Structure of finite abelian groups

Theorem

Structure of finite abelian groups

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Statement

Let \( G \) be a finite abelian group, written additively, with \( |G| = n \geq 1 \). Then there exist a unique integer \( k \geq 0 \) and a unique sequence of integers \( d_1, d_2, \ldots, d_k \geq 2 \) satisfying \( d_1 \mid d_2 \mid \cdots \mid d_k \) (the invariant factors) such that \[ G \cong \mathbb{Z}/d_1\mathbb{Z} \times \mathbb{Z}/d_2\mathbb{Z} \times \cdots \times \mathbb{Z}/d_k\mathbb{Z}, \qquad d_1 d_2 \cdots d_k = n. \] Equivalently, there is a unique multiset of prime powers \( p_1^{e_1}, \ldots, p_r^{e_r} \) (the elementary divisors, with repeats allowed across different primes) such that \[ G \cong \mathbb{Z}/p_1^{e_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_r^{e_r}\mathbb{Z}, \qquad p_1^{e_1}\cdots p_r^{e_r} = n. \]

Why it matters

This theorem is the complete classification of a natural and enormous class of objects: it says a finite abelian group carries no information beyond a finite list of numbers. Two finite abelian groups are isomorphic if and only if their invariant factor lists (or elementary divisor multisets) coincide, so classification questions ("how many abelian groups of order 72 are there, up to isomorphism?") become pure combinatorics — counting multiplicative partitions of the prime factorisation of \( n \).

It is also the prototype for a family of structure theorems: the same argument, run over the ring \( \mathbb{Z} \) replaced by any principal ideal domain, classifies finitely generated modules over a PID, which in turn yields both this theorem and the Jordan form of a linear operator as special cases. Understanding this proof is understanding the machine behind both results at once.

Hypotheses

\( G \) is a group.Without a group structure there is no notion of a generating relation to decompose; the statement is vacuous. \( G \) is abelian.The theorem is false for non-abelian groups: \( S_3 \) has order 6 but is not isomorphic to any product of cyclic groups, since every such product is abelian while \( S_3 \) is not (\( (12)(13) \neq (13)(12) \)). Non-abelian finite groups still decompose via composition series (Jordan–H\"older) but not as direct products of cyclic pieces. \( G \) is finite.Drop finiteness and the clean statement fails: \( (\mathbb{Q}, +) \) is a torsion-free abelian group that is not a direct sum of cyclic groups (it is not even finitely generated, and it has no minimal generating set behaving like a basis). The correct infinite analogue — the structure theorem for finitely generated abelian groups, \( G \cong \mathbb{Z}^r \times \mathbb{Z}/d_1\mathbb{Z}\times\cdots\times\mathbb{Z}/d_k\mathbb{Z} \) — needs finite generation as a replacement hypothesis, and even that excludes groups like \( \mathbb{Q} \) or the Pr\"ufer group \( \mathbb{Z}(p^\infty) \). \( |G| \) is finite so every element has finite order dividing \( |G| \).This is not an independent hypothesis but a consequence (Lagrange's theorem) used repeatedly in the proof; flagged because the proof implicitly relies on "torsion everywhere," which is exactly what finiteness buys.

Proof

We give the elementary-divisor form of the proof. Two steps do the real work: (1) split \( G \) into its primary (Sylow) components, and (2) show a finite abelian \( p \)-group is a direct product of cyclic \( p \)-groups. Existence and uniqueness are treated separately.

\text{Write } |G| = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r} \text{ with distinct primes } p_i.

Fundamental theorem of arithmetic applied to \( n = |G| \). A

G_{p_i} := \{ g \in G : p_i^{m} g = 0 \text{ for some } m \geq 0 \} \text{ is a subgroup of } G, \text{ the } p_i\text{-primary component.}

Closure: if \( p_i^m g = 0 \) and \( p_i^{m'} h = 0 \) then \( p_i^{\max(m,m')}(g-h) = 0 \) since \( G \) is abelian, so the sum/difference of two \( p_i \)-torsion elements is \( p_i \)-torsion. A

G \cong G_{p_1} \times G_{p_2} \times \cdots \times G_{p_r} \qquad(\textbf{primary decomposition}).

Since \( \gcd(|G|/p_i^{a_i}, p_i^{a_i}) = 1 \) for the appropriate cofactor, B\'ezout gives integers with \( 1 = u_i (|G|/p_i^{a_i}) + v_i p_i^{a_i} \); the map \( g \mapsto \big( u_1(|G|/p_1^{a_1}) g,\ldots \big)\) composed with projections realises an explicit isomorphism onto the internal direct product \( G_{p_1}+\cdots+G_{p_r} \), using that every element of \( G \) has order dividing \( |G| \) (Lagrange) and the Chinese Remainder idempotents \( e_i = u_i(|G|/p_i^{a_i}) \) satisfy \( \sum e_i \equiv 1 \) and \( e_i e_j \equiv 0 \pmod{|G|} \) for \( i\neq j \). This is the standard "CRT decomposition of a torsion abelian group" lemma. C

\text{It suffices to prove: every finite abelian } p\text{-group } P \text{ (i.e. } |P|=p^a\text{) is a direct product of cyclic } p\text{-groups.}

By Step 3, \( G \) is the direct product of its primary components \( G_{p_i} \), each of which has order a prime power; decomposing each \( G_{p_i} \) and concatenating the cyclic factors decomposes \( G \). B

\text{Let } P \text{ be abelian, } |P| = p^a \gt 1, \text{ and let } x \in P \text{ have maximal order } p^{e}, \, e = \max\{\text{ord}(y): y\in P\}.

\( P \) is finite and nonempty, so the set of orders of elements has a maximum; every order divides \( p^a \) (Lagrange), so is a power of \( p \). A

\text{Claim: } \langle x \rangle \text{ is a direct factor, i.e. there exists a subgroup } K \leq P \text{ with } P = \langle x \rangle \times K.

This is the key inductive step (\textbf{pure subgroup splitting lemma}), proved in Steps 7–9 below. B

\text{Consider } P/\langle x\rangle. \text{ By induction on } |P| \text{ it is a direct product of cyclic groups; pick } \bar y \in P/\langle x\rangle \text{ of maximal order } p^{f} \text{ generating a cyclic direct factor, and any preimage } y\in P.

Strong induction on group order: \( |P/\langle x\rangle| = p^{a-e} \lt p^a \) since \( e\geq 1 \) (as \( P\neq \{0\} \)), so the inductive hypothesis applies to the strictly smaller quotient. C

\text{Order of } \bar y \text{ in } P/\langle x\rangle \text{ is } \leq \text{ order of } x \text{ in } P, \text{ i.e. } p^{f}\leq p^{e}; \text{ moreover } y \text{ can be corrected to } y' = y - c x \text{ with } \text{ord}(y') = p^f \text{ exactly (not just dividing).}

Maximality of \( e \) (Step 5) forces \( p^f \mid p^e \) since the order of any element of \( P \) is at most \( p^e \) and orders can only shrink under the quotient map composed with lifting arguments; the explicit correction \( y' = y - c x \) killing the extra \( \langle x\rangle \)-component uses that \( p^{f} y \in \langle x \rangle \) (because \( \bar y \) has order \( p^f\) in the quotient) together with divisibility \( p^f \mid p^e\) to solve \( p^f y = p^f (cx) \) for \( c \) inside the cyclic group \( \langle x\rangle \) — this is the standard "lifting a maximal-order element" computation underlying Kaplansky's pure-subgroup argument. C

\text{Let } K = \langle y' \rangle \times (\text{preimages of the remaining cyclic factors of } P/\langle x\rangle, \text{ corrected the same way}). \text{ Then } P = \langle x \rangle \times K.

By construction each corrected generator has \( \langle x \rangle \cap \langle (\text{corrected generator}) \rangle = \{0\} \) and their images generate \( P/\langle x \rangle \), so together with \( x \) they generate \( P \); a counting argument on orders (\( |\langle x\rangle|\cdot |K| = |P| \)) confirms the product is direct and exhaustive, completing the induction begun at Step 5 (base case \( |P|=1\) trivial, or \( |P|=p \) cyclic). C

\text{Iterating Steps 5\textendash 9 on } K \text{ (which is strictly smaller than } P\text{) yields } P \cong \mathbb{Z}/p^{e_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{e_s}\mathbb{Z}.

Finite descent: \( |K| \lt |P| \), so repeating the construction terminates, giving \textbf{existence} of the elementary-divisor decomposition for \( p \)-groups, and via Step 4 for general finite abelian \( G \). B

\textbf{Uniqueness.} \text{ For each prime } p \text{ and each } j\geq 1, \text{ the rank } r_j(p) := \dim_{\mathbb{F}_p}\big( p^{j-1}G_p[p] / p^{j}G_p[p]\big) \text{ (equivalently, the number of cyclic factors } \mathbb{Z}/p^{e}\mathbb{Z} \text{ with } e \geq j\text{) is an isomorphism invariant of } G.

The subgroups \( p^{j-1}G_p \) and the subgroup of elements killed by \( p \) (the "\( p \)-socle") are defined purely group-theoretically (no reference to a chosen decomposition), so any two decompositions of the same \( G_p \) must produce the same sequence of ranks \( r_j(p) \); this is the standard invariant-subgroup argument for uniqueness in the structure theorem. C

\text{The multiset of pairs } (p, e) \text{ with multiplicity } r_e(p)-r_{e+1}(p) \text{ recovers exactly the elementary divisors } p^{e}; \text{ hence the decomposition of Step 10 is unique up to reordering the factors.}

Direct bookkeeping: the number of cyclic factors \( \mathbb{Z}/p^e\mathbb{Z} \) equals \( r_e(p) - r_{e+1}(p) \), a quantity depending only on \( G \) by Step 11, so it is the same for every decomposition of \( G \). Regrouping the elementary divisors by CRT (\( \mathbb{Z}/p^{e}\mathbb{Z}\times\mathbb{Z}/q^{f}\mathbb{Z}\cong \mathbb{Z}/p^eq^f\mathbb{Z} \) for \( p\neq q \)) then gives the unique invariant-factor form with the divisibility chain \( d_1\mid d_2\mid\cdots\mid d_k \), completing uniqueness. B

Result

G \cong \mathbb{Z}/d_1\mathbb{Z}\times\cdots\times\mathbb{Z}/d_k\mathbb{Z},\quad d_1\mid d_2\mid\cdots\mid d_k,\quad \text{uniquely}

Reading. Every finite abelian group is completely determined, up to isomorphism, by one short list of integers: either the chain of invariant factors, or equivalently the multiset of prime-power elementary divisors. There is no other information hiding in a finite abelian group.

Scope. Applies to every finite abelian group with no further restriction. It does not extend as stated to infinite or non-abelian groups; the correct generalisation to finitely generated abelian groups adds a free part \( \mathbb{Z}^r \), and the correct generalisation to modules is the structure theorem for finitely generated modules over a PID.

Corollaries & converses

The number of isomorphism classes of abelian groups of order \( n = \prod p_i^{a_i} \) equals \( \prod_i \pi(a_i) \), the product of the number of integer partitions of each exponent \( a_i \).
A finite abelian group is cyclic if and only if it has at most one elementary divisor for each prime, equivalently \( k = 1 \) in the invariant-factor form, equivalently \( \gcd \) of any two of its invariant factors beyond \( d_1 \) does not arise (there is only \( d_1 = n \)).
\( G \) is cyclic of order \( n \) iff \( G \) has an element of order \( n \) iff, for every prime \( p \mid n \), \( G \) has a unique subgroup of order \( p \) (this refines the classification into an easily checked criterion).
Converse direction: given any chain \( d_1 \mid d_2 \mid \cdots \mid d_k \) with \( d_i \geq 2 \), the product \( \mathbb{Z}/d_1\mathbb{Z}\times\cdots\times\mathbb{Z}/d_k\mathbb{Z} \) is a finite abelian group realising that data — existence of a group for any admissible invariant-factor sequence holds trivially by construction, so the classification is a genuine bijection between isomorphism classes and admissible sequences, not just an upper bound.
Sylow's theorems for abelian groups become redundant complexity: the Sylow \( p \)-subgroup is unique and equals \( G_p \) from Step 2/3, since in an abelian group every subgroup is normal.

Fails without

Abelian dropped: \( Q_8 \) (quaternion group, order 8) is not a direct product of cyclic groups: every proper subgroup of \( Q_8 \) is cyclic, but \( Q_8 \) itself has a unique element of order 2, whereas \( \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z} \) (the only order-8 abelian candidate with an order-4 element) has three elements of order 2 — so no cyclic decomposition matches \( Q_8 \)'s element-order profile.
Finite dropped: \( (\mathbb{Z}, +) \) is abelian and a direct product of "a cyclic group" trivially, but \( (\mathbb{Q}, +) \) is abelian, torsion-free, and NOT isomorphic to any direct sum of cyclic groups: it is divisible (for every \( q\in\mathbb{Q}, n\in\mathbb{N} \) there is \( q'\) with \( nq'=q \)) while no nontrivial direct sum of copies of \( \mathbb{Z} \) or \( \mathbb{Z}/m\mathbb{Z} \) is divisible.
Finite dropped (second example): the Pr\"ufer group \( \mathbb{Z}(p^\infty) = \{ z \in \mathbb{C}^\times : z^{p^n}=1 \text{ for some } n\} \) is an infinite abelian torsion group in which every proper subgroup is finite cyclic, yet the whole group is not a direct product of cyclic groups (it is not even finitely generated) — showing torsion alone cannot replace finiteness.

Common errors

Writing the elementary-divisor decomposition and calling it "the" decomposition without noting the invariant-factor form is a different (though equivalent) grouping — students often merge the two and get the divisibility chain wrong.
Assuming the theorem gives an isomorphism to \( \mathbb{Z}/n\mathbb{Z} \) whenever \( |G|=n \); this is only valid when \( \gcd \) conditions force a single invariant factor (e.g. \( n \) squarefree), not in general — \( \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} \not\cong \mathbb{Z}/4\mathbb{Z} \) despite both having order 4.
Applying the CRT merge \( \mathbb{Z}/a\mathbb{Z}\times\mathbb{Z}/b\mathbb{Z}\cong\mathbb{Z}/ab\mathbb{Z} \) when \( \gcd(a,b)\neq 1 \); this isomorphism requires coprimality and is false otherwise.
Forgetting that the theorem classifies up to isomorphism only — two different-looking generating sets or presentations of the same abstract group are not "different groups," so counting decompositions is about counting isomorphism classes, not counting subgroup lattices.
Misapplying the theorem to non-abelian groups of the same order (e.g. thinking every group of order 8 is one of the abelian ones); order alone never determines the group without the abelian hypothesis.

Discussion

The result traces to Kronecker's 1870 work on finite abelian groups (phrased then in terms of forms and congruences rather than groups explicitly), with the modern group-theoretic and module-theoretic proof crystallising through the late 19th and early 20th centuries alongside the general theory of finitely generated modules over Euclidean domains and PIDs. Its final, clean shape is inseparable from the language of module theory: \( \mathbb{Z}/n\mathbb{Z} \) modules are exactly finite abelian groups viewed as \( \mathbb{Z} \)-modules, and the theorem is the \( \mathbb{Z} \) case of the structure theorem for finitely generated modules over a PID.

The proof strategy generalises verbatim: replace \( \mathbb{Z} \) by any PID \( R \), replace "order" by an appropriate length/valuation argument, and the same primary decomposition plus maximal-order-splitting argument produces the structure theorem for finitely generated \( R \)-modules, \( M \cong R^r \oplus R/(d_1)\oplus\cdots\oplus R/(d_k) \) with \( d_1\mid\cdots\mid d_k \). Taking \( R = k[x] \) and \( M \) the underlying \( k[x] \)-module of a linear operator (via \( x\cdot v = Tv \)) recovers rational canonical form, and further specialising to algebraically closed \( k \) recovers Jordan canonical form — the invariant factors here play exactly the role of the invariant factors of a matrix (via Smith normal form of \( xI - A \)).

There is a third equivalent formulation worth knowing: the elementary-divisor form corresponds precisely to the Smith normal form of the relation matrix when \( G \) is presented as \( \mathbb{Z}^m / \mathrm{im}(A) \) for an integer matrix \( A \) — diagonalising \( A \) over \( \mathbb{Z} \) by row/column operations (which preserve the cokernel up to isomorphism) computes the invariant factors directly and algorithmically, giving a constructive route to the theorem that complements the existence/uniqueness proof given here.

Common misconceptions: that "cyclic group of order \( n \)" and "abelian group of order \( n \)" are the same thing (false whenever \( n \) is not squarefree); that the decomposition is literally unique as a subgroup decomposition of \( G \) (false — only the isomorphism type of each factor and their multiplicities are unique, the actual subgroups \( \langle x\rangle \) chosen in the proof are not canonical); and that the theorem requires computing anything about the group's automorphisms or presentation to apply (it does not — order and abelianness alone, via the prime factorisation of \( |G| \), determine every possible isomorphism type).

Worked examples

\text{Classify all abelian groups of order } 72.

\( 72 = 2^3\cdot 3^2 \); by the theorem, an abelian group of order 72 is determined by choosing a partition of the exponent 3 (for the prime 2) and a partition of the exponent 2 (for the prime 3), independently. A

\text{Partitions of } 3: \ 3,\ 2+1,\ 1+1+1 \quad(\pi(3)=3). \qquad \text{Partitions of } 2: \ 2,\ 1+1 \quad(\pi(2)=2).

Each partition of the exponent \( a_i \) of prime \( p_i \) lists the exponents \( e_1\geq e_2\geq\cdots \) of the elementary divisors \( p_i^{e_1}, p_i^{e_2},\ldots \) at that prime, by Steps 10–12 of the proof. A

\text{Total isomorphism classes} = \pi(3)\cdot \pi(2) = 3\cdot 2 = 6.

Corollary from the theorem: the choices at distinct primes are independent because elementary divisors for different primes combine freely (Step 3, primary decomposition). B

\text{Listing them (2-part} \times \text{3-part):} \quad \mathbb{Z}_8\!\times\!\mathbb{Z}_9,\ \ \mathbb{Z}_8\!\times\!\mathbb{Z}_3^2,\ \ \mathbb{Z}_4\!\times\!\mathbb{Z}_2\!\times\!\mathbb{Z}_9,\ \ \mathbb{Z}_4\!\times\!\mathbb{Z}_2\!\times\!\mathbb{Z}_3^2,\ \ \mathbb{Z}_2^3\!\times\!\mathbb{Z}_9,\ \ \mathbb{Z}_2^3\!\times\!\mathbb{Z}_3^2.

Direct enumeration of the \( 3\times 2 \) grid of partition pairs, writing \( \mathbb{Z}_m := \mathbb{Z}/m\mathbb{Z} \). A

\text{There are exactly } 6 \text{ abelian groups of order } 72, \text{ up to isomorphism.}

\text{Find the invariant-factor decomposition of } G = \mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/10\mathbb{Z}, \text{ and decide whether } G \text{ is cyclic.}

Starting data given in a "mixed" (non-elementary, non-invariant) presentation; we must first pass to elementary divisors, per the algorithm implicit in Steps 3, 10–12. A

\text{Elementary divisors: } 4=2^2 \Rightarrow 2^2;\quad 6=2\cdot 3 \Rightarrow 2,\,3;\quad 10 = 2\cdot 5 \Rightarrow 2,\,5.

CRT splitting of each cyclic factor into its primary components (Step 3 in reverse), valid since the factors of each \( \mathbb{Z}/m\mathbb{Z} \) are coprime. A

\text{Collected by prime: } \ p=2: \{2^2, 2, 2\}; \quad p=3: \{3\}; \quad p=5: \{5\}.

Regrouping the multiset of elementary divisors by prime, as in Step 11/12's rank-counting argument. A

\text{Invariant factors (largest exponent per prime goes into } d_k\text{, working down): } d_3 = 2^2\cdot3\cdot5=60,\ \ d_2 = 2,\ \ d_1 = 2.

Standard construction: sort each prime's elementary-divisor exponents descending, then \( d_k \) is the product of the largest exponent at each prime, \( d_{k-1} \) the product of the second largest (using 1, i.e. \( p^0 \), if a prime runs out), and so on, guaranteeing \( d_1\mid d_2\mid\cdots\mid d_k \) by construction. B

\text{Check divisibility: } 2 \mid 2 \mid 60. \quad G \cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/60\mathbb{Z}, \quad |G|=2\cdot2\cdot60=240=4\cdot6\cdot10.\ \checkmark

Sanity check against Step 12's uniqueness statement and order consistency; the chain condition holds as required for the invariant-factor form. A

\text{Since } k=3 \gt 1 \text{ (more than one invariant factor exceeding 1), } G \text{ is not cyclic.}

Corollary from the "Corollaries" section: \( G \) cyclic \( \iff k=1 \). Here there are two nontrivial factors sharing the prime 2, matching the presence of two elements of order 2 that cannot be merged. B

\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/10\mathbb{Z} \;\cong\; \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/60\mathbb{Z}, \text{ not cyclic.}

Problems

How many abelian groups of order \( 100 \) are there, up to isomorphism?
Solution
\( 100 = 2^2\cdot 5^2 \). The count is \( \pi(2)\cdot\pi(2) = 2\cdot 2 = 4 \), since \( \pi(2)=2 \) (partitions \( 2 \) and \( 1+1 \)). Explicitly: \( \mathbb{Z}_4\times\mathbb{Z}_{25} \), \( \mathbb{Z}_4\times\mathbb{Z}_5^2 \), \( \mathbb{Z}_2^2\times\mathbb{Z}_{25} \), \( \mathbb{Z}_2^2\times\mathbb{Z}_5^2 \). Answer: 4.
Determine, with justification, whether \( \mathbb{Z}/9\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z} \) and \( \mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z} \) are isomorphic.
Solution
Elementary divisors of the first: \( \{9,4\} = \{3^2, 2^2\} \). Elementary divisors of the second: \( 6=2\cdot3 \) twice, giving \( \{2,2,3,3\} \). These multisets differ (the first has one factor of \( 3^2 \), the second has two factors of \( 3^1 \)), so by the uniqueness clause (Step 11–12) the groups are NOT isomorphic. Confirmed by orders: both have order 36 but the first is cyclic (\( \gcd(9,4)=1\Rightarrow \mathbb{Z}_9\times\mathbb{Z}_4\cong\mathbb{Z}_{36}\)) while the second is not (it has invariant factors \( 6\mid 6 \), and an element of order dividing 6 everywhere, so no element of order 36).
Prove directly, without quoting the general theorem, that every abelian group of order \( p^2 \) (\( p \) prime) is isomorphic to \( \mathbb{Z}/p^2\mathbb{Z} \) or \( \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z} \).
Solution
Let \( |P|=p^2 \). If \( P \) has an element of order \( p^2 \), \( P \) is cyclic, done. Otherwise every non-identity element has order \( p \) (orders divide \( p^2 \) by Lagrange, and order \( p^2 \) was excluded). Pick \( x\neq 0 \); \( \langle x\rangle \) has order \( p \). Pick \( y\notin\langle x\rangle \) (exists since \( p^2\gt p \)); \( \langle y\rangle \) has order \( p \) and \( \langle x\rangle\cap\langle y\rangle=\{0\}\) since a nontrivial intersection would force \( \langle x\rangle=\langle y\rangle\) (both order \( p \), a prime, so any nontrivial subgroup of \( \langle x \rangle\) is all of it). Then \( \langle x\rangle+\langle y\rangle\) has order \( p\cdot p / |\langle x\rangle\cap\langle y\rangle| = p^2 = |P|\), so \( P=\langle x\rangle\times\langle y\rangle\cong\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}\).
Give an explicit isomorphism showing \( \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z} \cong \mathbb{Z}/210\mathbb{Z} \), and explain why this collapsing is NOT possible for \( \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} \).
Solution
Since \( 2,3,5,7 \) are pairwise coprime, the Chinese Remainder Theorem gives a ring (hence group) isomorphism \( \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}\to \mathbb{Z}/210\mathbb{Z} \) via \( (a,b,c,d)\mapsto \) the unique residue mod 210 congruent to \( a,b,c,d \) mod \( 2,3,5,7 \) respectively; equivalently \( 1\mapsto(1,1,1,1) \) generates the product, since \( (1,1,1,1) \) has order \( \mathrm{lcm}(2,3,5,7)=210 \). For \( \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z} \) the moduli are not coprime (\( \gcd(2,2)=2\)), so \( \mathrm{lcm}(2,2)=2\neq 4=|G| \): no element has order 4, so no such isomorphism to \( \mathbb{Z}/4\mathbb{Z}\) can exist (a cyclic group of order 4 must contain an order-4 element).
A finite abelian group \( G \) has exactly 7 elements \( x \) satisfying \( x+x=0 \) (including the identity). Show \( |G| \) is divisible by \( 8 \) and find the possible 2-primary elementary divisor patterns consistent with this.
Solution
The elements with \( 2x=0 \) form the "2-torsion subgroup" \( G[2] \), which by the elementary divisor decomposition of the 2-primary part \( G_2\cong \mathbb{Z}/2^{e_1}\mathbb{Z}\times\cdots\times\mathbb{Z}/2^{e_s}\mathbb{Z} \) satisfies \( G[2]\cong (\mathbb{Z}/2\mathbb{Z})^s \) (one copy of \( \mathbb{Z}/2\mathbb{Z} \) from each cyclic factor, since each \( \mathbb{Z}/2^{e_i}\mathbb{Z} \) contributes exactly one element of order dividing 2 besides identity... precisely one nonzero 2-torsion element). So \( |G[2]| = 2^s = 8 \Rightarrow s=3 \): the 2-primary part has exactly 3 elementary divisors \( 2^{e_1},2^{e_2},2^{e_3} \) with \( e_1,e_2,e_3\geq 1 \), so \( |G_2| = 2^{e_1+e_2+e_3}\geq 2^3=8 \), and since \( G_2 \) is a direct factor of \( G \) (Step 3), \( 8 \mid |G_2| \mid |G| \). The possible patterns (up to the exponents themselves, which are otherwise unconstrained by the given data) are any \( \{e_1,e_2,e_3\}\) with each \( e_i\geq1 \), e.g. \( \{1,1,1\}\) (giving \( |G_2|=8\)), \( \{2,1,1\}\) (giving 16), etc. — the count of 2-torsion elements alone pins down \( s=3 \) but not the individual exponents.