# Isomorphism between R-{0} and C-{0}

• Oct 14th 2011, 03:51 PM
Bernhard
Isomorphism between R-{0} and C-{0}
In the exercises of Section 1.6 Dummit and Foote (Abstract Algebra) - Exercise 4 reads as follows:

Prove that the multiplicative groups $\displaystyle \mathbb{R}$ - {0} and $\displaystyle \mathbb{C}$ - {0} are not isomorphic.

Would appreciate some help with this problem.

Peter
• Oct 14th 2011, 06:48 PM
Deveno
Re: Isomorphism between R-{0} and C-{0}
if they are isomorphic, then R* would have an element of order 4. show that this cannot happen.

(hint: x^4 = 1 --> x^4 - 1 = 0).
• Oct 14th 2011, 08:54 PM
Bernhard
Re: Isomorphism between R-{0} and C-{0}
Thanks Deveno

Now my proof (or demonstration) that $\displaystyle \mathbb{R}$ - {0} does not have an element of order 4 but $\displaystyle \mathbb{C}$ - {0} does!

Let $\displaystyle z^4$ = 1

Solving gives $\displaystyle z_k$ = cos(2k$\displaystyle \pi$/4) + i sin(2k$\displaystyle \pi$/4) for k = 0,1,2,3

Thus $\displaystyle z_0$ = cos 0 + i sin 0 = 1

But this element of $\displaystyle \mathbb{C}$ - {0} has order 1

So try $\displaystyle z_1$ = cos(2$\displaystyle \pi$/4) + i sin(2$\displaystyle \pi$/4) = i

This element has order 4 since $\displaystyle i^2$ = -1 and $\displaystyle i^3$ = -i and $\displaystyle i^4$ = 1 so |z| = 4

But for $\displaystyle \mathbb{R}$ - {0} we have $\displaystyle x^4$ = 1 implies x = 1 or -1 so |x| = 1

Therefore, no element in $\displaystyle \mathbb{R}$ - {0} has order 4 (is this a satisfactory proof? I wonder???)

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Is that a satisfactory argument? Can someone please check?

Furthermore, in the above (and in D&F's question in general)where does the exclusion of 0 come into it? That is, why deal with $\displaystyle \mathbb{R}$ - {0} and $\displaystyle \mathbb{C}$ - {0} - why not just deal with $\displaystyle \mathbb{R}$ and $\displaystyle \mathbb{C}$
[I am guessing that if we do not exclude zero we are not dealing with multiplicative groups - but why is this???]

Peter
• Oct 14th 2011, 09:39 PM
Deveno
Re: Isomorphism between R-{0} and C-{0}
actually, the idea is to factor x^4 -1 = (x^2+1)(x+1)(x-1), and show that x^2+1 has no solutions in R, but does in C.

what you have is "close" though. if x^4 = 1, then either x^2 = 1, or x^2 = -1. if x^2 = 1, we have x = 1 or x = -1, as you have correctly stated,

but both of these elements have order < 4 (slight correction to what you wrote, the order of -1 is 2, not 1).

however, you need to explain why x^2 = -1 has no solutions in R (it's a one-line statement: x^2 is always _____ for a real number).

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the reason we exclude 0 from the multiplicative groups, is that 0 has no multiplicative inverse, so if we included it, we would NOT have a group

(every group element has to have an inverse). to see this: note that if a+a = a, then a = 0:

a+a = a
(a+a)+(-a) = a+(-a)
a+(a+(-a)) = 0
a+0 = 0
a = 0 and now, for the coup de grace:

0a = (0+0)a = 0a+0a, so 0a = 0 for ALL a.

if 0 HAD an inverse b, then 0b = 1. but 0b = 0, so 0 has no inverse.
• Oct 14th 2011, 09:59 PM
Bernhard
Re: Isomorphism between R-{0} and C-{0}
Thanks Deveno

That was a REALLY helpful post

Peer