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Math Help - Direct Producrs and Quotient Groups

  1. #1
    Super Member Bernhard's Avatar
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    Direct Producrs and Quotient Groups

    In Beachy and Blair: Abstract Algebra, Section 3.8 Cosets, Normal Groups and Factor Groups, Exercise 17 reads as follows:

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    17. Compute the factor group  ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2)

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    Since I did not know the meaning of "Compute the factor group" I proceeded to try to list them members of  ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2)  but had some difficulties, when I realised that I was unsure of whether the group  ( \mathbb{Z}_6 \times \mathbb{Z}_4 )   was a group under multiplication or addition. SO essentially I did not know how to carry out group operations in  ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2)  .

    Reading Beachy and Blair, Chapter 3 Groups, page 118 (see attachment) we find the following definition:

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    3,3,3 Definition. Let  G_1 and  G_2 be groups. The set of all ordered pairs   (x_1, x_2) such that  x_1 \in G_1 and  x_2 \in G_2 is called the direct product of  G_1 and  G_2 , denoted by  G_1 \times  G_2 .

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    Then Proposition 3,3,4 reads as follows:

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    3,3,4 Proposition. Let  G_1 and  G_2 be groups.
    (a) The direct product  G_1 \times  G_2 is a group under the operation defined for all  (a_1, a_2) , (b_1, b_2)  \in  G_1 \times  G_2 by

     (a_1, a_2) (b_1, b_2) = (a_1b_1,  a_2b_2 ) .

    (b) etc etc

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    However in Example 3.3.3 on page 119 we find the group  ( \mathbb{Z}_2 \times \mathbb{Z}_2 )  dealt with as having addition as its operation.

    My question is - what is the convention on direct products of  ( \mathbb{Z}_n \times \mathbb{Z}_m )  - does one use addition or multiplication?

    Presumably, since the operations involve integers the matter is more than one of notation?

    Can someone please clarify this matter?

    Would appreciate some help.

    Peter
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  2. #2
    Member Haven's Avatar
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    Re: Direct Producrs and Quotient Groups

    In general \mathbb{Z}_n is not a group under multiplication. For example \mathbb{Z}_6 is not a group under multiplication, as 2*3=0. However the group of units mod 6 U(6) = \{1,5\} IS a group under multiplication.

    I feel that the book is using the same notation as Gallian's Contemporary Abstract Algebra. Where \mathbb{Z}_n denotes the group under addition.

    So in this case, it's useful to compute the order of (2,2). The order of 2 in \mathbb{Z}_6 is 3 and the order of 2 in \mathbb{Z}_4 is 2. So the order of (2,2) is the least common multiple of 2 and 3, which is 6. As \mathbb{Z}_6\times\mathbb{Z}_4 is a group of order 24, the factor group \mathbb{Z}_6\times\mathbb{Z}_4/\langle(2,2)\ranglewill be of order $24/6 = 4$. So just find a representative for each coset and you'll "compute the factor group"
    Thanks from Bernhard
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