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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.

There is no known closed formula for the number of partitions of a given natural α.
  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.

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.

|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…

Therefore, G ≋ ℤ4⊕ℤ4.

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:

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, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn (Abstract Algebra), and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. Andrew Misseldine: College Algebra and Abstract Algebra.
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