## Saturday, December 1, 2012

### Problem Solutions for Groups, Factor Groups

Problem solutions: Basic Groups (11/26 blog):

1. Show that {1,a2} is a proper subgroup of C4, the cyclic group of order 4.

To show a subgroup is a proper subset, we prepare the table for the operation, thus:

Solution Table 1:

1-/---1------- a2
-----------------
1  --- 1 ------ a2

a2   --- a2-- ---  1

From the table above we see that both elements are in the subgroup, so that it is closed under the operation, and therefore a proper subgroup.

2. Sketch the cyclic graphic associated with the cyclic group C8, the cyclic group of order 8. Prepare the table for this group with all elements indicated, and thence or otherwise identify the subgroups by order and indicate which are proper, which improper and list the respective elements of each.

Solution summary:

This is done simply by drawing a 4-5” diameter circle, then marking off eight equal spaces around its circumference. Label each of the marks from a0 to a7. Your derived table will show 1 to a7 across the top and 1 to a7 across the left margin. You then would simply have filled in the results for each line. For example, the top line would show:  1, a,  a2 ,   a3,    a4,    a5 ,   a   a7

E.g. 1 multiplied by each top table member

The second line would show:  a2 , a3, a4, a5 , a6 , a7, 1
And so on.

The improper subgroups are of order:

(1) : {1} Not a proper subgroup since only the identity element comprises it.

(8): {1,a, a2,a3,a4, a5, a6, a7} - Not closed under the operation •

The proper subgroups are of order:

(2): {1, a4}

(4): {1, a2 , a4 , a6}

3. Show that the largest order subgroup in C9 is a proper subgroup of C9.

From the preceding example, the largest order subgroup is evidently (4). We can write out the table for this subgroup:

1 ---a2 ---a4 ----a6
---------------------
1 ---   1---a2 --- a4 ----a6

a2  ---a2 ---a4 ---a6---- 1

a4 ---a4 ---a6--- 1 ----a2

a6--- a6 ---1 ---a2---- a4

Factor Group Problems (11/30 blog) :

1) Show that: dim Ho - dim Z1dim Co - dim C1

Solution:  We see that dim Ho =  dim H1 = 1 and dim Z1 = (5 – 4) = 1

Also: dim Co =  the number of nodes = n = 4

dim C1  =  the number of branches = b = 4

Then: dim Co - dim C1  = 4 – 4 = 0 and dim Ho - dim Z1 = 1 – 1 = 0

Therefore: dim Ho - dim Z1dim Co - dim C1

2) Show that: dim Z1 = dim  C1    -  dim Bo

Solution:

From the preceding: dim Z1 = 1  and dim C1  = 4

But:  dim Bo    = (n – 1) = 4 – 1 = 3

So: dim Z1 = 1 =   dim  C1    -  dim Bo   = (4 – 3) = 1

3) Show that: dim Ho   =   dim Co  - dim Bo   (This is the quotient space Ho = Co/ Bo )

Solution:
dim Co =  the number of nodes = n = 4

dim Bo    = (n – 1) = 4 – 1 = 3

And: dim Ho =  dim H1 = 1

But: dim Co  - dim Bo   =  4 – 3 = 1 So: dim Ho   =   dim Co  - dim Bo

4) Find: a,   b

Recall from blog:   (branch) = (node)final – (node) initial

Then:

a = B – A = [0, 1, 0, 0]T– [1, 0, 0, 0]T = [-1, 1, 0, 0]T

b = C – B =  [0, 0, 1, 0]T – [0, 1, 0, 0]T = [0, -1, 1, 0]T

5) Thence, find a  -  b :

Solution:

a  -  b  = [-1, 1, 0, 0]T - [0, -1, 1, 0]T = [-1, 0, -1, 0]T

Is it true that k =0 for the WHOLE figure?

YES, because k =0 = a + b +  g  + b

E.g. [(a + b)] + [(d + g) ]- [(a + b)] - [(d + g )]