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 $\displaystyle ( \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 $\displaystyle ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2) $ but had some difficulties, when I realised that I was unsure of whether the group $\displaystyle ( \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 $\displaystyle ( \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 $\displaystyle G_1 $ and $\displaystyle G_2 $ be groups. The set of all ordered pairs $\displaystyle (x_1, x_2) $ such that $\displaystyle x_1 \in G_1 $ and $\displaystyle x_2 \in G_2 $ is called the direct product of $\displaystyle G_1 $ and $\displaystyle G_2 $, denoted by $\displaystyle 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 $\displaystyle G_1 $ and $\displaystyle G_2 $ be groups.

(a) The direct product $\displaystyle G_1 \times G_2 $ is a group under the operation defined for all $\displaystyle (a_1, a_2) , (b_1, b_2) \in G_1 \times G_2 $ by

$\displaystyle (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 $\displaystyle ( \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 $\displaystyle ( \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