Date: 02/22/99 at 01:24:35 From: Mari Subject: Algebraic Structures A friend and I are stuck on two homework problems. Please help us. 1) Suppose that N and M are two normal subgroups of G and that (N intersect M) = (e). Show that for any n element of N, m element of M, nm = mn. We began by manipulating what we wanted and have e = m^-1 n^-1 m n, where we have the 2 forms of elements being in N and M. Is this a correct start? 2) Let G be defined as all formal symbols (x^i y^j), i = 0, j =0, 2, 3, ..., n-1, where we assume (x^i y^j) = (x^i prime y^j prime) iff i = i prime, j = j prime x^2 = y^n = e, n>2 xy = y^-1 x. a) Find the form of the product (x^i y^j)(x^k y^L) as (x^alpha y^beta). b) Using this, prove G is a non-abelian group of order 2n. c) If n is odd, prove the center of G is (e), while if n is even, the center of G is larger than (e). For a) do we did cases applying n > 2, n < 2 since we tried to substitute the given into what we wanted, but we could not get it. For b) and c) we really did not have a clue. We thank you for your time and consideration.
Date: 02/22/99 at 13:25:05 From: Doctor Rob Subject: Re: Algebraic Structures In 1), you have made a good start. Set z = m^(-1)*n^(-1)*m*n. You want to prove x = e. Now look at this formula like this: z = m^(-1)*[n^(-1)*m*n]. The quantity in brackets is something you should be able to say something about because M is a normal subgroup (review the definition of a normal subgroup). Use that to say something about x. Now look at the formula like this: x = [m^(-1)*n^(-1)*m]*n. Say something about the quantity in brackets using the fact that N is a normal subgroup. Use that to say something about z. Combine these two facts about z to conclude that z = e. In the second problem, the key to (a) is to use x*y = y^(-1)*x in the form x*y*x^(-1) = y^(-1), and so x*y^n*x^(-1) = y^(-n). Therefore, y^n*x = x*y^(-n). That will allow you to write y^j*x^k = x*y^(-j)*x^(k-1) = x^2*y^j*x^(k-2) = ... = x^k*y^([-1]^k*j) (proof by induction on k). That will allow you to complete part (a). To prove part (b), prove that the elements e, y, y^2, ..., y^(n-1), x, x*y, x*y^2, ..., x*y^(n-1) are all distinct, and that every element of the group is equal to one of them, and that the group is not an abelian group because the pair of elements x and y do not commute. To prove part (c), take the list of 2*n elements above, and see which ones commute with x, and which ones commute with y. Those which commute with both will constitute the center. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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