Re: Abstract Algebra Proof Using the First Isomorphism Theory

It means the Kernel can **have** the matrix you described as an element.

The kernel is a **set** containing matricies from your group G. The kernel **is not** a matrix.

If you take out one of the matricies in the set we call the Kernel, It has the property that under the mapping $\displaystyle \phi$ it becomes 0 (the real number)

So basically we pick up all the matricies in our group G which under the mapping defined, go to 0, then we throw all these elements into a set and call it the kernel.

Re: Abstract Algebra Proof Using the First Isomorphism Theory

So we've shown that it is onto, a homomorphism, and that ker(phi) = H. THAT'S IT, RIGHT!?

Re: Abstract Algebra Proof Using the First Isomorphism Theory

Now tell me why you think $\displaystyle \phi $ is onto ? and after this we are indeed done.

Re: Abstract Algebra Proof Using the First Isomorphism Theory

Phi is said to map G onto H if for each element x in G there exists an element y in H with Phi(x) = y. NO?

Re: Abstract Algebra Proof Using the First Isomorphism Theory

You got it backwards, Phi is said to map G onto H if for each element y in H, there exists an element x in G with Phi(x) = y.

Basically its saying, no matter which element you pick up in H, i can always find an element in G which maps to that said element. So there can be no element in H which is not hit by some element in G. So G hits every element in H. Thus G is onto H

Re: Abstract Algebra Proof Using the First Isomorphism Theory

Thank you, thank you, thank you so much! You have a way of explaining things. Wish I hadn't waited so long to have someone explain everything so thorough. You are awesome. Hey, I'm willing to give you my first born (if I ever have one) to repay my debt! Thanks!

Do you mind me bugging you a bit more?

EDIT: I don't need to write that it is onto in a fancy way? Just saying it in layman's terms? Would that get my point across?

Re: Abstract Algebra Proof Using the First Isomorphism Theory

Not sure what fancy way is. Just say, here look, You say, give me any element in the real numbers, call it x. Now i say, look i'll take the same x you gave me (just a real number) and i'll use it as a component in my matrix, i certainly have the matrix $\displaystyle A = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} $ in my group G (Yes?) Then $\displaystyle \phi(A) = x + 0 + 0 + 0 = x $. There you go i found a matrix, an element, in my group which maps to the element you gave me.

Re: Abstract Algebra Proof Using the First Isomorphism Theory

Thanks again. Hopefully, I'll impress my professor. You're awesome!