisms of G is again an automorphism. You can show this directly by proving bijectivity and the homomorphism property, which both behave nicely under composition.
For associativity, isn't it true that function composition in general is associative?
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Re: An odd group theory questions
I guess I'm stuck on how to demonstrate that it is generally associative meaning how to show that the phi's behave properly in closure. It seems like it should be easy, but I guess its these stupid little details that trip me up.

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Re: An odd group theory questions
Could a sufficient proof go as follows:
Suppose we have two isomorphisms of G, [math]\phi[/math]_{1} and [math]\phi[/math]_{2} both sending G > G. We know that given any g_{1} and g_{2} in G, g_{1}g_{2} is in G. So, since all these isomorphisms in A simply move elements from G > G, composing two isomorphisms with a given element in G will simply send the element to G once again. In class we demonstrated this somewhat using the transitivity of an isomorphisms. Since G is a group, and J is all the isomorphisms from G > G, we won't be able to pick and element g such that phi_1 compose phi_2 will end up outside of G, and thus causing the composition to be outside of J.
Suppose we have two isomorphisms of G, [math]\phi[/math]_{1} and [math]\phi[/math]_{2} both sending G > G. We know that given any g_{1} and g_{2} in G, g_{1}g_{2} is in G. So, since all these isomorphisms in A simply move elements from G > G, composing two isomorphisms with a given element in G will simply send the element to G once again. In class we demonstrated this somewhat using the transitivity of an isomorphisms. Since G is a group, and J is all the isomorphisms from G > G, we won't be able to pick and element g such that phi_1 compose phi_2 will end up outside of G, and thus causing the composition to be outside of J.
Last edited by chaotixmonjuish on Tue Oct 13, 2009 3:33 am UTC, edited 1 time in total.
Re: An odd group theory questions
You've proved that phi1 composed with phi2 is indeed a function from G to G. But your set J isn't just a set of functions from G to G. It's a set of isomorphisms from G to G. So for closure, you need phi1 composed with phi2 to be an isomorphism as well.
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Re: An odd group theory questions
Since composition is the operation (I think I forgot to say that) wouldn't it already be proved. Since composing them still generates an isomorphism, it is still in J. Would that work as a conclusion that it is infact closed?
Re: An odd group theory questions
The fact that the composition of two isomorphisms is itself an isomorphism is part of what you're being asked to prove, yes.
If you've already proven it (or your professor has shown it in lecture or whatever), good for you.
If you've already proven it (or your professor has shown it in lecture or whatever), good for you.
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Re: An odd group theory questions
Associativity: So if you have F := (AB)C and G := A(BC), you have to show that they are equal.
A function is determined by it's range, and where it sends elements in the range.
So to show that they are equal, you just have to show that F and G have the same range (can you show that?), and that they send arbitrary elements to the same value.
A function is determined by it's range, and where it sends elements in the range.
So to show that they are equal, you just have to show that F and G have the same range (can you show that?), and that they send arbitrary elements to the same value.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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Re: An odd group theory questions
I think perhaps you mean domain...
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Re: An odd group theory questions
skeptical scientist wrote:I think perhaps you mean domain...
Yes, yes I do. *sheepish grin*.
Always get those two mixed up.
So to rewrite:
Yakk, minus dumbassness wrote:Associativity: So if you have F := (AB)C and G := A(BC), you have to show that they are equal.
A function is determined by it's domain, and where it sends elements in the domain.
So to show that they are equal, you just have to show that F and G have the same domain (can you show that?), and that they send arbitrary elements to the same value.
This ends up being really trivial; but if you don't know that function composition in general is associative, then proving this kind of trivial thing is what you are expected to do. And if you cannot toss off a proof why this is true, then it would be a good idea to write out the proof for practice.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
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Re: An odd group theory questions
I suspect that they're allowed to assume that function composition is associative, but not allowed to assume that the composition of two isomorphisms is an isomorphism. Doing this two line definition fiddle is what this question is about imo.
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