1
Finitely Generated Abelian Groups
Def. The cartesian product of sets 𝑆1 , … , 𝑆𝑛 is the set of all ordered n-tuples
(𝑎1 , … , 𝑎𝑛 ), where 𝑎𝑖 ∈ 𝑆𝑖 for 𝑖 = 1, 2, … , 𝑛. We write:
𝑆1 × 𝑆2 × … × 𝑆𝑛 or ∏𝑛𝑖=1 𝑆𝑖
Theorem: Let 𝐺1 , 𝐺2 , … , 𝐺𝑛 be groups. For (𝑎1 , 𝑎2 , … , 𝑎𝑛 ) and
(𝑏1 , 𝑏2 , … , 𝑏𝑛 ) in ∏𝑛𝑖=1 𝐺𝑖
define (𝑎1 , 𝑎2 , … , 𝑎𝑛 )(𝑏1 , 𝑏2 , … , 𝑏𝑛 ) = (𝑎1 𝑏1 , 𝑎2 𝑏2 , … , 𝑎𝑛 𝑏𝑛 ).
Then ∏𝑛
𝑖=1 𝐺𝑖 is a group, the direct product of the groups 𝑮𝒊 , under this
multiplication.
Proof: The ∏𝑛 𝑖=1 𝐺𝑖 is closed under this multiplication and the
multiplication is associative because each component is.
(𝑒1 , 𝑒2 , … , 𝑒𝑛 ) is the identity element of ∏𝑛𝑖=1 𝐺𝑖 , where 𝑒𝑖 is the
identity element of 𝐺𝑖 .
(𝑎1 −1 , 𝑎2 −1 , … , 𝑎𝑛 −1 ) is the inverse of (𝑎1 , 𝑎2 , … , 𝑎𝑛 ).
When all of the 𝐺𝑖 ’s are abelian groups, additive notation is sometimes used and
∏𝑛𝑖=1 𝐺𝑖 is referred to as the direct sum of the groups 𝐺𝑖 and is written
𝐺1 ⊕ 𝐺2 ⊕ … ⊕ 𝐺𝑛 . The direct product (or sum) of abelian groups is also
abelian.
Notice that if |𝐺𝑖 | = 𝑟𝑖 for 𝑖 = 1, … , 𝑛 then |∏𝑛
𝑖=1 𝐺𝑖 | = (𝑟1 )(𝑟2 ) … (𝑟𝑛 ).
, 2
Ex. Let 𝐺 = ℤ3 × ℤ2 which has (3)(2) = 6 elements:
(0,0), (0, 1), (1, 0), (1, 1), (2, 0) and (2, 1).
Notice that ℤ3 × ℤ2 is a cyclic group because (1, 1) generates the
group:
1(1, 1) = (1, 1)
2(1, 1) = (1, 1) + (1, 1) = (2, 0)
3(1, 1) = (1, 1) + (1, 1) + (1, 1) = (0, 1)
4(1, 1) = (1, 0)
5(1, 1) = (2, 1)
6(1, 1) = (0, 0).
Up to an isomorphism there is only one cyclic group of order 𝑛, ℤ𝑛 . So
ℤ3 × ℤ2 is isomorphic to ℤ6 . This isomorphism, 𝜙, can be generated by
𝜙(1,1) = 1 (since (1,1) generates ℤ3 × ℤ2 and 1 generates ℤ6 ).
Ex. Show 𝐺 = ℤ4 × ℤ4 is not isomorphic to the cyclic group ℤ16 .
It is true that |𝐺 | =16= |ℤ16 | , but for 𝐺 to be cyclic we would need to
find an element of ℤ4 × ℤ4 which has order 16.
But for any 𝑎 ∈ ℤ4 , 𝑎 + 𝑎 + 𝑎 + 𝑎 = 0 in ℤ4 .
So any element of ℤ4 × ℤ4 , (𝑎, 𝑏) has at most order 4.
Thus ℤ4 × ℤ4 is not a cyclic group. In particular, ℤ4 × ℤ4 is not
isomorphic to ℤ16 .
Finitely Generated Abelian Groups
Def. The cartesian product of sets 𝑆1 , … , 𝑆𝑛 is the set of all ordered n-tuples
(𝑎1 , … , 𝑎𝑛 ), where 𝑎𝑖 ∈ 𝑆𝑖 for 𝑖 = 1, 2, … , 𝑛. We write:
𝑆1 × 𝑆2 × … × 𝑆𝑛 or ∏𝑛𝑖=1 𝑆𝑖
Theorem: Let 𝐺1 , 𝐺2 , … , 𝐺𝑛 be groups. For (𝑎1 , 𝑎2 , … , 𝑎𝑛 ) and
(𝑏1 , 𝑏2 , … , 𝑏𝑛 ) in ∏𝑛𝑖=1 𝐺𝑖
define (𝑎1 , 𝑎2 , … , 𝑎𝑛 )(𝑏1 , 𝑏2 , … , 𝑏𝑛 ) = (𝑎1 𝑏1 , 𝑎2 𝑏2 , … , 𝑎𝑛 𝑏𝑛 ).
Then ∏𝑛
𝑖=1 𝐺𝑖 is a group, the direct product of the groups 𝑮𝒊 , under this
multiplication.
Proof: The ∏𝑛 𝑖=1 𝐺𝑖 is closed under this multiplication and the
multiplication is associative because each component is.
(𝑒1 , 𝑒2 , … , 𝑒𝑛 ) is the identity element of ∏𝑛𝑖=1 𝐺𝑖 , where 𝑒𝑖 is the
identity element of 𝐺𝑖 .
(𝑎1 −1 , 𝑎2 −1 , … , 𝑎𝑛 −1 ) is the inverse of (𝑎1 , 𝑎2 , … , 𝑎𝑛 ).
When all of the 𝐺𝑖 ’s are abelian groups, additive notation is sometimes used and
∏𝑛𝑖=1 𝐺𝑖 is referred to as the direct sum of the groups 𝐺𝑖 and is written
𝐺1 ⊕ 𝐺2 ⊕ … ⊕ 𝐺𝑛 . The direct product (or sum) of abelian groups is also
abelian.
Notice that if |𝐺𝑖 | = 𝑟𝑖 for 𝑖 = 1, … , 𝑛 then |∏𝑛
𝑖=1 𝐺𝑖 | = (𝑟1 )(𝑟2 ) … (𝑟𝑛 ).
, 2
Ex. Let 𝐺 = ℤ3 × ℤ2 which has (3)(2) = 6 elements:
(0,0), (0, 1), (1, 0), (1, 1), (2, 0) and (2, 1).
Notice that ℤ3 × ℤ2 is a cyclic group because (1, 1) generates the
group:
1(1, 1) = (1, 1)
2(1, 1) = (1, 1) + (1, 1) = (2, 0)
3(1, 1) = (1, 1) + (1, 1) + (1, 1) = (0, 1)
4(1, 1) = (1, 0)
5(1, 1) = (2, 1)
6(1, 1) = (0, 0).
Up to an isomorphism there is only one cyclic group of order 𝑛, ℤ𝑛 . So
ℤ3 × ℤ2 is isomorphic to ℤ6 . This isomorphism, 𝜙, can be generated by
𝜙(1,1) = 1 (since (1,1) generates ℤ3 × ℤ2 and 1 generates ℤ6 ).
Ex. Show 𝐺 = ℤ4 × ℤ4 is not isomorphic to the cyclic group ℤ16 .
It is true that |𝐺 | =16= |ℤ16 | , but for 𝐺 to be cyclic we would need to
find an element of ℤ4 × ℤ4 which has order 16.
But for any 𝑎 ∈ ℤ4 , 𝑎 + 𝑎 + 𝑎 + 𝑎 = 0 in ℤ4 .
So any element of ℤ4 × ℤ4 , (𝑎, 𝑏) has at most order 4.
Thus ℤ4 × ℤ4 is not a cyclic group. In particular, ℤ4 × ℤ4 is not
isomorphic to ℤ16 .
