Math Is Fun Forum

Yes it made sense.

So then if I want to prove the following:

Sorry for not being useful, but I read the whole thing and really liked it. It kept me interested the whole time I was reading.

If I have more time later I will criticize it and give you more useful feedback!

By the way, I love the fact you are writing this for an English class

Execellent!

Thanks.

Edit:

Can A5 contain a 4-cycle or a 2-cycle? Given alpha a 4-cycle or a 2-cycle then its signum will be -1 and hence it is an odd permutation, not in A5?
Also, it cannot contain a combination of 2 and 3 such as (12)(345) because of the above argument as well?

I guess you could upload the .doc document in any free upload site and provide the link here?

Are you sure you copied the problem correctly? The denominator doesnt factor. It doesnt have any real roots.

Please make a new post and we will try to help you.

But I want to prove that G is Abelian, so dont I have to pick the arbitrary elements x,y in G and not in G/Z(G) ?

Which one is it?

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.

All lines have the form y = mx + b where m is the gradient and b is the y-intercept. You have the m (gradient) and you have a point (x,y). Substituting yields you the b and hence the equation of the line.

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.