Homomorphisms help.

LuisRodg · 2009-03-22 04:33:49

I been stuck in this question for a long time and would appreciate some help.

JaneFairfax · 2009-03-22 05:58:55

Answer hidden - Ricky

Ricky · 2009-03-22 06:05:47

I'm going to assume the question is stating that V is a group. A quick check on the Cayley Table will verify this if not. There is also a much nicer construction for what I'm about to say, but it requires use of Van Dyck's theorem, which I don't believe you would have seen yet.

V is a group of order 4, and so it is either Z/4Z or Z/2Z x Z/2Z. Taking a glance at the structure of the elements will indeed tell you that it is Z/2Z x Z/2Z. With the isomorphism type found, what we are really doing is finding another copy of the Klein-4 group in S4.

The Klein-4 group is described by 4 elements where all nonidentity elements have order 2. But looking at elements in S4, an element has order two if and only if it can be represented by disjoint 2-cycles (called transpositions). So let's look at all the elements that have 2 disjoint transpositions:

(12)(34), (13)(24), (14)(23)

Certainly an element of S4 can't have more than 2 disjoint transpositions. So there is only one other possibility. Figure this out, and it will give you a very good idea of what the elements of H have to be.

LuisRodg · 2009-03-22 11:10:10

Ricky, I didnt really understand your explanation.

I tried using Jane's H group but could not find a function to define the isomorphism.

Here is what I used:

I then was able to come up with a function from H to V which I proved to be a bijection and a homomorphism.

Thanks both.

Last edited by LuisRodg (2009-03-22 11:12:27)

Ricky · 2009-03-22 12:10:41

Ricky, I didnt really understand your explanation.

That's rather vague. If you consider the problem solved and don't care for the method of finding it, then we can leave it be. But if you want clarification, please be more specific.

I tried using Jane's H group but could not find a function to define the isomorphism.

There are only two groups of order 4 up to isomorphism: Z/2Z x Z/2Z and Z/4Z. If it's not the latter, it must be the first. There is no isomorphism necessary. But the isomorphism would be given by:

(12) -> (12)(34)
(34) -> (13)(24)

LuisRodg · 2009-03-22 12:31:04

Im sorry for being vague, I will surely come back in a hour or so and go over the post carefully.

However, I just wanted to make sure if what I did was wrong.

Last edited by LuisRodg (2009-03-22 12:32:25)

JaneFairfax · 2009-03-22 13:30:15

Why do you really want to define an isomorphism? The question isnt asking for one.

Just use the fact that any two groups of order 4 in which are isomorphic.

Ricky · 2009-03-22 14:02:29

Why do you really want to define an isomorphism? The question isnt asking for one.

There is nothing wrong with going above and beyond what a question asked. When I was first doing group theory, I did all sorts of computations like this with symmetric groups. It really helps you get a feel for what a group is.

Edit: Ok, there's nothing wrong as long as you know it isn't needed. If you don't know, then you're missing out. I suppose that was your point...

As to Luis's question:

I haven't checked your definition, as it seems overly complicated. As I said in the other thread you started, define isomorphisms by generators. In this case, (13) and (24) generate your group H. Define where these go, and it defines the entire isomorphism.

LuisRodg · 2009-03-23 07:05:31

Hello Ricky.

It seems im not understanding groups well enough as I feel lost. For example, in this same question, when they asked to find H subgroup of S4 and isomorphic to V, I thought the only way to establish the isomorphism is to find a function that is bijective and a homomorphism. Looking at your post as well as Jane's comment, you have mentioned there is no need for a function. And this completely confused me. I would REALLY appreciate it if you could explain this.

Im doing really well in my Real Analysis class as well as Axiomatic Set Theory, but Abstract Algebra is just beating me around.

In your post, when you mentioned:

"V is a group of order 4, and so it is either Z/4Z or Z/2Z x Z/2Z."

Why can you make that conclusion and what is "Z/4Z or Z/2Z x Z/2Z." ? Z mod out by 4Z? But what is 4Z? Etc. I dont think im understanding this.

Hopefully with this post you get a better grasp of where I feel lost, and it would really help if you could explain. I apologize for my earlier vague post.

Thanks.

Last edited by LuisRodg (2009-03-23 07:06:25)

Ricky · 2009-03-23 09:11:04

For example, in this same question, when they asked to find H subgroup of S4 and isomorphic to V, I thought the only way to establish the isomorphism is to find a function that is bijective and a homomorphism. Looking at your post as well as Jane's comment, you have mentioned there is no need for a function. And this completely confused me.

The definition of an isomorphism is that there exists a bijective homomorphism, yes. However, this does not mean one has to find such a map. If you can prove the map exists (without finding it), then that is enough.

Let me give you a simpler example. Prove to me that 189518123599594883469824968269829864264098262498494898487654654186546542 is even. Well, an even number by definition is a number n where n = k + k for some integer k. But you can conclude the above number is even without ever finding what k is by just looking at the 1's digit: it's 2, and therefore the above number is indeed even. It would be a complete disaster to try to find k without an infinite-precision calculator to help you.

So if we can show the existence of a map without actually finding a map, then we're done. It turns out that a lot of times, showing the existence of a map is actually easier than finding the map itself.

Why can you make that conclusion and what is "Z/4Z or Z/2Z x Z/2Z." ? Z mod out by 4Z? But what is 4Z? Etc. I dont think im understanding this.

You may have heard of them as "the integers modulo 4", also written as:

When I say Z/2Z x Z/2Z, I mean:

Where addition is component-wise modulo 2 (i.e (0, 1) + (1, 0) = (1, 1) and (1, 1) + (1, 0) = (0, 1)).

Now as for the claim I made, perhaps that is too advanced for you. Let me offer an alternate solution to your problem:

You should know that isomorphisms preserves the order of elements: if f is an isomorphism from G to H, and f(g) = h, then the order of g is equal to the order of h.

So if H were isomorphic to V, there would exist an isomorphism f from V to H. Now f((12)(34)) is an element of H, and by the above it must have order 2. So must all the other nonidentity elements of V.

What are the elements of order 2 in S_4? As I said before, they are all elements that you can represent by disjoint transpositions:

(12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23)

Well we want H to not be equal to V, so it's best to start by choosing something such as (12). From here, you have severely limited choices for what the other elements are mapped to. A little playing around and you should come across a solution in no time (considering that multiple solutions exist).

Math Is Fun Forum

#1 2009-03-22 04:33:49

Homomorphisms help.

#2 2009-03-22 05:58:55

Re: Homomorphisms help.

#3 2009-03-22 06:05:47

Re: Homomorphisms help.

#4 2009-03-22 11:10:10

Re: Homomorphisms help.

#5 2009-03-22 12:10:41

Re: Homomorphisms help.

#6 2009-03-22 12:31:04

Re: Homomorphisms help.

#7 2009-03-22 13:30:15

Re: Homomorphisms help.

#8 2009-03-22 14:02:29

Re: Homomorphisms help.

#9 2009-03-23 07:05:31

Re: Homomorphisms help.

#10 2009-03-23 09:11:04

Re: Homomorphisms help.

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