“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.
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_{pknk} where the p_{i}’s are not necessarily distinct primes. Moreover, the list of prime powers appearing {p_{1}^{n1}, p_{2}^{n2}, ···, p_{k}^{nk}} 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 n_{1} + n_{2} + ··· + n_{t}, where n_{i} ∈ ℤ, n_{i} > 0, then ℤ_{pn1}⊕ℤ_{pn2}⊕···⊕ℤ_{pnk} is an Abelian group of order p^{k}.
Order of G | Partitions of k | Possible direct products for G |
---|---|---|
p | 1 | ℤ_{p} |
p^{2} | 2 (+ 0) | ℤ_{p2} |
1 + 1 | ℤ_{p} ⊕ ℤ_{p} | |
p^{3} | 3 | ℤ_{p3} |
2 + 1 | ℤ_{p2} ⊕ ℤ_{p} | |
1 + 1 + 1 | ℤ_{p} ⊕ ℤ_{p} ⊕ ℤ_{p} | |
p^{4} | 4 | ℤ_{p4} |
3 +1 | ℤ_{p3} ⊕ ℤ_{p} | |
2 +2 | ℤ_{p2} ⊕ ℤ_{p2} | |
2 +1 + 1 | ℤ_{p2} ⊕ ℤ_{p} ⊕ ℤ_{p} | |
1 + 1 + 1 + 1 | ℤ_{p} ⊕ ℤ_{p} ⊕ ℤ_{p} ⊕ ℤ_{p} |
Notice that distinct partitions of k yield distinct isomorphism classes, e.g., ℤ_{9}⊕ℤ_{3} is not isomorphic to ℤ_{3}⊕ℤ_{3}⊕ℤ_{3}. Let G, |G| = 1008 = 2^{4}·3^{2}·7, what are the option, G ≋ G_{16} ⊕ G_{9} ⊕ G_{7}?
Therefore, ℤ_{16} ⊕ ℤ_{9} ⊕ ℤ_{7}, ℤ_{8} ⊕ ℤ_{2} ⊕ ℤ_{9} ⊕ ℤ_{7}, ℤ_{4} ⊕ ℤ_{4} ⊕ ℤ_{9} ⊕ ℤ_{7}, ··· ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{3} ⊕ ℤ_{3} ⊕ ℤ_{7}.
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 ≋ ℤ_{2}, 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:
{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.
⟨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).
eH =[ord(gH) = p] (gH)^{p} = g^{p}H ⇒ [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 ⇒ g^{1} = g^{mx + py} = g^{mx}g^{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| = p^{n} 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| = p^{n}, ∀g ∈ G ⇒[By Lagrange’s Theorem] |⟨g⟩| | |G| and ord(g)| p^{n} ⇒[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| = p_{1}^{r1}p_{2}^{r2}···p_{k}^{rk} where p_{1}, p_{2}, ···, 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 ≋ G_{1} x G_{2} x···x G_{k} with |G_{i}| = p_{i}^{ri}
Proof.
Their intersection is trivial ↭ G_{i} ∩ G_{j} = {e}. Suppose g ∈ G_{i} ∩ G_{j}, i ≠ j ⇒ ord(g) = p_{i}^{m} = p_{j}^{n} for some m, n ⇒[By assumption, p_{i} and p_{j} are distinct primes] m = n = 0 ⇒ ord(g) = 1 ⇒ g = {e}.
G = G_{1}G_{2}···G_{k}, that is, every element of G can be expressed as g_{1}g_{2}···g_{k}, and therefore G is a direct product ⇒[Recall Theorem. If G is the internal direct product of a finite number of subgroups H_{1}, H_{2}, ··· H_{n}, then G is isomorphic to the external direct product of H_{1}, H_{2}, ··· H_{n}. More concisely, H_{1} x H_{2} x ··· x H_{n} ≋ H_{1} ⊕ H_{2} ⊕ ··· ⊕ H_{n}]
∀g ∈ G ⇒[By Lagrange] ord(g) | p_{1}^{r1}p_{2}^{r2}···p_{k}^{rk}.
Let’s define a_{i} = p_{1}^{r1}p_{2}^{r2}··p_{i-1}^{ri-1}p_{i+1}^{ri+1}··p_{k}^{rk}, |a_{i}| = |G|/p_{i}^{ri}, and consider (a_{1}, a_{2}, ···, a_{k}) = 1 ⇒ [Bézout’s identity can be extended to more than two integers] ∃b_{i} ∈ ℤ: a_{1}b_{1} + a_{2}b_{2} + ··· + a_{k}b_{k} = 1.
g = g^{1} = g^{a1b1 + a2b2 + ··· + akbk} = g^{a1b1} + g^{a2b2} + ··· + g^{akbk} = g_{1}g_{2}···g_{k} where each g_{i} = g^{aibi}, and notice that $g_i^{p_i^{r_i}}=g^{(a_i·p_i^{r_i})b_i}$ = [Recall |a_{i}| = |G|/p_{i}^{ri}] (g^{|G|})^{bi} =[By Lagrange’s Theorem] e, and therefore, $g_i^{p_i^{r_i}}=e$ ⇒ g_{i} ∈ G_{i} ⇒ g = g_{1}g_{2}···g_{k} ∈ G_{1}G_{2}···G_{k} ∎