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Modern Abstract

Posted by admin on August 10th, 2011

Modern Abstract Algebra Proof – any help would be greatly appreciated?

Let H be the subset of M2(R) consisting of all matrices of the form [ a-b]
[b a]
for a,b belong to R

a.) show that is isomorphic to
b.) show that is isomorphic to

For ease, I’m going to denote that matrix [a b] instead of it’s normal 2×2 representation.

You want to use the isomorphism:

φ( [a b] ) = a + b i

This is clearly a bijection. All you have to do is show that it is a homeomorphism, and to do that you need to prove:

Part a: φ( [a b] + [c d] ) = φ( [a b] ) + φ( [c d] )

Part b: φ( [a b] * [c d] ) = φ( [a b] ) φ( [c d] )

PART A:

This will be pretty trivial. First find [a b] + [c d]

[a -b] + [c -d] = [a+c -b-d ]
[b a] [d c] [b+d a+c ]

This means [a b] + [c d] = [a+c b+d].

And so φ( [a b] + [c d] ) = φ( [a+c b+d] ) = a+c+(b+d)i

Now look at φ( [a b] ) + φ( [c d])

= a + bi + c + di = a+c+(b+d)i

So we conclude φ( [a b] + [c d] ) = φ( [a b] ) + φ( [c d])

Thus this is a homeomorphism and since it’s bijective, we know this is an isomorphism.

PART B:

First find [a b] * [c d]

[a -b] [c -d] = [ac-bd -bc-ad ]
[b a] [d c] [bc+ad ac-bd ]

Which means [a b] * [c d] = [ac-bd ad+bc]

Thus φ( [a b] * [c d] ) = ac-bd + (ad+bc) i

Now we look at φ( [a b] ) φ( [c d] )

= (a+bi)(c+di) = ac + bci + adi + bdi²

= ac + (bc+ad)i – bd = ac-bd + (ad+bc)i

Thus φ( [a b] * [c d] ) = φ( [a b] ) φ( [c d] )

And we know it’s an isomorphism (bijective homeomorphism).

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