TIPS FOR EXAM: MATH DISCUSSION-GROUPS

THIS WAS CONTRIBUTED BY:

Shri Ganachari M, Lecturer in Mathematics. GPUC,CHKM

Question: Show that G = {0,1,2,3} is an abelian group under addition mod 4 .

Explanation: First you have to construct the composition table. Here it is.


The correct table and closure law carries one mark. Sometimes wrong table carries zero mark. So it is necessary to construct the correct table.

 

While showing the given set forms an Abelian group under addition modulo 4, many students will commit mistakes.

Here is clarification:

CLOSURE LAW:

Some students are showing the Closure law like   therefore closure law satisfied.But it is not enough. You have to verify for all the elements of given set G. Now how to verify this?  This can be verified by composition table and you have to write like this

All the entries in the table are in G hence closure law satisfied.

Associative law:

Again some students fail to get marks in this. Some time if you write this

therefore associative law satisfied. But it wrong

you must write like….

For all a, b and c in G, a+(b+c) and (a+b)+c leave the same remainder when divided by 4, therefore

. Hence associative law satisfied.

Existence of Identity element: You may write: clearly 0 is the identity element in G. It’s OK. But for more accuracy it is essential to write as…

The row headed by 0 is same as the top most row, hence 0 is the identity element.

Existence of inverse element : If you write this “ Every row has the identity element, hence every element

has its inverse in G. Inverse law satisfied.” With this it is better to write the inverse of 0, 1, 2, 3 are 0, 3, 2, 1 respectively.

In verifying Inverse law, It is must to write the inverses of all the given elements seperately

Commutative law : Many will write like  . Hence commutative law is satisfied.

It is not sufficient to declare commutative law satisfied. What can you say about other elements? So it is better write like this…

A Following single line is enough to declare commutative law,

The table is symmetrical about the principal diagonal, hence commutative satisfies.

Hence G is an abelian group under addition mod 4.

0

1

2

3

0

0

1

2

3

1

1

2

3

0

2

2

3

0

1

3

3

0

1

2

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