Would you tell me, please, which way I ought to go from here?” “That depends a good deal on where you want to get to,” said the Cat. “I don’t much care where—” said Alice. “Then it doesn’t matter which way you go,” said the Cat.“—so long as I get somewhere,” Alice added as an explanation. “Oh, you’re sure to do that,” said the Cat, “if you only walk long enough", Alice in Wonderland, Lewis Carroll.

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 p
^{2}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., ℤ
_{4}⊕ℤ_{4}≆ ℤ_{4}⊕ℤ_{2}⊕ℤ_{2}because ℤ_{4}≆ ℤ_{2}⊕ℤ_{2}.

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.

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).

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}

**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.

Lemma 4. Let G be a finite Abelian p-group (a prime-power order group) and let g be an element of maximal order in G. Then, G can be written in the form ⟨g⟩ x K for some K ≤ G, G ≋ ⟨g⟩ x K.

Proof.

We denote |G| = p^{n} and induct on n.

Base Case. If n = 1, then |G| = p, p prime ⇒[**All groups of prime order are cyclic** groups and they are isomorphic to ℤ_{p}] ℤ_{p} ≋[G is cyclic ⇒ ∃g ∈ G: G = ⟨g⟩] ⟨g⟩ x {e}

Induction Hypothesis. Suppose ∀m, 1 ≤ m < n, the result holds for all groups of order p^{m}, and |G| = p^{n} and let g ∈ G such that ord(g) is maximal.

In other words, among all the elements of G, choose “g” of maximal order, |g| = ord(p) = p^{k}.

Since the order of an element divides the order of a group, p is prime, and therefore ord(g) = p^{k} where k is some integer as large as possible, but obviously it could never be bigger than n.

- If
**k = n**⇒ ord(g) = ord(G) ⇒ g is the generator of the group,**G = ⟨g⟩ ≋ ⟨g⟩ x {e}**. - Otherwise,
**k < n**. Next, among all the elements of G, choose h of smallest order such that h ∉ ⟨g⟩, that is, h ∈ G \ ⟨g⟩. Notice that h ≠ e, since e ∈ ⟨g⟩, the cyclic subgroup generated by g.

Since the order of an element divides the order of a group, p is prime, and therefore **ord(h) = p ^{r} and p^{r} ≤ p^{k}** ↭ r ≤ k and that’s the case because g is an element of maximal order in G and ord(g) = p

$(g^a)^{p^{k-1}} = (h^p)^{p^{k-1}} = h^{p^k}$ = [ord(h) = p^{r}, r ≤ k] e ⇒ $(g^a)^{p^{k-1}} =e$ ⇒ ord(g^{a}) ≤ p^{k-1} ⇒[g ∈ G has order p^{k} maximal, ord(g^{a}) = $\frac{ord(g)}{gcd(ord(g),a)}$. If gcd(ord(g), a) = 1, then ord(g^{a}) = ord(g)⊥ ord(g) = p^{k} and ord(g^{a}) ≤ p^{k-1}] gcd(ord(g), a) ≠ 1, that is, gcd(p^{k}, a) ≠ 1, and p prime ⇒ p | a ⇒ ∃s∈ ℤ: a = ps ⇒ h^{p} = g^{a} = g^{ps} ⇒ h^{p} = g^{ps} ⇒ (g^{ps})^{-1}h^{p} = e ⇒ (g^{-s})^{p}h^{p} = e ⇒ (g^{-s}h)^{p} = e.

Let’s consider this recently constructed element **x = g ^{-s}h, x ≠ e** (x = e ⇒ g

Let’s study the intersection ⟨g⟩ ∩ ⟨h⟩, ∀y ∈ ⟨g⟩ ∩ ⟨h⟩ ⇒ [ord(h)=p ⇒ ⟨h⟩ = {e, h, h^{2}, ···, h^{p-1}}] ∃a∈ ℤ: y = h^{a}, 0 ≤ a ≤ p-1.

If a ≠ 0 ⇒[p prime, 0 < a ≤ p-1] gcd(a, p) = 1 ⇒[Bezout’s identity] ∃r, s∈ ℤ: ar + ps = 1 ⇒ y^{r} = $h^{a^r}=h^{ar} = h^{1-ps}=h·(h^p)^{-s}$ =[ord(h)=p ⇒h^{p} = e] h. Consider ∀y ∈ ⟨g⟩ ∩ ⟨h⟩ ⇒[Intersection of two subgroups of a group is again a subgroup] y^{r} = h ∈ ⟨g⟩ ∩ ⟨h⟩ ⇒ h ∈ ⟨g⟩⊥ ⇒ a = 0 ⇒ y = h^{a} = h^{0} = e, that is, ∀y ∈ ⟨g⟩ ∩ ⟨h⟩, y = e ⇒ ⟨g⟩ ∩ ⟨h⟩ = {e}.

Set H = ⟨h⟩, consider the coset gH ∈ G/H (|G/H| = $\frac{p^n}{p} = p^{n-1}$), and let’s calculate the order of gH. Set r = ord(gH), r = ord(gH) | ord(g) 💡. Futhermore, g^{r}H = (gH)^{r} = eH ⇒[aH = bH ↭ a ∈ bH] g^{r} ∈ H ⇒ g^{r} ∈ ⟨g⟩ ∩ ⟨h⟩ = {e} ⇒ g^{r} = e ⇒ ord(g) | r ⇒ [We have previously stated that r | ord(g)] r = ord(g) = ord(gH) = p^{k}.

💡Order of Coset in Quotient group divides order of element. Suppose ord(g) = m ⇒ g^{m} = e. (gH)^{m} = g^{m}H = eH = H, which is the identity of G/H ⇒ ord(gH) ≤ m. Hence, m is one positive integer with (gH)^{m}, but there may be smaller ones.

Since m ≥ n = ord(gH) ⇒ m = qn + r, 0 ≤ r < n. H = (gH)^{m} = (gH)^{qn+r} = (gH)^{qn}(gH)^{r} = ((gH)^{n})^{q}(gH)^{r} =[H is the identity on the quotient group G/H, that is, G/H under the multiplication of cosets] H(gH)^{r} = (gH)^{r}.

If r ≠ 0 ⊥ because ord(gH) = n, r < n ⇒ r = 0 ⇒ m = qn ⇒ **n | m**

Therefore, **ord(gH) = p ^{k} and this is maximal** because ∀x ∈ G, ord(xH) | ord(x) and by construction, g has maximal order (|g| = p

Claim: G ≋ ⟨g⟩ x K.

**⟨g⟩ ∩ K = {e}**. Suppose x ∈ ⟨g⟩ ∩ K ⇒ xH ∈ ⟨gH⟩ ∩ K/H = {eH} because of the induction hypothesis**G/H ≋ ⟨gH⟩ x K/H**(internal direct product) so these two groups intersect trivially ⇒ xH ∈ eH = H ⇒ x ∈ H ⇒[x ∈⟨g⟩∩K ⇒ x∈ ⟨g⟩, ⟨g⟩ ∩ H = {e}] x = e.**G = ⟨g⟩K**. Suppose x ∈ G, xH ∈ G/H ⇒ [**G/H ≋ ⟨gH⟩ x K/H**, internal direct product, so G/H = ⟨gH⟩(K/H)] xH = (g^{r}k)H where g^{r}∈⟨g⟩ and k ∈ K ⇒[aH = bH ↭ a ∈ bH] x ∈ g^{r}kH ⇒ [H ≤ K ≤ G] x ∈ ⟨g⟩K∎

**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 = p_{1}^{r1}p_{2}^{r2}···p_{k}^{rk}. Then,

- G ≋ G
_{1}x G_{2}x···x G_{k}with |G_{i}| = p_{i}^{ri}, 1 ≤ i ≤ k. - For each G
_{i}, G_{i}≋ ℤ_{pin1}⊕ℤ_{pin2}⊕···⊕ℤ_{pink}, n_{1}+ n_{2}+ ··· + n_{k}= r_{i} - 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.

Proof.

By 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}] G ≋ G_{1} x G_{2} x ··· G_{i} where p_{1}, p_{2},···, p_{k} are distinct primes that divide m, |G_{i}| = p_{i}^{ri}.

By Lemma 4, each G_{i} can be written in the form ⟨g_{i}⟩ x H_{i} for some H_{i} ≤ G_{i} ⇒[⟨g_{i}⟩ ≋ ℤ_{pi}^{si}] G_{i} ≋ ℤ_{pi}^{si} x H_{i}, and we proceed by induction on H_{i}, and we will end up on the desired result ℤ_{pin1}⊕ℤ_{pin2}⊕···⊕ℤ_{pink}, n_{1} + n_{2} + ··· + n_{k} = r_{i} ∎

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.