1. Show that {1,a

^{2}} 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------- a

^{2}

-----------------

1 --- 1 ------ a

^{2}

a

^{2 }--- a

^{2}-- --- 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 a

^{7}across the top and 1 to a

^{7}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, a

^{2 }, a

^{3}, a

^{4}, a

^{5 }, a

^{6 },

^{ }a

^{7}

E.g. 1 multiplied by each top table member

The second line would show: a

^{2}, a

^{3}, a

^{4}, a

^{5}, a

^{6}, a

^{7}, 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, a

^{2},a

^{3},a

^{4}, a

^{5}, a

^{6}, a

^{7}} - Not closed under the operation •

The proper subgroups are of order:

(2): {1, a

^{4}}

(4): {1, a

^{2}, a

^{4}, a

^{6}}

3. Show that the largest order subgroup in C

_{9}is a proper subgroup of C

_{9}.

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

1 ---a

^{2}---a

^{4}----a

^{6}

---------------------

1 --- 1---a

^{2}--- a

^{4}----a

^{6}

a

^{2}---a

^{2}---a

^{4}---a

^{6}---- 1

a

^{4}---a

^{4}---a

^{6}--- 1 ----a

^{2}

a

^{6}--- a

^{6}---1 ---a

^{2}---- a

^{4}

Factor Group Problems (11/30 blog) :

1) Show that: dim H

_{o}- dim Z

_{1}= dim C

_{o}- dim C

_{1}

Solution: We see that dim H

_{o}= dim H

_{1}= 1 and dim Z

_{1 }= (5 – 4) = 1

Also: dim C

_{o}= the number of nodes = n = 4

dim C

_{1}= the number of branches = b = 4Then: dim C

_{o}- dim C

_{1}= 4 – 4 = 0 and dim H

_{o}- dim Z

_{1}= 1 – 1 = 0

Therefore: dim H

_{o}- dim Z

_{1}= dim C

_{o}- dim C

_{1}

2) Show that: dim Z

_{1}= dim C

_{1 }- dim B

_{o}

Solution:

From the preceding: dim Z

_{1}= 1 and dim C_{1}= 4But: dim B

_{o}= (n – 1) = 4 – 1 = 3

So: dim Z

_{1}= 1 = dim C_{1 }- dim B_{o}= (4 – 3) = 13) 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 H_{1}= 1
But: dim C

_{o }- dim B_{o}= 4 – 3 = 1 So: dim H_{o}= dim C_{o }- dim B_{o}_{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 )]**
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