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 = p₁^r₁p₂^r₂···p_k^r_k. Then,

1. G ≋ G₁ x G₂ x···x G_k with |G_i| = p_i^r_i, 1 ≤ i ≤ k.
2. For each G_i, G_i ≋ ℤ_piⁿ¹⊕ℤ_piⁿ²⊕···⊕ℤ_pi^nk, n₁ + n₂ + ··· + n_k = r_i
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: ℤ_5³, ℤ_5² ⊕ ℤ₅ or ⊕ ℤ₅⊕ ℤ₅⊕ ℤ₅. 💡Notice there is one Abelian group of order p^k 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 p^k 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 -ℤ_2⁴-, 3+1 -ℤ_2³⊕ℤ₂–, 2+2, 2+1+1, 1+1+1+1+1), and two ways of partitioning 2 (2 -ℤ_3²-, 1 +1 -ℤ_3²⊕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.

• 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: ℤ_2³, ℤ_2² ⊕ 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…

• ℤ₁₆ = ⟨1⟩, |1| = 16. There is no element of order 16 in G ⊥
• ℤ₈⊕ℤ₂. (1, 0) has order 8, but there is no element of order 8 in G ⊥
• ℤ₂⊕ℤ₂⊕ℤ₂⊕ℤ₂. All elements except the identity (0, 0, 0, 0) have order 2 ⊥ |8| = 4.
• ℤ₄⊕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:

• ℤ₄⊕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.
• ℤ₈⊕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.
• ℤ₈⊕ℤ₉, {(a, b)| a ∈ {0, 2, 4, 6}, b ∈ {0, 3, 6}}.
• ℤ₄⊕ℤ₂⊕ℤ₃⊕ℤ₃ ≋[ℤ_m ⊕ ℤ_n ≋ ℤmn if (m, n) = 1] ℤ₁₂⊕ℤ₆, {(a, 0)| a ∈ ℤ₁₂}.
• ℤ₂⊕ℤ₂⊕ℤ₂⊕ℤ₉ ≋[ℤ_m ⊕ ℤ_n ≋ ℤmn if (m, n) = 1] ℤ₂⊕ℤ₂⊕ℤ₁₈, {(a, 0, b), a ∈ ℤ₂, b ∈ {0, 3, 6, 9, 12, 15}}
• ℤ₂⊕ℤ₂⊕ℤ₂⊕ℤ₃⊕ℤ₃ ≋[ℤ_m ⊕ ℤ_n ≋ ℤmn if (m, n) = 1] ℤ₂⊕ℤ₆⊕ℤ₆, {(a, b, 0), a ∈ ℤ₂, b ∈ ℤ₆}.

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