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Algebraic Structures

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   
Associated Topics:
College Modern Algebra

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