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### Isomorphic Groups and Subrings

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Date: 04/15/98 at 15:15:07
From: Leonie Williams
Subject: abstract or modern algebra

I have a few problems in my abstract algebra class that I just can't
figure out. The problem is that I don't even know how to get started
on these.

1) Let G = {a+b*sqrt[2]: a,b element of Q (rationals)},
H = {Matrix((a 2b) (b a): a,b element Q}
Show that G is isomorphic to H as additive groups.

2) Let M = {Matrix ((a -b) (b a))/ a,b element R (reals).
Show that C (complex numbers) is isomorphic to M as additive
groups. Also, delete 0 + 0i from C and Matrix ((0 0) (0 0)) from M
and show that C* is isomorpic to M* under multiplication where *
indicates deletion of "zero."

3) Show that {a + b*2^(2/3)+ c*4^(2/3): a,b element Q(rationals)} is
a subring of R(reals).
```

```
Date: 04/15/98 at 16:31:49
From: Doctor Rob
Subject: Re: abstract or modern algebra

To show two additive groups A and B are isomorphic, you have to
construct a map f:A -> B, from one to the other. This map has to have
several properties, which you would have to prove:

(1) f is one-to-one: f(x) = f(y) implies x = y
(2) f is onto: for every y in B, there is an x in A with y = f(x)
(3) f is an additive homomorphism: for every x and x' in A,
f(x+x') = f(x) + f(x')

In the first problem, A is G, B is H, and the map is:

f(a+b*sqrt[2]) = ((a 2b) (b a))

You have to show that the three properties are true for this
function f.

In the second problem, A is M and B is C, and the map is:

f(((a -b)(b a))) = a + b*i

Again you have to show that the three properties are true. To show
that A = M* and B = C* are isomorphic as *multiplicative* groups, you
need to replace (3) above with:

(3') f is a multiplicative homomorphism: for every x and x' in A,
f(x*x') = f(x)*f(x')

For the third problem, you have to show that the set, call it S, is a
subset of the real numbers (that's pretty easy!), and that the ring
axioms hold in S. There is no problem with commutativity,
associativity, or distributivity, because those properties are
inherited from the real numbers. (If they are true for any real
numbers, they must be true for these special ones.) The main things
you have to show are that addition and multiplication of numbers in S
give you results which are again in S (this is called closure under
addition and multiplication), that the real number 0 is in S (the
element in S is also in S. To do this, you need to use the fact that
r = 2^(2/3) and s = 4^(2/3) satisfy the equations r*r = s, r*s = 4,
and s*s = 4*r.  You also need to use the fact that Q is closed under

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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