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 ,   a⁶ ,    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 H_o - dim Z₁ = dim C_o - dim C₁ 


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

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

dim C₁  =  the number of branches = b = 4

Then: dim C_o - dim C₁  = 4 – 4 = 0 and dim H_o - dim Z₁ = 1 – 1 = 0

Therefore: dim H_o - dim Z₁ = dim C_o - dim C₁ 


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

Solution: 

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

But:  dim B_o    = (n – 1) = 4 – 1 = 3

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

3) Show that: dim H_o   =   dim C_o  - dim B_o   (This is the quotient space H_o = C_o/ B_o )

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

dim B_o    = (n – 1) = 4 – 1 = 3

And: dim H_o =  dim H₁ = 1

But: dim C_o  - dim B_o   =  4 – 3 = 1 So: dim H_o   =   dim C_o  - dim B_o  

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 )]