Classification of Finite Abelian Groups

We verify the group axioms for \(G = G_1 \times G_2 \times \cdots \times G_n\).

Closure: Since each \(G_i\) is closed under its operation, the product \(g_i h_i \in G_i\) for all \(i\). Thus \((g_1 h_1, \ldots, g_n h_n) \in G\).

Associativity: Let \(a = (a_1, \ldots, a_n)\), \(b = (b_1, \ldots, b_n)\), \(c = (c_1, \ldots, c_n)\). Then: \[ \begin{align*} (ab)c &= (a_1 b_1, \ldots, a_n b_n)(c_1, \ldots, c_n) \\\\ &= ((a_1 b_1)c_1, \ldots, (a_n b_n)c_n) \\\\ &= (a_1(b_1 c_1), \ldots, a_n(b_n c_n)) && \text{(by associativity in each } G_i \text{)}\\\\ &= a(bc). \end{align*} \]

Identity: Let \(e = (e_1, e_2, \ldots, e_n)\), where \(e_i\) is the identity of \(G_i\). For any \((g_1, \ldots, g_n) \in G\), \[ (g_1, \ldots, g_n)(e_1, \ldots, e_n) = (g_1 e_1, \ldots, g_n e_n) = (g_1, \ldots, g_n), \] and similarly \((e_1, \ldots, e_n)(g_1, \ldots, g_n) = (g_1, \ldots, g_n)\). Hence \(e\) is the identity of \(G\).

Inverses: For \((g_1, g_2, \ldots, g_n) \in G\), define \((g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1}) \in G\) (this lies in \(G\) since each \(g_i^{-1} \in G_i\)). Then \[ (g_1, \ldots, g_n)(g_1^{-1}, \ldots, g_n^{-1}) = (g_1 g_1^{-1}, \ldots, g_n g_n^{-1}) = (e_1, \ldots, e_n) = e, \] and analogously on the left. Thus every element of \(G\) has an inverse.

Consider \(\mathbb{Z}_2 \times \mathbb{Z}_3\). The elements are all pairs \((a, b)\) where \(a \in \{0, 1\}\) and \(b \in \{0, 1, 2\}\): \[ \mathbb{Z}_2 \times \mathbb{Z}_3 = \{(0,0), (0,1), (0,2), (1,0), (1,1), (1,2)\}. \] The operation is addition modulo 2 in the first component and modulo 3 in the second: \[ (1, 2) + (1, 1) = (1+1 \bmod 2, \, 2+1 \bmod 3) = (0, 0). \] This group has order \(|\mathbb{Z}_2 \times \mathbb{Z}_3| = 2 \cdot 3 = 6\).

By the componentwise definition, an element of \(G_1 \times \cdots \times G_n\) is uniquely determined by an independent choice of one element from each \(G_i\). The \(i\)th component admits \(|G_i|\) choices. By the multiplication principle of counting, the total number of such tuples is \(|G_1| \cdot |G_2| \cdots |G_n|\).

\((\Rightarrow)\) Suppose \(G = G_1 \times \cdots \times G_n\) is Abelian. For any \(a_i, b_i \in G_i\), consider the elements \((e_1, \ldots, a_i, \ldots, e_n)\) and \((e_1, \ldots, b_i, \ldots, e_n)\) in \(G\), where \(a_i\) and \(b_i\) sit in the \(i\)th slot and all other slots hold the identity. Their product in \(G\) is \((e_1, \ldots, a_i b_i, \ldots, e_n)\); commuting them gives \((e_1, \ldots, b_i a_i, \ldots, e_n)\). Since \(G\) is Abelian, these tuples are equal, which forces \(a_i b_i = b_i a_i\) in \(G_i\).

\((\Leftarrow)\) Suppose each \(G_i\) is Abelian. For any \((a_1, \ldots, a_n), (b_1, \ldots, b_n) \in G\): \[ \begin{align*} (a_1, \ldots, a_n)(b_1, \ldots, b_n) &= (a_1 b_1, \ldots, a_n b_n) \\\\ &= (b_1 a_1, \ldots, b_n a_n) \\\\ &= (b_1, \ldots, b_n)(a_1, \ldots, a_n). \end{align*} \]

Let \(s = |(g_1, g_2, \ldots, g_n)|\) and \(t = \operatorname{lcm}(|g_1|, |g_2|, \ldots, |g_n|)\).

Since \(t\) is a multiple of each \(|g_i|\), write \(t = q_i |g_i|\) for each \(i\); then \(g_i^t = (g_i^{|g_i|})^{q_i} = e_i^{q_i} = e_i\). Thus: \[ (g_1, g_2, \ldots, g_n)^t = (g_1^t, g_2^t, \ldots, g_n^t) = (e_1, e_2, \ldots, e_n). \] By order dividing any annihilating power, this gives \(s \mid t\).

Conversely, \((g_1, g_2, \ldots, g_n)^s = (e_1, e_2, \ldots, e_n)\) implies \(g_i^s = e_i\) for all \(i\). Applying the same theorem componentwise, \(|g_i| \mid s\) for all \(i\). Since \(s\) is then a common multiple of all the \(|g_i|\), the definition of lcm yields \(t \mid s\).

Since \(s\) and \(t\) are positive integers with \(s \mid t\) and \(t \mid s\), we conclude \(s = t\).

Consider the element \((2, 3) \in \mathbb{Z}_4 \times \mathbb{Z}_6\).

In \(\mathbb{Z}_4\): the element \(2\) has order \(|2| = 2\) since \(2 + 2 = 4 \equiv 0 \pmod{4}\).

In \(\mathbb{Z}_6\): the element \(3\) has order \(|3| = 2\) since \(3 + 3 = 6 \equiv 0 \pmod{6}\).

Therefore: \(|(2, 3)| = \operatorname{lcm}(2, 2) = 2\).

Now consider \((1, 1) \in \mathbb{Z}_4 \times \mathbb{Z}_6\). In \(\mathbb{Z}_4\), the element \(1\) has order \(4\); in \(\mathbb{Z}_6\), the element \(1\) has order \(6\).

Therefore: \(|(1, 1)| = \operatorname{lcm}(4, 6) = 12\).

\((\Rightarrow)\):
Suppose \(G \times H\) is cyclic with \(|G| = m\) and \(|H| = n\). By order of direct product, \(|G \times H| = mn\). We must show that \(\operatorname{gcd}(m, n) = 1\).

Let \(d = \operatorname{gcd}(m, n)\) and let \((g, h)\) be a generator of \(G \times H\). By the definition of cyclic group, \(|(g, h)| = |G \times H| = mn\). Since \(G\) and \(H\) are finite, Lagrange's theorem gives \(|x| \mid |G|\) for every \(x \in G\), hence \(x^{|G|} = e\); in particular \(g^m = e\), and similarly \(h^n = e\). Computing: \[ (g, h)^{mn/d} = (g^{mn/d}, h^{mn/d}) = ((g^m)^{n/d}, (h^n)^{m/d}) = (e, e), \] where we used \(mn/d = m \cdot (n/d) = (m/d) \cdot n\). By order dividing any annihilating power, \(|(g, h)| \mid mn/d\), i.e., \(mn \mid mn/d\). Since both are positive integers, \(mn \leq mn/d\), forcing \(d \leq 1\). Combined with \(d \geq 1\), this gives \(d = 1\), so \(m\) and \(n\) are relatively prime.

\((\Leftarrow)\):
Let \(G = \langle g \rangle\) and \(H = \langle h \rangle\), so \(|g| = m\) and \(|h| = n\). Suppose \(\operatorname{gcd}(m, n) = 1\). By order in direct products, \(|(g, h)| = \operatorname{lcm}(m, n)\). When \(\operatorname{gcd}(m, n) = 1\), elementary number theory gives \(\operatorname{lcm}(m, n) = mn / \gcd(m, n) = mn\). Combined with order of direct product (\(|G \times H| = mn\)), the cyclic subgroup \(\langle (g, h) \rangle\) has order \(mn = |G \times H|\), so \(\langle (g, h) \rangle = G \times H\). Therefore \((g, h)\) generates \(G \times H\), making it cyclic.

By induction on \(n\). The base case \(n = 2\) is the previous theorem.

For the inductive step, suppose the corollary holds for any collection of \(n - 1\) cyclic groups, and let \(G_1, \ldots, G_n\) be \(n\) cyclic groups with \(|G_i|, |G_j|\) relatively prime for all \(i \neq j\). By the inductive hypothesis applied to \(G_1, \ldots, G_{n-1}\), the product \(G_1 \times \cdots \times G_{n-1}\) is cyclic, with order \(|G_1| \cdots |G_{n-1}|\) by order of direct product. Since \(\gcd(|G_n|, |G_i|) = 1\) for each \(i < n\), an integer coprime to each of \(|G_1|, \ldots, |G_{n-1}|\) is coprime to their product (elementary number theory), giving \(\gcd(|G_n|, |G_1| \cdots |G_{n-1}|) = 1\). Applying the previous theorem to the two cyclic groups \(G_1 \times \cdots \times G_{n-1}\) and \(G_n\) shows that \((G_1 \times \cdots \times G_{n-1}) \times G_n = G_1 \times \cdots \times G_n\) is cyclic.

For the converse, suppose some pair \(|G_i|, |G_j|\) (\(i \neq j\)) shares a common factor greater than \(1\). The maximum order of any element of \(G_1 \times \cdots \times G_n\) is \(\operatorname{lcm}(|G_1|, \ldots, |G_n|)\) (achieved by a tuple of generators, by order in direct products). However, \(\operatorname{lcm}(|G_1|, \ldots, |G_n|) \leq |G_1| \cdots |G_n|\) with equality if and only if the \(|G_i|\) are pairwise coprime (elementary number theory). Hence when some pair fails to be coprime, the maximum element order is strictly less than \(|G_1 \times \cdots \times G_n|\), so \(G_1 \times \cdots \times G_n\) cannot be cyclic.

Isomorphic: Since \(\gcd(2, 3) = 1\): \[ \mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6. \] Both are cyclic groups of order 6.

Not Isomorphic: Since \(\gcd(2, 4) = 2 \neq 1\): \[ \mathbb{Z}_2 \times \mathbb{Z}_4 \not\cong \mathbb{Z}_8. \] The maximum order of an element in \(\mathbb{Z}_2 \times \mathbb{Z}_4\) is \(\operatorname{lcm}(2, 4) = 4\), but \(\mathbb{Z}_8\) has an element of order 8.

Three factors: Since \(\gcd(2, 3) = \gcd(2, 5) = \gcd(3, 5) = 1\): \[ \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{30}. \] Moreover, we can express the same group in different forms: \[ \begin{align*} \mathbb{Z}_{30} &\cong \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5 \\\\ &\cong \mathbb{Z}_5 \times \mathbb{Z}_6 \\\\ &\cong \mathbb{Z}_3 \times \mathbb{Z}_{10}. \end{align*} \]

Define \(\phi: U(st) \to U(s) \times U(t)\) by \(\phi(x) = (x \bmod s, \, x \bmod t)\).

Well-defined: Let \(x \in U(st)\), so \(\operatorname{gcd}(x, st) = 1\). Since \(s \mid st\), any common divisor of \(x\) and \(s\) also divides \(st\), so \(\operatorname{gcd}(x, s) = 1\); similarly \(\operatorname{gcd}(x, t) = 1\). In particular \(x \bmod s \neq 0\) (otherwise \(s \mid x\), contradicting \(\operatorname{gcd}(x, s) = 1\)) and likewise \(x \bmod t \neq 0\), so \((x \bmod s) \in U(s)\) and \((x \bmod t) \in U(t)\).

Bijection: By the Chinese Remainder Theorem of elementary number theory, for each pair \((a, b)\) with \(a \in \mathbb{Z}_s\) and \(b \in \mathbb{Z}_t\), there exists a unique \(x \in \mathbb{Z}_{st}\) with \(x \equiv a \pmod{s}\) and \(x \equiv b \pmod{t}\).

• Surjectivity onto \(U(s) \times U(t)\): given \((a, b) \in U(s) \times U(t)\), the corresponding \(x\) satisfies \(\operatorname{gcd}(x, s) = \operatorname{gcd}(a, s) = 1\) (since \(x \equiv a \pmod{s}\)) and similarly \(\operatorname{gcd}(x, t) = 1\). Combined with \(\operatorname{gcd}(s, t) = 1\), this gives \(\operatorname{gcd}(x, st) = 1\), so \(x \in U(st)\) and \(\phi(x) = (a, b)\).
• Injectivity: if \(\phi(x) = \phi(y)\), then \(x \equiv y \pmod{s}\) and \(x \equiv y \pmod{t}\); uniqueness in CRT forces \(x \equiv y \pmod{st}\), i.e., \(x = y\) in \(U(st)\).

Homomorphism: The group operation in \(U(st)\) is multiplication modulo \(st\). Since \(s \mid st\), reduction modulo \(s\) is unaffected by reduction modulo \(st\): \((xy \bmod st) \bmod s = xy \bmod s\), and similarly for \(t\). Therefore \[ \phi(xy) = (xy \bmod s, \, xy \bmod t) = ((x \bmod s)(y \bmod s) \bmod s, \, (x \bmod t)(y \bmod t) \bmod t) = \phi(x) \phi(y). \]

Combining bijection and homomorphism, \(\phi\) is a group isomorphism, so \(U(st) \cong U(s) \times U(t)\).

The "Moreover" claim: under \(\phi\), the subgroup \(U_s(st) = \{x \in U(st) : x \equiv 1 \pmod{s}\}\) maps to \(\{1\} \times U(t)\), which is isomorphic to \(U(t)\) via the projection \((1, y) \mapsto y\). Similarly \(U_t(st) \cong U(s)\).

We follow three steps: (i) elements from different \(H_i\) commute, (ii) each element of \(G\) has a unique factorization, (iii) the map sending the external tuple to its product is the desired isomorphism.

Step 1 (Commutativity): Let \(h_i \in H_i\) and \(h_j \in H_j\) with \(i \neq j\), and consider the commutator \([h_i, h_j] = h_i h_j h_i^{-1} h_j^{-1}\). Since \(H_j \trianglelefteq G\), we have \(h_i h_j h_i^{-1} \in H_j\), so \([h_i, h_j] = (h_i h_j h_i^{-1}) h_j^{-1} \in H_j\). Since \(H_i \trianglelefteq G\), we have \(h_j h_i^{-1} h_j^{-1} \in H_i\), so \([h_i, h_j] = h_i (h_j h_i^{-1} h_j^{-1}) \in H_i\). Hence \([h_i, h_j] \in H_i \cap H_j\). Without loss of generality assume \(i < j\); then any \(h \in H_i\) can be written as a product \(h = e \cdot e \cdots h \cdots e\) where \(h\) appears at position \(i\) and the identity \(e\) appears at all other positions \(1, 2, \ldots, j-1\) (each factor lying in the corresponding \(H_k\)), so \(h \in H_1 H_2 \cdots H_{j-1}\). Hence \(H_i \cap H_j \subseteq (H_1 H_2 \cdots H_{j-1}) \cap H_j = \{e\}\) by condition 2 of the definition. Thus \([h_i, h_j] = e\), i.e., \(h_i h_j = h_j h_i\).

Step 2 (Unique factorization): Existence of a factorization \(g = h_1 h_2 \cdots h_n\) with \(h_i \in H_i\) is condition 1. For uniqueness, suppose \[ h_1 h_2 \cdots h_n = h'_1 h'_2 \cdots h'_n, \quad h_i, h'_i \in H_i. \] By Step 1, elements from different \(H_i\)'s commute. Starting from \(h_1 h_2 \cdots h_n = h'_1 h'_2 \cdots h'_n\), multiply both sides on the right by \(h_n^{-1}\) and on the left by \((h'_1 h'_2 \cdots h'_{n-1})^{-1} = (h'_{n-1})^{-1} \cdots (h'_1)^{-1}\) to obtain \[ (h'_{n-1})^{-1} \cdots (h'_1)^{-1} \cdot h_1 h_2 \cdots h_{n-1} = h'_n h_n^{-1}. \] By Step 1, each \((h'_i)^{-1}\) (in \(H_i\)) commutes with every \(h_j\) and every \((h'_k)^{-1}\) for \(j, k \neq i\); reordering the \(2(n-1)\) factors on the left to pair \((h'_i)^{-1}\) with \(h_i\) gives \[ h'_n h_n^{-1} = (h'_1)^{-1} h_1 \, (h'_2)^{-1} h_2 \, \cdots \, (h'_{n-1})^{-1} h_{n-1}. \tag{*} \] The left side \(h'_n h_n^{-1}\) lies in \(H_n\). For the right side, each factor \((h'_i)^{-1} h_i\) belongs to \(H_i\) (as \(H_i\) is a subgroup), so the right side is a product of one element from each of \(H_1, H_2, \ldots, H_{n-1}\) and hence lies in \(H_1 H_2 \cdots H_{n-1}\). Therefore \[ h'_n h_n^{-1} \in (H_1 H_2 \cdots H_{n-1}) \cap H_n = \{e\} \] by condition 2 with \(i = n-1\). So \(h'_n = h_n\). Cancelling \(h_n\) from both sides of the original equation gives \(h_1 \cdots h_{n-1} = h'_1 \cdots h'_{n-1}\); applying the same argument inductively yields \(h_i = h'_i\) for all \(i\).

Step 3 (Isomorphism): Define \(\phi: H_1 \times H_2 \times \cdots \times H_n \to G\) by \(\phi(h_1, h_2, \ldots, h_n) = h_1 h_2 \cdots h_n\), where the left side is the external direct product. By Step 2, \(\phi\) is a bijection (surjectivity from existence, injectivity from uniqueness). For the homomorphism property: \[ \begin{align*} \phi\big((h_1, \ldots, h_n)(h'_1, \ldots, h'_n)\big) &= \phi(h_1 h'_1, \ldots, h_n h'_n) \\\\ &= (h_1 h'_1)(h_2 h'_2) \cdots (h_n h'_n) \\\\ &= h_1 h_2 \cdots h_n \cdot h'_1 h'_2 \cdots h'_n \\\\ &= \phi(h_1, \ldots, h_n) \, \phi(h'_1, \ldots, h'_n), \end{align*} \] where the third equality uses Step 1: factors with distinct \(H_i\) indices commute, allowing all \(h_k\) to be moved before all \(h'_l\) while preserving the relative order of \(h_k\) and \(h'_k\) within each \(H_k\). Hence \(\phi\) is a group isomorphism.

Recall that \(U(105) \cong \mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_6\). This isomorphism arises because \(U(105)\) is the internal direct product of three subgroups: \[ U(105) = U_{35}(105) \times U_{21}(105) \times U_{15}(105), \] where \[ U_{35}(105) \cong U(3), \quad U_{21}(105) \cong U(5), \quad U_{15}(105) \cong U(7), \] recovering the external decomposition \(U(105) \cong U(3) \times U(5) \times U(7)\) from earlier. Concretely, \[ U_{35}(105) = \{1, 71\}, \quad U_{21}(105) = \{1, 22, 43, 64\}, \quad U_{15}(105) = \{1, 16, 31, 46, 61, 76\}. \] These subgroups satisfy the three conditions: each is normal (since \(U(105)\) is Abelian), their product gives all of \(U(105)\), and they have pairwise trivial intersections.

This decomposition makes it easy to find subgroups of \(U(105)\) with a specific order. Since subgroups of a direct product come from subgroups of the factors, we just need to find factor orders that multiply to the desired order.

For example, to find subgroups of order 12 in \(U(105)\), we look for divisors \(d_1 | 2\), \(d_2 | 4\), \(d_3 | 6\) with \(d_1 \cdot d_2 \cdot d_3 = 12\):

• \(1 \times 2 \times 6 = 12\): \[ \{1\} \times \{1, 64\} \times U_{15}(105) = \{1,4,16,19,31,34,46,61,64,76,79,94\} \]
• \(2 \times 2 \times 3 = 12\): \[ U_{35}(105) \times \{1,64\} \times \{1,16,46\} = \{1,4,11,16,29,44,46,64,71,74,79,86\} \]
• \(1 \times 4 \times 3 = 12\): \[ \{1\} \times U_{21}(105) \times \{1,16,46\} = \{1,4,16,22,37,43,46,58,64,67,79,88\} \]
• \(2 \times 1 \times 6 = 12\): \[U_{35}(105) \times \{1\} \times U_{15}(105) = U_5(105) = \{1,11,16,26,31,41,46,61,71,76,86,101\} \]

Each combination gives a distinct subgroup of order 12. Without the decomposition, finding these subgroups in the 48-element group \(U(105)\) would be far more tedious.

We proceed by strong induction on \(|G|\). The base case \(|G| = 1\) is vacuous, since no prime divides \(1\). For \(|G| = 2\), the only prime dividing \(2\) is \(p = 2\), and the unique non-identity element has order \(2\).

For the inductive step, suppose the theorem holds for all finite Abelian groups of order less than \(|G|\). Take any non-identity element \(x \in G\), and let \(m = |x|\).

Case 1 (\(p \mid m\)): Write \(m = pn\). Then \(x^n\) has order \(p\): indeed \((x^n)^p = x^m = e\), and if \((x^n)^k = e\) for some \(0 < k < p\) then \(x^{nk} = e\) gives \(m \mid nk\), i.e., \(pn \mid nk\), so \(p \mid k\) — impossible since \(0 < k < p\). Hence \(|x^n| = p\).

Case 2 (\(p \nmid m\)): Form the factor group \(\overline{G} = G / \langle x \rangle\), well-defined since \(G\) is Abelian (so every subgroup is normal). Then \(|\overline{G}| = |G| / m\) by Lagrange's theorem; since \(p \mid |G|\) and \(p \nmid m\), we have \(p \mid |\overline{G}|\). Also \(|\overline{G}| < |G|\) (as \(m \geq 2\)), so by the induction hypothesis \(\overline{G}\) contains an element \(\overline{y}\) of order \(p\); choose a representative \(y \in G\).

Now \(\overline{y}^p = \overline{e}\) in \(\overline{G}\), so \(y^p \in \langle x \rangle\), and therefore \((y^p)^m = y^{pm} = e\). By order dividing any annihilating power, \(|y|\) divides \(pm\). On the other hand, since \(\overline{y}\) has order \(p\) in \(\overline{G}\) and \(\overline{y}^{|y|} = \overline{e}\), the same theorem applied in \(\overline{G}\) gives \(p \mid |y|\). Combining \(p \mid |y|\) and \(|y| \mid pm\): write \(|y| = pa\) with \(a \geq 1\); then \(pa \mid pm\) gives \(a \mid m\). Setting \(k = a\), we obtain \(|y| = pk\) with \(k \mid m\).

Finally, \(y^k\) has order \(p\): \((y^k)^p = y^{pk} = y^{|y|} = e\), so \(|y^k|\) divides \(p\) and is therefore \(1\) or \(p\). If \(|y^k| = 1\) then \(y^k = e\), giving \(|y| \mid k\); but \(|y| = pk > k\), a contradiction. Hence \(|y^k| = p\).

It is straightforward to verify that \(H\) and \(K\) are subgroups of \(G\) using the one-step subgroup test. Both are nonempty (\(e^{p^n} = e^m = e\), so \(e \in H \cap K\)), and since \(G\) is Abelian, for \(x, y \in H\) we have \((xy^{-1})^{p^n} = x^{p^n}(y^{-1})^{p^n} = x^{p^n}(y^{p^n})^{-1} = e \cdot e^{-1} = e\), so \(xy^{-1} \in H\); similarly for \(K\).

Because \(G\) is Abelian, every subgroup is normal, so condition 1 of the internal direct product definition holds automatically. Hence to prove \(G = H \times K\), we need only show that \(G = HK\) and \(H \cap K = \{e\}\).

Showing \(G = HK\):
Since \(\gcd(m, p^n) = 1\), there exist integers \(s\) and \(t\) such that \(1 = sm + tp^n\). For any \(x \in G\), we have: \[ x = x^1 = x^{sm + tp^n} = x^{sm} \cdot x^{tp^n}. \] Now, since \(|G| = p^n m\), by Lagrange's theorem we have \(x^{|G|} = e\) for all \(x \in G\). Therefore: \[ (x^{sm})^{p^n} = x^{smp^n} = x^{s |G|} = (x^{|G|})^s = e, \] so \(x^{sm} \in H\). Similarly, \[ (x^{tp^n})^m = x^{tp^n m} = x^{t |G|} = (x^{|G|})^t = e, \] so \(x^{tp^n} \in K\). Thus \(x = x^{sm} \cdot x^{tp^n} \in HK\), and we conclude \(G = HK\).

Showing \(H \cap K = \{e\}\):
Suppose \(x \in H \cap K\). Then \(x^{p^n} = e\) and \(x^m = e\), so by order dividing any annihilating power, \(|x|\) divides both \(p^n\) and \(m\). Since \(\operatorname{gcd}(p^n, m) = 1\) (as \(p \nmid m\)), the only positive integer dividing both is \(1\), forcing \(|x| = 1\) and \(x = e\).

Showing \(|H| = p^n\):
Since \(H \cap K = \{e\}\), the order of \(HK\) gives \[ |HK| = |H||K| / |H \cap K| = |H||K|. \] Since \(G = HK\), we get \(p^n m = |H||K|\).

Now we determine what \(|H|\) and \(|K|\) must be. Consider any element \(x \in H\). By definition of \(H\), we have \(x^{p^n} = e\), so \(|x|\) divides \(p^n\). This means every element of \(H\) has order that is a power of \(p\). By Lagrange's theorem, \(|x|\) divides \(|H|\) for all \(x \in H\), and since all element orders are powers of \(p\), the group order \(|H|\) must also be a power of \(p\). (If a prime \(q \neq p\) divided \(|H|\), then by Cauchy's theorem for Abelian groups, \(H\) would contain an element of order \(q\), contradicting the fact that all elements of \(H\) have orders dividing \(p^n\).)

Similarly, every element \(x \in K\) satisfies \(x^m = e\), so \(|x|\) divides \(m\). Since \(p\) does not divide \(m\), no element of \(K\) can have order divisible by \(p\). By the same theorem, \(p\) does not divide \(|K|\).

Now we determine \(|H|\) and \(|K|\) explicitly. Since \(|H|\) is a power of \(p\) (from above), write \(|H| = p^k\) for some \(k \geq 0\). Since \(p\) does not divide \(|K|\), the prime factorization of \(|K|\) consists only of primes other than \(p\); in particular, \(|K|\) divides \(m\) (whose prime factorization is also \(p\)-free). Write \(m = |K| \cdot \ell\) for some positive integer \(\ell\). Substituting into \(|H||K| = p^n m\): \[ p^k \cdot |K| = p^n \cdot |K| \cdot \ell, \quad \text{so} \quad p^k = p^n \ell. \] Since \(\ell \geq 1\) is coprime to \(p\) (else \(p \mid m\)) and the left side is a pure power of \(p\), we must have \(\ell = 1\) and \(k = n\). Therefore \(|H| = p^n\) and \(|K| = m\).

Let \(|G| = p^n\). We proceed by induction on \(n\). If \(n = 1\), then \(G = \langle a \rangle \times \{e\}\), and we are done.

Now assume the result holds for all Abelian groups of order \(p^k\) where \(k < n\). Let \(a\) be an element of maximum order \(p^t\) in \(G\). By Lagrange's Theorem, every element of \(G\) has order dividing \(p^n\), hence a power of \(p\); since \(p^t\) is the maximum such order, every \(|x|\) divides \(p^t\), and therefore \(x^{p^t} = e\) for all \(x \in G\).

Case 1: If \(G = \langle a \rangle\), then \(G = \langle a \rangle \times \{e\}\), and we are done.

Case 2: Assume \(G \neq \langle a \rangle\). The set of elements of \(G\) not in \(\langle a \rangle\) is nonempty (by assumption) and finite (since \(G\) is finite), so it contains an element of smallest order; choose \(b\) to be such an element.

Claim: \(\langle a \rangle \cap \langle b \rangle = \{e\}\).

Proof of Claim. Since \(G\) is a \(p\)-group and \(b \neq e\) (as \(b \notin \langle a \rangle\) while \(e \in \langle a \rangle\)), the order \(|b|\) is a positive power of \(p\); in particular \(p\) divides \(|b|\). The proof proceeds in two stages: first we establish the Sub-claim that \(|b| = p\), and then we deduce \(\langle a \rangle \cap \langle b \rangle = \{e\}\) from it.

Sub-claim: \(|b| = p\).

Proof of Sub-claim. We first show \(b^p \in \langle a \rangle\). If \(|b| = p\), then \(b^p = e \in \langle a \rangle\) trivially. Otherwise \(|b| \geq p^2\), and by the Order of a Power formula, \[ |b^p| = \frac{|b|}{\gcd(|b|, p)} = \frac{|b|}{p} < |b|. \] In particular \(|b^p| \geq p > 1\), so \(b^p \neq e\). If \(b^p \notin \langle a \rangle\), then \(b^p\) would be an element outside \(\langle a \rangle\) with strictly smaller order than \(b\), contradicting the minimality of \(|b|\). Hence \(b^p \in \langle a \rangle\) in both cases. Say \(b^p = a^i\) for some integer \(i\).

Now observe: \[ e = b^{p^t} = (b^p)^{p^{t-1}} = (a^i)^{p^{t-1}} = a^{i \cdot p^{t-1}}, \] so \(|a| = p^t\) divides \(i \cdot p^{t-1}\). This means \(|a^i| \leq p^{t-1}\), so \(a^i\) is not a generator of \(\langle a \rangle\). By the Order of a Power formula applied to \(a\), we have \(|a^i| = p^t / \gcd(p^t, i)\), and \(|a^i| \leq p^{t-1}\) forces \(\gcd(p^t, i) \geq p\), hence \(p\) divides \(i\). Write \(i = pj\) for some integer \(j\).

Consider the element \(c = a^{-j}b\). Since \(b \notin \langle a \rangle\), certainly \(c \notin \langle a \rangle\) (for if \(c \in \langle a \rangle\), then \(b = a^j c \in \langle a \rangle\), a contradiction). In particular \(c \neq e\) (as \(e \in \langle a \rangle\) and \(c \notin \langle a \rangle\)). Also: \[ c^p = (a^{-j})^p \cdot b^p = a^{-pj} \cdot b^p = a^{-i} \cdot b^p = b^{-p} \cdot b^p = e. \] So \(c\) is an element of order dividing \(p\) that lies outside \(\langle a \rangle\); since \(c \neq e\), in fact \(|c| = p\). By the choice of \(b\) (smallest order among elements outside \(\langle a \rangle\)), we conclude \(|b| \leq |c| = p\); combined with \(p \mid |b|\), this gives \(|b| = p\), proving the Sub-claim.

Deducing the Claim. With \(|b| = p\) established, \(\langle a \rangle \cap \langle b \rangle\) is a subgroup of both \(\langle a \rangle\) and \(\langle b \rangle\). Since \(|\langle b \rangle| = p\) is prime, either \(\langle a \rangle \cap \langle b \rangle = \{e\}\) or \(\langle a \rangle \cap \langle b \rangle = \langle b \rangle\). The latter would mean \(b \in \langle a \rangle\), contradicting \(b \notin \langle a \rangle\). Therefore \(\langle a \rangle \cap \langle b \rangle = \{e\}\), proving the Claim.

Completing the proof:
Consider the factor group \(\overline{G} = G/\langle b \rangle\). Let \(\overline{x}\) denote the coset \(x\langle b \rangle\) in \(\overline{G}\).

We claim that \(|\overline{a}| = |a| = p^t\). If \(|\overline{a}| < |a| = p^t\), then \(\overline{a}^{p^{t-1}} = \overline{e}\). This means that \((a\langle b \rangle)^{p^{t-1}} = a^{p^{t-1}} \langle b \rangle = \langle b \rangle\), so that \(a^{p^{t-1}} \in \langle a \rangle \cap \langle b \rangle = \{e\}\) by the Claim above, contradicting the fact that \(|a| = p^t\). Thus, \(|\overline{a}| = |a| = p^t\), and therefore \(\overline{a}\) is an element of maximum order in \(\overline{G}\).

By induction, we know that \(\overline{G}\) can be written in the form \(\langle \overline{a} \rangle \times \overline{K}\) for some subgroup \(\overline{K}\) of \(\overline{G}\).

Let \(K\) be the pullback of \(\overline{K}\) under the natural homomorphism \(\pi: G \to \overline{G}\) (i.e., \(K = \pi^{-1}(\overline{K}) = \{x \in G \mid \overline{x} \in \overline{K}\}\)); \(K\) is a subgroup of \(G\) by the Properties of Subgroups under Homomorphism (the preimage statement). Note that \(\langle b \rangle \subseteq K\) since \(\overline{b} = \overline{e} \in \overline{K}\).

We claim \(\langle a \rangle \cap K = \{e\}\). For if \(x \in \langle a \rangle \cap K\), then \(\overline{x} \in \langle \overline{a} \rangle \cap \overline{K} = \{\overline{e}\}\), which means \(x \in \langle b \rangle\). Thus \(x \in \langle a \rangle \cap \langle b \rangle = \{e\}\).

Now we show \(G = \langle a \rangle K\). The natural projection \(\pi: G \twoheadrightarrow \overline{G}\) has kernel \(\langle b \rangle\), and by the Properties of Subgroups under Homomorphism (the \(n\)-to-\(1\) statement), every fiber of \(\pi\) has size \(|\langle b \rangle| = p\). Since \(K = \pi^{-1}(\overline{K})\) is a union of \(|\overline{K}|\) such fibers, \[ |K| = |\overline{K}| \cdot p. \] Combining \(\langle a \rangle \cap K = \{e\}\) with \(|\langle a \rangle| = |\langle \overline{a} \rangle|\) (proven above), the Order of \(HK\) formula gives: \[ |\langle a \rangle K| = \frac{|\langle a \rangle| \cdot |K|}{|\langle a \rangle \cap K|} = |\langle a \rangle| \cdot |K| = |\langle \overline{a} \rangle| \cdot |\overline{K}| \cdot p = |\overline{G}| \cdot p = |G|. \] Thus \(\langle a \rangle K = G\), and since \(\langle a \rangle \cap K = \{e\}\), we conclude \(G = \langle a \rangle \times K\).

Let \(|G| = p^n\) with \(n \geq 1\) (the case \(|G| = 1\) is vacuous, as there is nothing to decompose). We proceed by induction on \(n\).

Base case (\(n = 1\)): \(G\) has prime order \(p\), hence is cyclic (generated by any non-identity element). Thus \(G = \langle a \rangle\) is itself a cyclic group, giving the trivial one-factor decomposition.

Inductive step: Suppose the statement holds for all Abelian groups of prime-power order \(p^k\) with \(1 \leq k < n\), and let \(|G| = p^n\) with \(n \geq 2\). Choose an element \(a \in G\) of maximum order. Since \(G \neq \{e\}\), we have \(a \neq e\), and since \(G\) is a \(p\)-group, \(|a|\) is a positive power of \(p\); in particular \(|a| \geq p\). By the Extraction Lemma, there exists a subgroup \(K \leq G\) such that \(G = \langle a \rangle \times K\) (internal direct product).

From \(|G| = |\langle a \rangle| \cdot |K|\) and \(|\langle a \rangle| \geq p\), we obtain \(|K| = p^n / |\langle a \rangle| \leq p^{n-1} < p^n\). Hence \(|K|\) is a positive power of \(p\) (possibly \(p^0 = 1\)) strictly less than \(p^n\); moreover \(K\), as a subgroup of an Abelian group, is itself Abelian of prime-power order. If \(K = \{e\}\), then \(G = \langle a \rangle\) is already cyclic and we are done. Otherwise the inductive hypothesis applies to \(K\), yielding cyclic subgroups \(C_2, \ldots, C_r \leq K\) such that \(K = C_2 \times \cdots \times C_r\) (internal direct product).

Setting \(C_1 = \langle a \rangle\), we obtain \[ G = C_1 \times C_2 \times \cdots \times C_r, \] an internal direct product of cyclic subgroups.

We proceed by induction on \(|G|\).

Base case (\(|G| = p\)): \(G\) is cyclic of prime order, so its only decomposition as a direct product of nontrivial cyclic subgroups is \(G\) itself; hence \(m = n = 1\) and \(|H_1| = |K_1| = p\).

Inductive step. Suppose the statement is true for all Abelian groups of order less than \(|G|\), with \(|G| > p\) (so any decomposition into nontrivial cyclic subgroups has at least one factor, i.e. \(m, n \geq 1\)). For any Abelian group \(L\), define \[ L^p = \{x^p \mid x \in L\}. \] We first record two properties of this construction.

Proof of Remark. (1) \(e = e^p \in L^p\), and for \(x^p, y^p \in L^p\), commutativity gives \(x^p (y^p)^{-1} = x^p y^{-p} = (xy^{-1})^p \in L^p\); the One-Step Subgroup Test applies.

(2) By the Internal \(\cong\) External theorem applied to \(L = M_1 \times \cdots \times M_k\), every \(x \in L\) has a unique factorization \(x = m_1 m_2 \cdots m_k\) with \(m_i \in M_i\), and elements of distinct factors commute. Hence \(x^p = m_1^p m_2^p \cdots m_k^p\), which shows \(L^p \subseteq M_1^p M_2^p \cdots M_k^p\). Conversely, any product \(m_1^p m_2^p \cdots m_k^p\) equals \((m_1 m_2 \cdots m_k)^p \in L^p\), giving the reverse inclusion. Therefore \(L^p = M_1^p M_2^p \cdots M_k^p\), establishing condition 1 of the n-fold internal direct product.

For condition 2, we must show \((M_1^p \cdots M_i^p) \cap M_{i+1}^p = \{e\}\) for \(i = 1, \ldots, k-1\). Since \(M_j^p \leq M_j\) for each \(j\), we have the containments \(M_1^p \cdots M_i^p \subseteq M_1 \cdots M_i\) and \(M_{i+1}^p \subseteq M_{i+1}\). Intersecting: \[ (M_1^p \cdots M_i^p) \cap M_{i+1}^p \subseteq (M_1 \cdots M_i) \cap M_{i+1} = \{e\}, \] where the last equality is condition 2 for the \(M_j\)'s. Each \(M_j^p\) is also normal in \(L\) (since \(L\) is Abelian, every subgroup is normal). Thus \(L^p\) is the internal direct product \(M_1^p \times \cdots \times M_k^p\).

(3) Write \(L = \langle h \rangle\) with \(|h| = p^s\). Then \(L^p = \{h^{kp} \mid k \in \mathbb{Z}\} = \langle h^p \rangle\), and by the Order of a Power formula, \(|h^p| = p^s / \gcd(p^s, p) = p^{s-1}\). \(\blacksquare\)

We now apply the Remark to both decompositions of \(G\). By part (2) of the Remark, \[ G^p = H_1^p \times H_2^p \times \cdots \times H_m^p = K_1^p \times K_2^p \times \cdots \times K_n^p. \] These expressions include trivial factors: whenever \(|H_j| = p\), part (3) of the Remark gives \(H_j^p = \{e\}\), which contributes nothing to the product. To apply the inductive hypothesis — which requires decompositions into nontrivial cyclic subgroups — we let \(m'\) be the largest integer \(j\) such that \(|H_j| > p\) (so \(H_j^p\) is nontrivial precisely for \(j \leq m'\)), and define \(n'\) analogously for the \(K_i\). Dropping the trivial \(\{e\}\) factors: \[ G^p = H_1^p \times H_2^p \times \cdots \times H_{m'}^p = K_1^p \times K_2^p \times \cdots \times K_{n'}^p, \] now a genuine decomposition into nontrivial cyclic subgroups.

By part (3) of the Remark, \(|H_j^p| = |H_j|/p\) for \(j \leq m'\) and \(|K_i^p| = |K_i|/p\) for \(i \leq n'\); the orderings \(|H_1| \geq \cdots \geq |H_{m'}|\) and \(|K_1| \geq \cdots \geq |K_{n'}|\) are therefore preserved under this division. To see \(|G^p| < |G|\), we use the full \(m\)-factor form (where \(|H_j^p| = |H_j|/p\) holds uniformly, including \(|H_j^p| = 1 = p/p\) for \(j > m'\)): \[ |G^p| = \prod_{j=1}^{m} |H_j^p| = \prod_{j=1}^{m} \frac{|H_j|}{p} = \frac{|G|}{p^m}. \] Since \(m \geq 1\) (the decomposition of \(G\) involves at least one nontrivial factor), \(|G^p| < |G|\). Moreover \(G^p\) is Abelian of prime-power order, and the two decompositions \(G^p = H_1^p \times \cdots \times H_{m'}^p = K_1^p \times \cdots \times K_{n'}^p\) are into nontrivial cyclic subgroups of (weakly) decreasing order; hence the inductive hypothesis applies. It gives \(m' = n'\) and \(|H_i^p| = |K_i^p|\) for \(i = 1, \ldots, m'\). Multiplying each equality by \(p\): \[ |H_i| = |K_i| \quad \text{for } i = 1, \ldots, m'. \]

It remains to show \(m - m' = n - n'\), i.e., that the two decompositions have the same number of factors of order exactly \(p\). Splitting the product \(|G| = \prod_{j=1}^{m} |H_j|\) at index \(m'\), we have \(\prod_{j > m'} |H_j| = p^{m-m'}\) (since each such \(|H_j| = p\)), so \[ |H_1| \cdots |H_{m'}| \cdot p^{m-m'} = |G| = |K_1| \cdots |K_{n'}| \cdot p^{n-n'} \] by the analogous splitting for the \(K_i\). Since \(|H_i| = |K_i|\) for \(i = 1, \ldots, m' = n'\), cancellation gives \(p^{m-m'} = p^{n-n'}\), hence \(m - m' = n - n'\). Combined with \(m' = n'\), this gives \(m = n\) and \(|H_i| = |K_i|\) for all \(i\).

Let \(G\) be a finite Abelian group of order 72. Since \(72 = 2^3 \cdot 3^2\), Lemma 1 (Primary Decomposition) decomposes \(G\) as \(G = G(2) \times G(3)\), where \(|G(2)| = 8\) and \(|G(3)| = 9\). We must therefore pair a decomposition of the \(2\)-primary component with a decomposition of the \(3\)-primary component.

The partitions of \(3\) are \(3\), \(2+1\), and \(1+1+1\). Via the correspondence sending a partition \(b_1 + b_2 + \cdots + b_s\) (with \(b_1 \geq b_2 \geq \cdots \geq b_s \geq 1\)) to the cyclic-factor structure \(\mathbb{Z}_{2^{b_1}} \times \mathbb{Z}_{2^{b_2}} \times \cdots \times \mathbb{Z}_{2^{b_s}}\), these correspond to \[ \mathbb{Z}_8, \quad \mathbb{Z}_4 \times \mathbb{Z}_2, \quad \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2. \] The partitions of \(2\) are \(2\) and \(1+1\), giving (under the analogous correspondence with \(p = 3\)) \[ \mathbb{Z}_9, \quad \mathbb{Z}_3 \times \mathbb{Z}_3. \] Taking external direct products across primes yields \(3 \times 2 = 6\) non-isomorphic Abelian groups of order 72:

• \(\mathbb{Z}_8 \times \mathbb{Z}_9 \cong \mathbb{Z}_{72}\)
• \(\mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_3\)
• \(\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9\)
• \(\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3\)
• \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9\)
• \(\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3\)

Two such groups are pairwise non-isomorphic because their decompositions differ in cyclic-factor structure: by Lemma 1, the primary components \(G(2)\) and \(G(3)\) are intrinsically determined by \(G\) (as the sets of elements whose orders are powers of \(2\) and \(3\), respectively); and by Lemma 4 (Uniqueness) applied to each primary component, the cyclic-factor list within each component is itself unique. Hence the six combinations above yield six non-isomorphic groups. Conversely, by Lemma 1 (applied iteratively to separate all prime factors) and Lemma 3 (Existence) (to decompose each primary component into cyclic factors), every Abelian group of order 72 is isomorphic to one of them. Therefore there are exactly six non-isomorphic Abelian groups of order 72.