Fundamental Theorem of Finite Abelian Groups. Exercises.

So let me tell you, as you prepare to go off into the world, remember those six rules: Trust yourself, Break some rules, Don’t be afraid to fail, Ignore the naysayers, Work like hell, and Give something back, Arnold Schwarzenegger.

Recall. The Fundamental Theorem of Finite Abelian Groups. Every finite Abelian group G is the direct sum of cyclic groups, each of prime power order. |G| = n = p1r1p2r2···pkrk. Then,

1. G ≋ G1 x G2 x···x Gk with |Gi| = piri, 1 ≤ i ≤ k.
2. For each Gi, Gi ≋ ℤpin1⊕ℤpin2⊕···⊕ℤpink, n1 + n2 + ··· + nk = ri
3. The decomposition in (1) and (2) is unique (it is left as an exercise).

The number of terms in the direct product, as well as the order of the cyclic groups, are uniquely determined by the group.

• How many elements of order 6 are there in ℤ6 ⊕ ℤ9? The order of (a, b) is the least common multiple of the order of a and that of b, |(a, b)| = lcm(|a|, |b|). |(a, b)| = 6 ↭ the order of a is 6 (1, 5) and that of b is 1 (0) or 3 (3, 6), or the order of a is 2 (3) and that of b is 3 (3, 6). The desired elements of order 6 are: (1, 0), (5, 0), (1, 3), (1, 6), (5, 3), (5, 6), (3, 3), (3, 6).

• How many Abelian groups of order 125? (up to isomorphism) Let |G| = 125 = 53, so there are three possibilities: ℤ53, ℤ52 ⊕ ℤ5 or ⊕ ℤ5⊕ ℤ5⊕ ℤ5. 💡Notice there is one Abelian group of order pk for each partition of k, i.e., each way of writing k as a sum of positive integers. These integers become the exponents in the power of p.

• Let |G| = 12 = 22·3, there are two possibilities, namely ℤ4 ⊕ ℤ3 and ℤ2 ⊕ ℤ2 ⊕ ℤ3.

• Let |G| = 540 = 22·33·5, there are six Abelian groups of order 540 up to isomorphism: ℤ4⊕ℤ27⊕ℤ5, ℤ2⊕ℤ2⊕ℤ27⊕ℤ5, ℤ4⊕ℤ9⊕ℤ3⊕ℤ5, ℤ2⊕ℤ2⊕ℤ9⊕ℤ3⊕ℤ5, ℤ4⊕ℤ3⊕ℤ3⊕ℤ3⊕ℤ5, ℤ2⊕ℤ2⊕ℤ3⊕ℤ3⊕ℤ3⊕ℤ5.

• Let |G| = 1176 = 23·3·72, then the list of the distinct isomorphism classes of Abelian groups is {ℤ8⊕ℤ3⊕ℤ49, ℤ4⊕ℤ2⊕ℤ3⊕ℤ49, ℤ2⊕ℤ2⊕ℤ2⊕ℤ3⊕ℤ49, ℤ8⊕ℤ3⊕ℤ7⊕ℤ7, ℤ4⊕ℤ2⊕ℤ3⊕ℤ7⊕ℤ7, ℤ2⊕ℤ2⊕ℤ2⊕ℤ3⊕ℤ7⊕ℤ7}

• How many Abelian groups are there of order 144? There is one Abelian group of order pk for each partition of k, i.e., each way of writing k as a sum of positive integers. Since 144 = 24·32, there are five ways of partitioning 4 (4 -ℤ24-, 3+1 -ℤ23⊕ℤ2–, 2+2, 2+1+1, 1+1+1+1+1), and two ways of partitioning 2 (2 -ℤ32-, 1 +1 -ℤ32⊕ℤ3-), hence there are ten such groups. Similarly, there are 28 = 2·2·7 non-isomorphic Abelian groups of order 3781575 = 32·52·75 since the number of partitions of 2 and 5 are 2 (2 and 1+1) and 7 (5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1+ 1, 1+1+1+1+1) respectively.

There is no known closed formula for the number of partitions of a given natural α.
• Could you recognize this group?
* o b d p q c n u
o o b d p q c n u
b b o p d c q u n
d d p o b n u q c
p p d b o u n c q
q q c n u o b d p
c c q u n b o p d
n n u q c d p o b
u u n c q p d b o
1. |G| = 8, o is the neutral element, and G is Abelian (it is symmetric about the main diagonal).
2. From the Fundamental Theorem of Finite Abelian group, there are three options: ℤ23, ℤ22 ⊕ ℤ2 or ℤ2⊕ ℤ2⊕ ℤ2, that is, ℤ8, ℤ4 ⊕ ℤ2 or ⊕ ℤ2⊕ ℤ2⊕ ℤ2
3. We use the fact that two finite Abelian groups are isomorphic if and only if they have the same number of elements of any order. Every element that is not the identity has order 2, so the only possibility is ℤ2⊕ ℤ2⊕ ℤ2.
• Let G = U(13) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} with multiplication mod 13. Since U(13) is Abelian and |U(13)| = 12 ⇒[By the Fundamental theorem of Finite Abelian Groups] U(13) ≋ ℤ4 ⊕ ℤ3 or U(13) ≋ ℤ2 ⊕ ℤ2 ⊕ ℤ3.

Which one? The elements in ℤ4 ⊕ ℤ3 can have order 1, 2, 3, 4, 6, 12 and the elements in ℤ2 ⊕ ℤ2 ⊕ ℤ3 can have order 1, 2, 3, 6, 12. Thus, if U(13) has an element of order 4 then it must be isomorphic to ℤ4 ⊕ ℤ3 and we are done. 5 ∈ U(13), |5| = 4 (50 = 1, 51 = 5, 52 = 25 =13 12, 53 = 125=13 8, 54 = 625 =13 1) and so U(13) ≋ Z4 ⊕ Z3. Futhermore, U(13) ≋ ℤ4 ⊕ ℤ3 ≋ ℤ12 = ⟨1⟩, this also tells us that U(13) is cyclic.

• Let G = {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64}, · (mod 65). Their orders are {1, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, 4, 2} respectively, e.g., 82 = 64, 83 = 512 (mod 65) = 57, 84 = 1, |8| = 4.

|G| = 16 = 24 ⇒ Partitions of 4 = 1 + 1 + 1 + 1, 2 + 1 + 1, 2 + 2, 3 + 1, 4, so we have 5 classes. G is isomorphic to one of these classes: ℤ2⊕ℤ2⊕ℤ2⊕ℤ2, ℤ4⊕ℤ2⊕ℤ2, ℤ4⊕ℤ4, ℤ8⊕ℤ2, ℤ16… but which one?

We can rule out…

• 16 = ⟨1⟩, |1| = 16. There is no element of order 16 in G ⊥
• 8⊕ℤ2. (1, 0) has order 8, but there is no element of order 8 in G ⊥
• 2⊕ℤ2⊕ℤ2⊕ℤ2. All elements except the identity (0, 0, 0, 0) have order 2 ⊥ |8| = 4.
• 4⊕ℤ2⊕ℤ2. G has only three elements of order 2 but ℤ4⊕ℤ2⊕ℤ2 has more, e.g., (0, 0, 1), (0, 1, 0), (0, 1, 1), (2, 0, 0).

Therefore, G ≋ ℤ4⊕ℤ4.

• Suppose G is an Abelian group of order 1008 and that G has no elements of order 24. |G| = 1008 = 24·32·7. Then, by the Fundamental Theorem of Abelian Groups, the options are: ℤ16 ⊕ ℤ9 ⊕ ℤ7, ℤ8 ⊕ ℤ2 ⊕ ℤ9 ⊕ ℤ7, ℤ4 ⊕ ℤ4 ⊕ ℤ9 ⊕ ℤ7, ··· ℤ2 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ3 ⊕ ℤ3 ⊕ ℤ7.

Which elements have order of elements of order 24 = lcm{8, 3}. Elements of order 8 could come from ℤ16 and ℤ8. Elements of order 3 could come from ℤ9 and ℤ3. If G has no elements of order 24, it narrows the possibilities to the following groups, ℤ4 ⊕ ℤ4 ⊕ ℤ9 ⊕ ℤ7, ℤ4 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ9 ⊕ ℤ7, ℤ2 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ9 ⊕ ℤ7, ℤ4 ⊕ ℤ4 ⊕ ℤ3 ⊕ ℤ3 ⊕ ℤ7, ℤ4 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ3 ⊕ ℤ3 ⊕ ℤ7, ℤ2 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ2 ⊕ ℤ3 ⊕ ℤ3 ⊕ ℤ7.

Example. Let’s G be an Abelian group, |G| = 72 = 23·32. Construct a subgroup of order 12. By the Fundamental Theorem of Finite Abelian Groups, we have the following options:

• 4⊕ℤ2⊕ℤ9, {(a, 0, b)| a ∈ ℤ4, b ∈ {0, 3, 6}} is such a subgroup because we have just put together ⟨3⟩ = {0, 3, 6} ≤ ℤ9 and |⟨3⟩| = 3, |ℤ4| = 4.
• 8⊕ℤ3⊕ℤ3, {(a, b, 0)| a ∈ {0, 2, 4, 6}, b ∈ ℤ3} is such a subgroup because we have just put together |ℤ3| = 3 and |⟨2⟩| = 4, ⟨2⟩ = {0, 2, 4, 6} ≤ ℤ8.
• 8⊕ℤ9, {(a, b)| a ∈ {0, 2, 4, 6}, b ∈ {0, 3, 6}}.
• 4⊕ℤ2⊕ℤ3⊕ℤ3 ≋[m ⊕ ℤn ≋ ℤmn if (m, n) = 1] ℤ12⊕ℤ6, {(a, 0)| a ∈ ℤ12}.
• 2⊕ℤ2⊕ℤ2⊕ℤ9 ≋[m ⊕ ℤn ≋ ℤmn if (m, n) = 1] ℤ2⊕ℤ2⊕ℤ18, {(a, 0, b), a ∈ ℤ2, b ∈ {0, 3, 6, 9, 12, 15}}
• 2⊕ℤ2⊕ℤ2⊕ℤ3⊕ℤ3 ≋[m ⊕ ℤn ≋ ℤmn if (m, n) = 1] ℤ2⊕ℤ6⊕ℤ6, {(a, b, 0), a ∈ ℤ2, b ∈ ℤ6}.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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