Assignment No. 2 MTH 633 (Spring 2022) Solution file

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Assignment No.  2           MTH 633 (Spring 2022)

Total Marks 10                                                                        Due Date: August 22, 2022

 

 

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Q2: Write down the elements of 2 +7ℤ.                                                    Marks = 3

Solution:-

Given that:-

Elements of 2 + 7Z

Now,

Z = \left\{ {0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5,.........} \right\}

 Elements of 2 + 7Z = {....., -33, -26, -19, -12, -5, 2, 9, 16, 23, 30, 37,….}

Q3: Prove that a factor group of a cyclic group is cyclic.                               Marks = 3

Solution:-

Prove:

A factor group of a cyclic group is cyclic:

Now,

Let, G be cyclic with generator a, and let, N be a normal subgroup of G.

We claim the co-set aN generates G/N.

We must compute all power of aN. But this amounts to computing, in G all powers of the representative a and all these powers given all elements in G.

 Hence the powers of aN certainly given all co-sets of N and G/N is cyclic. 

Q1: Let S be a subgroup of a group G. Then show that S is normal if and only if  

                (aS)(bS) = (ab)S, for all a, b in G                                                  Marks = 4 

Solution:-

Given that:-

                 (aS)(bS) = (ab)S, for all a, b in G

Now,

We show that:               

                                      aS = Sa, for all a in G.

We do this by showing:   

                                     aS \subseteq Sa\,\,and\,\,Sa \subseteq aS,  for all a in G.

aS \subseteq Sa

Firs observe that:

aS{a^{ - 1}} \subseteq Sa{a^{ - 1}} = S\left( {a{a^{ - 1}}} \right) = S\left( e \right) = S.
Let,
x be in aS.

Then,

x = ah, for some h in S.

Then,

Xa-1 = aha-1 , which is in = asa-1, thus in S.

Thus xa-1 is in S.

Thus x is in Sa.

aS \subseteq aS:Sa \subseteq SaS = \left( {eS} \right)\left( {aS} \right) = \left( {ea} \right)S = aS.

This establishes normality.

For he converse, assume S is normal.

\left( {aS} \right)\left( {bS} \right) \subseteq \left( {ab} \right)S:\,\,for\,\,a\,\,and\,\,b\,\,\,in\,\,G.

\begin{array}{l} x\,\,in\,\,\left( {aS} \right)\left( {bS} \right)\,\,implies\,\,that\,\,x = \,a{h_1}b{h_2},\,\,for\,\,some\,\,{h_1}\,\,and\,\,{h_2}\,\,in\,\,S.\\ \,\, \end{array}

But,

h1b is in Sb, thus bS.

Thus,

h1b = bh3 for some h3 in S.

Thus,

x = abh3h2 is in abS.

\left( {ab} \right)S \subseteq \left( {aS} \right)\left( {bS} \right):\,x\,\,in\,\,\,\left( {ab} \right)S \Rightarrow \,x = aebh,\,\,for\,\,some\,\,h\,\,in\,\,S.






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