Mathematics 581   Problem Set 1   Due Monday, April 8

Textbook Abstract Algebra, by W. E. Deskins. Dover, 1992. ISBN: 0-486-68888-7 (pbk.) The library copy is on closed reserve in the Sci/Eng Library.

1. (a) Write as a product of disjoint cycles the permutation given in 2-row notation by

1 2 3 4 5 6 7 8 9

3 7 9 4 8 2 6 5 1

***
(1 3 9)(2 7 6)(5 8)
***

(b) Write as a product of disjoint cycles the permutation

(1 2 7)(2 8 9)(9 7 3 4)(4 7)

***
(1  8  4  9  2  3  7)
***

(c) Write in 2-row notation the 4-cycle (1 3 2 4)
***
1  2  3  4
3  4  2  1
***

2. (a) Show that the inverse of the permutation  (1 2 7) is  (7 2 1).
***
(1 2 7)(7 2 1) = (1).  Just multiply them together.
***

(b) Explain why the inverse of any n-cycle can be found by simply writing the entries in the cycle in reverse order.
***
Suppose that the permutation  F  is an  n-cycle, where we might as well assume that  n  is at least 2.  Suppose that iF = j.  Unless   i  is at one end of the cycle and  j  is at the other end, the  i  and  j  appear consecutively as  ...i j...  within the cycle.  When you write the cycle in reverse order,  the  i  and  j  appear as    ...j i...  which exactly expresses that fact that
jF-1 = i.
On the other hand, if  i  and  j  are at opposite ends of the cycle, then  j  is at the left and  i  at the right, because  iF = j.   When you reverse the order of the entries,  j will be at the right and  i  at the left, which expresses the fact the  jF-1 = i.
***

(c) Is part (b) true for products of cycles? Explain. Hint: Whatever you think is going to be your answer, test it on the permutation  (1 2)(2 3 4).
***Because in a group  (ab)' = b'a', you will need to reverse the order in which the individual cycles apper as well: for example, the inverse of (1 2)(2 3 4)  would
be  (4 3 2)(2 1).
***
 

3. Textbook, p. 197, #1:  List all permutations of the set  {a, b}  and using composition of functions as the multiplication, write out the table of this group of permutations.
***
Let  I  be the identity, and  J be the transposition that interchanges  a  and  b.  that is all the permutationbs that there are for a set with two elements.
The multiplication table is  II = J,   IJ = J,   JI = J,   JJ = I.
***

4. Textbook, p. 197, #4. (Hint: In Example 3 every group element is its own inverse.  Is the same true in Example 5?)
***
Example 5 with  m = 5.
               [1]    [2]    [3]    [4]

[1]          [1]    [2]    [3]    [4]
[2]          [2]    [4]    [1]    [3]
[3]          [3]    [1]    [4]    [2]
[4]          [4]    [3]    [2]    [1]

In Example  3,  a2 = e  for  every  a  in the group.  But this is not true in Example 5 with  m = 5.  So the two groups are not isomorphic.
***

5. Prove that if  a  is an element of a group  G  and  a2 = a, then  a  is the identity element of  G.  (Hint: If e  is the identity element, then  a2 = a = ae.  Now multiply on the left by the inverse of  a.)
***
a2 = ae.  Therefore,  a'(a2) = a'(ae),  and so  (a'a)a = (a'a)e, or  ea = ee.  Therefore,  a = e.
***

6. (a) Let  a  be an element of a group G. Let  a'  be the inverse of a.  Prove that the inverse of  an  is(a')n, for any positive integer  n.
***Induction on n.  If  n = 1, the statement says that the inverse of  a  is  a', which is true.  Suppose that the inverse of  an is(a')for some positive integer  n.  Then  the inverse of
an + 1 is the inverse of  ana, which is  a'(an)', which is  a'(a')n , which is (a')n + 1.
***

(b) Do you think that part (a) is true if  n  is negative?  Explain.
***Yes.  If  n  is negative, then  n = -k for some positive  k.  Then the inverse of ais the inverse of   a-k.  But  a-k  itself is the inverse of  ak, so that the inverse of  a-k  must be  ak, which is  [(a')']k =   (a')-k  which is    (a')n.
***