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Math Help - Abstract Algebra Proof Using the First Isomorphism Theory

  1. #1
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    Abstract Algebra Proof Using the First Isomorphism Theory

    Please look at attached file. I am having trouble using the first isomorphism theorem to prove a proof. Anyone care to help? Thanks!

    EDIT: Reason for the attached file is that I couldn't get all of the math symbols to copy over correctly. Sorry for the hassle.

    Abstract Algebra Proof Using the First Isomorphism Theory-number-4.png
    Last edited by bondvalencebond; November 26th 2012 at 05:52 PM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Can you find a homomorphism from G to R in which H is the kernel and the homomorphism is surjective
    Last edited by jakncoke; November 26th 2012 at 06:13 PM.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Could you please dumb it down for me? Pretend I'm a first grader. =P
    This is the fist roadblock I reach in abstract algebra and I can't seem to grasp at all any of this.
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Can you verify if  \phi(X) = x_{1, 1} + x_{1,2} + x_{2,1} + x_{2, 2} is a homomorphism from  G \to \mathbb{R} where the x_{i, j} are the ith row and jth col of the matrix.

    You need to do two verifications to ensure this is a valid homomorphism

    First verify that  \phi is well defined. Meaning if X = Y in  M_{2, 2}(\mathbb{R}) then  \phi(X) = \phi(Y) in  \mathbb{R}

    Next verify that  \phi is a group homomorphism meaning  \phi(X + Y ) = \phi(X) + \phi(Y)

    If  \phi is indeed a homomorphism then what is the Ker(\phi) (what elements from G does  \phi make 0 in  \mathbb{R}

    Also if  \phi is a homomorphism, then notice that the image of  \phi is  \mathbb{R} because elements of the form  \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} are in G. and these elements under  \phi map to x + 0 + 0 + 0 = x
    Last edited by jakncoke; November 26th 2012 at 06:24 PM.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Jak, you probably will think I'm the dumbest chick in the world and that I should switch majors but all you're saying looks like Japanese. My professor did not explain any of this into detail. How would I check if it's a homomorphism from G --> R?
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Okay, let me put it this way. I had never used matrices before this example. Linear Algebra wasn't a pre-req for this class and we haven't talked about a homomorphism being surjective. I can somewhat understand how to prove this if it were a natural group or Sn but I don't even know how to start doing what you've told me to do. I'm so overwhelmed.
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    Senior Member jakncoke's Avatar
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Take a deep breath.

    So lets start with the basics, what is a group homomorphism?

    First and foremost it is a well function. Just like every other function you have encountered in your life. You feed it an element in the domain and it pops out an element in the codomain. In this case the domain is the elements of G, and the codomain is  \mathbb{R}
    What is the fundamental property for something to be a function? It has to be well defined.


    Well what is well definition? It basically means that if you have x = y in the domain then the f(x) = f(y) in the range. Basically how could something be a function if it takes the same element (x = y), to two different values in the range?

    I made up a function \phi: G \to \mathbb{R} by saying if you input a 2x2 matrix into the function then this adds up all the components of the matrix to give us a real number (co domain). since i made up this functioni, i must show it is well defined, other wise it wouldnt be a function at all.
    So i said, our function  \phi: G \to \mathbb{R} is well defined.

    Now if X = Y, where  X, Y \in G then basically two matricies are equal if and only if their components are equal. so if  X = \begin{bmatrix} x_{1,1} & x_{1, 2} \\ x_{2, 1} & x_{2, 2} \end{bmatrix} and  Y = \begin{bmatrix} y_{1,1} & y_{1, 2} \\ y_{2, 1} & y_{2, 2} \end{bmatrix} then  x_{1,1} = y_{1,1}, x_{1, 2}  = y_{1, 2} etc...

    Now  \phi(X) just adds all the components of the matrix together, surely we get a number in the real number line. Since X = Y, and their components are equal, it is easy to see that the components of X added together, which is  \phi(X) , give us the same value as the components of Y added together, which is  \phi(Y)

    So we established that \phi is a well defined function

    Now to show that  \phi is a group homomorphism, we need to show that  \phi(X+Y) = \phi(X) + \phi(Y) Basically, we need to show that if we added the matricies together and then mapped it to the reals, we get the same result as first mapping X to the reals and mapping Y to the reals and then performing the addition of the two afore mentioned values.

     \phi(X+Y) = \phi(\begin{bmatrix} x_{1,1} + y_{1,1} & x_{1, 2} + y_{1,2}  \\ x_{2, 1} + y_{2,1}  & x_{2, 2} + y_{2,2}  \end{bmatrix}) = x_{1,1} + y_{1,1} + x_{1, 2} + y_{1,2} + x_{2, 1} + y_{2,1} + x_{2, 2} + y_{2,2} .

    Now  \phi(X) = x_{1,1} + x_{1, 2} + x_{2, 1} + x_{2, 2}  \phi(Y) = y_{1,1} + y_{1, 2} + y_{2, 1} + y_{2, 2}

    Now  \phi(X) + \phi(Y) = x_{1, 1} + x_{1, 2} + x_{2, 1} + x_{2, 2} + y_{1,1} + y_{1, 2} + y_{2, 1} + y_{2, 2}

    which is the same thing as  \phi(X+Y)

    Thus we proved  \phi(X+Y) = \phi(X) + \phi(Y) .

    Thus we showed  \phi : G \to \mathbb{R} is a homomorphism

    If you are good on this, i'll tell you the next step.
    Last edited by jakncoke; November 26th 2012 at 07:24 PM.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    YES! I understand! I'm so excited to finally understand all of this. Okay, so the next step is to find the kernel, which since this is a group homomorphism then the kernel is a subgroup of G? Do we need to show that the identity is in the kernel? Which would be the identity matrix? Or am I going the wrong way with this?
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Yes, the identity of the domain is always mapped to the identity of the codomain. Since you said the Kernel is a subgroup of G (which is true!), a subgroup always contains the identity no? so indeed the identity is in the Kernel.

    So what is a kernel of a homomorphism? It is basically the set of all elements in G (2x2 matricies) which go to 0 in the real numbers under the homomorphism. So for our homomorphism  \phi the kernel is the elements  X \in M_{2x2}(\mathbb{R} ) (just our good ol 2x2 matricies) such that \phi(X) =  \phi(\begin{bmatrix} x_{1,1} & x_{1, 2} \\ x_{2,1} & x_{2, 2} \end{bmatrix} = 0 (0 as in the 0 in the real numbers). Which means that for a matrix to be in the kernel of  \phi .  \phi(\begin{bmatrix} x_{1,1} & x_{1, 2} \\ x_{2,1} & x_{2, 2} \end{bmatrix}) = x_{1,1} + x_{1, 2} + x_{2, 1} + x_{2, 2} = 0 So the subgroup  Ker(\phi) contains all matricies whose components add up to 0. Does remind you of something you might have mentioned earlier? (Ding Ding, your subgroup H, that you defined earlier).

    So H = \ker(\phi) .

    Now the isomorphism theorem says

    G / Ker(\phi) = G / H = \phi(G)

    So what is the image of G, what are all the elements that G could possibly hit in the reals?
    Last edited by jakncoke; November 26th 2012 at 08:03 PM.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    So the zero matrix? That would make the elements from G by phi 0 in R.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    So the ker(phi) = { n in G| phi(n) = (0 0; 0 0)} = {0}?
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Well the matrix  X =  \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} (an element of G) is also in the kernel of  \phi
    Why? \phi(X) = -1 + 0 + 0 + 1 = 0 . So i just showed you a non zero element in the kernel.

    Yes the zero matrix is also in the Kernel but there might exist other elements too. Like the one above

    Also the map  \phi(X) takes a Matrix and gives our a real number. Use the definition of \phi that i gave earlier
    Last edited by jakncoke; November 26th 2012 at 08:08 PM.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Why can't it be the zero matrix?
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    The Kernel of  \phi is NOT an element of G. It is a set of elements of G. And the zero matrix is a member of this special set we call the Kernel of  \phi

    So Ker(\phi) = \{ \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}, .... , But every matrix in the kernel satisfies the property that if you add all the components together, which is what we defined  \phi(X) to mean, you get 0.
    Last edited by jakncoke; November 26th 2012 at 08:13 PM.
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    Re: Abstract Algebra Proof Using the First Isomorphism Theory

    Okay, I'm sorry I keep confusing things. Does this mean that the kernel can be x = [-n 0; 0 n] for n in R?

    Gotcha! As long as it satifies the property that when all components are added together = 0.
    Last edited by bondvalencebond; November 26th 2012 at 08:16 PM.
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