Abelian Groups. Fundamental Theorem of Finite Abelian Groups

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A group is called Abelian if ab = ba ∀a, b ∈ G. Every group of prime order (every group of prime order is cyclic), order five or smaller, or cyclic is Abelian. $\mathbb{S}_3$ is the smallest non-commutative group.

• Every group of prime order p is isomorphic to ℤ_p.
• Every group of order p² is isomorphic to ℤ_p² or ℤ_p⊕ℤ_p.
• ℤ_m ⊕ ℤ_n ≋ ℤmn if (m, n) = 1
• A finite, then A ⊕ B ≋ A ⊕ C iff B ≋ C., e.g., ℤ₄ ⊕ℤ₄ ≆ ℤ₄⊕ℤ₂⊕ℤ₂ because ℤ₄ ≆ ℤ₂⊕ℤ₂.

The Fundamental Theorem of Finite Abelian Groups. Every finite Abelian group G is the direct sum of cyclic groups, each of prime power order.

If G is a cyclic group of order n ⇒ G ≋ ℤ_n. Therefore, the Fundamental Theorem of Finite Abelian Groups states that every finite Abelian group is isomorphic to ℤ_p1ⁿ¹⊕ℤ_p2ⁿ²⊕···⊕ℤ_pk^nk where the p_i’s are not necessarily distinct primes. Moreover, the list of prime powers appearing {p₁^n₁, p₂^n₂, ···, p_k^n_k} is unique, up to re-ordering.

Suppose |G| = p^k where p is prime and k ≤ 4. There is one group of order p^k for each set of positive integers whose sum is k. Let k be equals to n₁ + n₂ + ··· + n_t, where n_i ∈ ℤ, n_i > 0, then ℤ_pⁿ¹⊕ℤ_pⁿ²⊕···⊕ℤ_p^nk is an Abelian group of order p^k.

Notice that distinct partitions of k yield distinct isomorphism classes, e.g., ℤ₉⊕ℤ₃ is not isomorphic to ℤ₃⊕ℤ₃⊕ℤ₃. Let G, |G| = 1008 = 2⁴·3²·7, what are the option, G ≋ G₁₆ ⊕ G₉ ⊕ G₇?

• 2⁴, 5 options: ℤ₁₆ (4), ℤ₈ ⊕ ℤ₂ (3 +1), ℤ₄ ⊕ ℤ₄ (2 + 2), ℤ₄ ⊕ ℤ₂ ⊕ ℤ₂ (2 + 1 +1), ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂ (1 + 1 + 1 + 1)
• 3², 2 options: ℤ₉ (2), ℤ₃ ⊕ ℤ₃ (1 + 1).
• 7¹, 1 option: ℤ₇.

Therefore, ℤ₁₆ ⊕ ℤ₉ ⊕ ℤ₇, ℤ₈ ⊕ ℤ₂ ⊕ ℤ₉ ⊕ ℤ₇, ℤ₄ ⊕ ℤ₄ ⊕ ℤ₉ ⊕ ℤ₇, ··· ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂ ⊕ ℤ₃ ⊕ ℤ₃ ⊕ ℤ₇.

Cauchy’s Theorem for Abelian Groups. Lemma 1. If p prime divides the order of a finite Abelian group G, then G has an element of order p.

Proof. (Another proof is provided for the sake of completeness, based on YouTube, MathMajor)

We are going to proceed by induction on n, the order of the finite Abelian group.

Base Case. n = 1 ⇒ G = {e}. It is not possible that a prime divides 1. n = 2 ⇒ G ≋ ℤ₂, then there is an element of order 2 = |G|, namely 1 (the non-identity element).

Induction Hypothesis. Let’s suppose that the statement holds ∀m, 1 ≤ m < n, and G such that |G| = n. There are two possibilities:

• G has no proper subgroups. Let g ∈ G be an element distinct from the identity, g ≠ e, and consider the following chain of subgroups: {e} ≤ ⟨g^n/p⟩ ≤ ⟨g⟩ =[By assumption, G has no proper subgroups & g ≠ e] G. Since G has no proper subgroups, there are only two options:

1. {e} = ⟨g^n/p⟩ ⇒ g^n/p = e ⇒[G = ⟨g⟩ ⇒ |g| = n, Recall, a^m = e ↭ |a| | m] n/p is a multiple of n ⇒ p = 1 ⊥ p is prime.

2. ⟨g^n/p⟩ = ⟨g⟩ ⇒ ord(g^n/p) = org(g) = n. org(g^n/p) =[|a| = n, $|⟨a^k⟩|=\frac{n}{gcd(n, k)}$] n/gcd(n, n/p) =[p prime, p|n] n/(n/p) = p ⇒ n = p and we have found an element of order p, namely g (ord(g) = n = p).

• G has a proper subgroup, say H ≤ G, 1 < |H| < n. Besides, since G is an Abelian group, H ◁ G. There are two cases.

1. If p| |H| ⇒[By the induction hypothesis, p | |H| < n] There exists h ∈ H, ord(h) = p and we are done.
2. If p ɫ |H| ⇒ [By assumption p|n and, by Lagrange, n = |G| = |H|·|G/H|] p| |G/H|. We can apply the induction hypothesis to G/H [1 < |H| < n ⇒|G/H| < n] there exist g ∈ G, gH ∈ G/H, such that ord(gH) = p and p ɫ |H| ⇒ gH ≠ eH = H ⇒[aH = bH ↭ a ∈ bH] g ∉ eH = H.

eH =[ord(gH) = p] (gH)^p = g^pH ⇒ [aH = bH ↭ a ∈ bH] g^p ∈ H (🪡₁). If |H| = m ⇒ (g^p)^m = e ⇒ (g^m)^p = e.

💥 Claim: g^m is the element that we are looking for, that is, ord(g^m) = p. Suppose for the sake of contradiction, ord(g^m) ≠ p ⇒[(g^m)^p = e ⇒ ord(g^m) divides p, p prime, so there are only two options, 1 and p] Suppose for the sake of contradiction that ord(g^m) = 1 ⇒ g^m = e (🪡₂)

Besides, gcd(m, p) = 1 (p ɫ |H| = m, i.e., p ɫ m, and p is prime) ⇒ ∃x, y: mx + py = 1 ⇒ g¹ = g^{mx + py} = g^mxg^py = (g^m)^x·(g^p)^y =[We have previously stated that g^m = e (🪡₂)] e(g^p)^y = (g^p)^y ⇒[g^p ∈ H (🪡₁)] g = (g^p)^y ∈ H ⊥ Therefore, ord(g^m) = p ∎

Definition. Let p be a prime and G a finite group. G is a p-group if every element in G has order a power of p ↭ ∀g ∈ G, ord(g) = p^a for some a ∈ ℤ.

Lemma 2. G is a finite Abelian p-group if and only if its order is a power of p (|G| = pⁿ for some positive integer n).

Proof.

⇒) Suppose for the sake of contradiction, p, q are distinct primes such that p | |G| and q | |G| ⇒[Lemma 1] ∃x, y ∈ G: ord(x) = p and ord(y) = q ⊥ G is not a p group (ord(y) = q).

⇐) Suppose |G| = pⁿ, ∀g ∈ G ⇒[By Lagrange’s Theorem] |⟨g⟩| | |G| and ord(g)| pⁿ ⇒[p prime] ord(g) = p^m for some m, 0 ≤ m ≤ n ⇒ G is a p-group ∎

Lemma 3. Let G be a finite Abelian group, and let m = |G| = p₁^r₁p₂^r₂···p_k^r_k where p₁, p₂, ···, p_k are distinct primes that divide m. Then, G is an internal direct product of cyclic groups of prime-power order, that is, G ≋ G₁ x G₂ x···x G_k with |G_i| = p_i^r_i

Proof.

• First, let’s define G_i = {g ∈ G| ord(g) = p_iⁿ for some positive integer n} and prove that G_i are subgroups.

1. e ∈ G_i because ord(e) = 1 = p_i⁰.
2. ∀ x, y ∈ G_i ⇒[By definition G_i] ord(x) = p_i^m, ord(y) = p_iⁿ for some m and n ⇒[Let G be an Abelian group, ∀a, b ∈ G ⇒ |ab| | lcm{|a|, |b|}] ord(xy) | lcm(p_i^m, p_iⁿ) = p_i^{max(m, n)} ⇒ ord(xy) | p_i^{max(m, n)}, p_i prime ⇒ ord(xy) = p_i^l for some positive integer l ⇒ xy ∈ G_i.
   |ab| | lcm{|a|, |b|} but |ab| ≠ lcm{|a|, |b|}, e.g., ℤ₈, |3·1| =_{Additive notation} |3+1| = |4| = 2 | lcm{|a|, |b|} =[|3| = 8 since (3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 0), |1| = 8 since (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 0)] lcm{ 8, 8} = 8
3. ∀ x ∈ G_i ⇒ ord(x) = p_iⁿ ⇒ ord(x^-1) =[x^p_in = e, (x^a)^b = x^ab = (x^b)^a ⇒ (x^-1)^p_in = (x^p_in)^-1 = e] p_iⁿ ⇒ x^-1 ∈ G_i.

• Their intersection is trivial ↭ G_i ∩ G_j = {e}. Suppose g ∈ G_i ∩ G_j, i ≠ j ⇒ ord(g) = p_i^m = p_jⁿ for some m, n ⇒[By assumption, p_i and p_j are distinct primes] m = n = 0 ⇒ ord(g) = 1 ⇒ g = {e}.

• G = G₁G₂···G_k, that is, every element of G can be expressed as g₁g₂···g_k, and therefore G is a direct product ⇒[Recall Theorem. If G is the internal direct product of a finite number of subgroups H₁, H₂, ··· H_n, then G is isomorphic to the external direct product of H₁, H₂, ··· H_n. More concisely, H₁ x H₂ x ··· x H_n ≋ H₁ ⊕ H₂ ⊕ ··· ⊕ H_n]

∀g ∈ G ⇒[By Lagrange] ord(g) | p₁^r₁p₂^r₂···p_k^r_k.

Let’s define a_i = p₁^r₁p₂^r₂··p_i-1^r_i-1p_i+1^r_i+1··p_k^r_k, |a_i| = |G|/p_i^r_i, and consider (a₁, a₂, ···, a_k) = 1 ⇒ [Bézout’s identity can be extended to more than two integers] ∃b_i ∈ ℤ: a₁b₁ + a₂b₂ + ··· + a_kb_k = 1.

g = g¹ = g^{a₁b₁ + a₂b₂ + ··· + a_kb_k} = g^a₁b₁ + g^a₂b₂ + ··· + g^a_kb_k = g₁g₂···g_k where each g_i = g^a_ib_i, and notice that $g_i^{p_i^{r_i}}=g^{(a_i·p_i^{r_i})b_i}$ = [Recall |a_i| = |G|/p_i^r_i] (g^|G|)^b_i =[By Lagrange’s Theorem] e, and therefore, $g_i^{p_i^{r_i}}=e$ ⇒ g_i ∈ G_i ⇒ g = g₁g₂···g_k ∈ G₁G₂···G_k ∎

Bibliography

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.