In the exercises of Section 1.6 Dummit and Foote (Abstract Algebra) - Exercise 4 reads as follows:
Prove that the multiplicative groups - {0} and - {0} are not isomorphic.
Would appreciate some help with this problem.
Peter
In the exercises of Section 1.6 Dummit and Foote (Abstract Algebra) - Exercise 4 reads as follows:
Prove that the multiplicative groups - {0} and - {0} are not isomorphic.
Would appreciate some help with this problem.
Peter
Thanks Deveno
Now my proof (or demonstration) that - {0} does not have an element of order 4 but - {0} does!
Let = 1
Solving gives = cos(2k /4) + i sin(2k /4) for k = 0,1,2,3
Thus = cos 0 + i sin 0 = 1
But this element of - {0} has order 1
So try = cos(2 /4) + i sin(2 /4) = i
This element has order 4 since = -1 and = -i and = 1 so |z| = 4
But for - {0} we have = 1 implies x = 1 or -1 so |x| = 1
Therefore, no element in - {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 - {0} and - {0} - why not just deal with and
[I am guessing that if we do not exclude zero we are not dealing with multiplicative groups - but why is this???]
Peter
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.