    # Abelian Groups. Fundamental Theorem of Finite Abelian Groups

“Doesn’t it look awesome on me” [···] “You look like a monkey with the sexual sophistication of a donkey, and belong to a zoo. I shake with fear and tremble in horror at the mere thought of what your mom looks like,” Fluffy Dick, retorted. “You, little dicksucker, remind me of a severe migraine. You should’ve been immediate peeled, anally impaled by a blunt object, and disposed, and in this very order,” Goofy Fingers replied.

The chief commanded, “Shut up or I’ll cut your balls and make sushi with them!” Apocalypse, Anawim, #justtothepoint.

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 ℤp.
• Every group of order p2 is isomorphic to ℤp2 or ℤp⊕ℤp.
• m ⊕ ℤn ≋ 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 ≋ ℤn. Therefore, the Fundamental Theorem of Finite Abelian Groups states that every finite Abelian group is isomorphic to ℤp1n1⊕ℤp2n2⊕···⊕ℤpknk where the pi’s are not necessarily distinct primes. Moreover, the list of prime powers appearing {p1n1, p2n2, ···, pknk} is unique, up to re-ordering.

Suppose |G| = pk where p is prime and k ≤ 4. There is one group of order pk for each set of positive integers whose sum is k. Let k be equals to n1 + n2 + ··· + nt, where ni ∈ ℤ, ni > 0, then pn1⊕ℤpn2⊕···⊕ℤpnk is an Abelian group of order pk.

Order of G Partitions of k Possible direct products for G
p 1 p
p2 2 (+ 0) p2
1 + 1 p ⊕ ℤp
p3 3 p3
2 + 1 p2 ⊕ ℤp
1 + 1 + 1 p ⊕ ℤp ⊕ ℤp
p4 4 p4
3 +1 p3 ⊕ ℤp
2 +2 p2 ⊕ ℤp2
2 +1 + 1 p2 ⊕ ℤp ⊕ ℤp
1 + 1 + 1 + 1 p ⊕ ℤp ⊕ ℤp ⊕ ℤp

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} ≤ ⟨gn/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} = ⟨gn/p ⇒ gn/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. ⟨gn/p⟩ = ⟨g⟩ ⇒ ord(gn/p) = org(g) = n. org(gn/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 = gpH ⇒ [aH = bH ↭ a ∈ bH] gp ∈ H (🪡1). If |H| = m ⇒ (gp)m = e ⇒ (gm)p = e.

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

Recall, a^m = e ↭ |a| | m

Besides, gcd(m, p) = 1 (p ɫ |H| = m, i.e., p ɫ m, and p is prime) ⇒ ∃x, y: mx + py = 1 ⇒ g1 = gmx + py = gmxgpy = (gm)x·(gp)y =[We have previously stated that gm = e (🪡2)] e(gp)y = (gp)y ⇒[gp ∈ H (🪡1)] g = (gp)y ∈ H ⊥ Therefore, ord(gm) = 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) = pa 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) = pm for some m, 0 ≤ m ≤ n ⇒ G is a p-group ∎

Lemma 3. Let G be a finite Abelian group, and let m = |G| = p1r1p2r2···pkrk where p1, p2, ···, pk 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 Gk with |Gi| = piri

Proof.

• First, let’s define Gi = {g ∈ G| ord(g) = pin for some positive integer n} and prove that Gi are subgroups.
1. e ∈ Gi because ord(e) = 1 = pi0.
2. ∀ x, y ∈ Gi ⇒[By definition Gi] ord(x) = pim, ord(y) = pin for some m and n ⇒[Let G be an Abelian group, ∀a, b ∈ G ⇒ |ab| | lcm{|a|, |b|}] ord(xy) | lcm(pim, pin) = pimax(m, n) ⇒ ord(xy) | pimax(m, n), pi prime ⇒ ord(xy) = pil for some positive integer l ⇒ xy ∈ Gi.
|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 ∈ Gi ⇒ ord(x) = pin ⇒ ord(x-1) =[xpin = e, (xa)b = xab = (xb)a ⇒ (x-1)pin = (xpin)-1 = e] pin ⇒ x-1 ∈ Gi.
• Their intersection is trivial ↭ Gi ∩ Gj = {e}. Suppose g ∈ Gi ∩ Gj, i ≠ j ⇒ ord(g) = pim = pjn for some m, n ⇒[By assumption, pi and pj are distinct primes] m = n = 0 ⇒ ord(g) = 1 ⇒ g = {e}.

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

∀g ∈ G ⇒[By Lagrange] ord(g) | p1r1p2r2···pkrk.

Let’s define ai = p1r1p2r2··pi-1ri-1pi+1ri+1··pkrk, |ai| = |G|/piri, and consider (a1, a2, ···, ak) = 1 ⇒ [Bézout’s identity can be extended to more than two integers] ∃bi ∈ ℤ: a1b1 + a2b2 + ··· + akbk = 1.

g = g1 = ga1b1 + a2b2 + ··· + akbk = ga1b1 + ga2b2 + ··· + gakbk = g1g2···gk where each gi = gaibi, and notice that $g_i^{p_i^{r_i}}=g^{(a_i·p_i^{r_i})b_i}$ = [Recall |ai| = |G|/piri] (g|G|)bi =[By Lagrange’s Theorem] e, and therefore, $g_i^{p_i^{r_i}}=e$ ⇒ gi ∈ Gi ⇒ g = g1g2···gk ∈ G1G2···Gk

# 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.
Bitcoin donation 