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Every finite abelian group is a quotient of $(Z / n)^{k}$ for some $n$ and $k$

Statement

For every finite abelian group $G$ there are positive integers $n$ and $k$ and a surjective group homomorphism

$(Z / n)^{k} ⟶ G,$

where $(Z / n)^{k}$ denotes the set of $k$ -tuples of classes in $Z / n$ , with componentwise addition, for the additive group $Z / n$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold). Equivalently, $G ≅ (Z / n)^{k} / H$ for a subgroup $H$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

Facts & Assumptions

Given: A finite abelian group $G$ .

[L1]

For every finite abelian group $G$ there is a unique list $1 < n_{1} ∣ n_{2} ∣ \dots ∣ n_{r}$ with $G ≅ C_{n_{1}} \times \dots \times C_{n_{r}}$ ; the trivial group corresponds to the empty list (Fundamental theorem of finite abelian groups: invariant-factor form, Invariant-factor data for a finite abelian group).

[L2]

A cyclic group of finite order $m$ is isomorphic to $(Z / m, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[L3]

For every natural number $m$ , one has $[a]_{m} = [b]_{m}$ exactly when $m ∣ a - b$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ), addition of classes is represented by addition of representatives, and $(Z / m, +)$ is an abelian group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L4]

For a homomorphism $f : A \to B$ , the rule $a \ker f \mapsto f (a)$ is an isomorphism $A / \ker f \to im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ , The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · cases

1.1

In the case that the invariant-factor list of [L1] is empty, $G$ is trivial; take $n = 1$ and $k = 1$ , so that $(Z / 1)^{1}$ is a one-element group and the unique map to $G$ is a surjective homomorphism.

assume-case trivialL1

1.2

In the case $r \geq 1$ , put $n : = n_{r}$ and $k : = r$ . Each $n_{i}$ divides $n$ , the list being a divisibility chain by [L1].

assume-case nontrivialL1

2.1

For each $i$ the rule $[a]_{n} \mapsto [a]_{n_{i}}$ is a well-defined surjective group homomorphism $Z / n \to Z / n_{i}$ : if $[a]_{n} = [b]_{n}$ then $n ∣ a - b$ , hence $n_{i} ∣ a - b$ because $n_{i} ∣ n$ , so $[a]_{n_{i}} = [b]_{n_{i}}$ by [L3]; it respects addition by [L3]; and every class $[a]_{n_{i}}$ is the image of $[a]_{n}$ .

step 1.2L3

3.1

Let $P : = (Z / n)^{r}$ and $Q : = (Z / n_{1}) \times \dots \times (Z / n_{r})$ , both with componentwise addition. By [L3] each coordinate $(Z / n, +)$ and $(Z / n_{i}, +)$ is an abelian group, so $P$ and $Q$ are abelian groups under these coordinatewise operations. Define $π ⁣ : P \to Q$ by $π ([a_{1}]_{n}, \dots, [a_{r}]_{n}) : = ([a_{1}]_{n_{1}}, \dots, [a_{r}]_{n_{r}})$ . Step 2.1 gives each coordinate map $Z / n \to Z / n_{i}$ as a well-defined surjective homomorphism, so $π$ is a well-defined surjective group homomorphism. Fact [L2] identifies each coordinate group $(Z / n_{i}, +)$ with $C_{n_{i}}$ , and [L1] identifies $C_{n_{1}} \times \dots \times C_{n_{r}}$ with $G$ up to isomorphism. Composing with that isomorphism gives a surjective homomorphism $(Z / n)^{k} \to G$ .

step 1.2step 2.1L1L2L3

4.1

The two cases are exhaustive, the invariant-factor list being empty or not, so such $n$ and $k$ exist for every finite abelian $G$ ; and [L4] turns any such surjection into an isomorphism $G ≅ (Z / n)^{k} / H$ with $H$ its kernel.

step 1.1step 3.1L4cases-exhaustive∎

Remarks

Why the invariant-factor form is convenient. The invariant factors form a divisibility chain, so the single modulus $n = n_{r}$ works immediately. A primary decomposition also proves the statement: take $n$ to be the least common multiple of the finitely many prime-power orders and reduce $Z / n$ onto each cyclic factor. The invariant-factor form simply avoids that extra choice of modulus.

Depends on

Used by

Every finite abelian group is the Galois group of some finite Galois extension of ℚ Theorem

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. L. Clark, Field Theory (course notes/monograph), Lemma 9.11 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 5 (standard reference, not scraped)

Every finite abelian group is a quotient of (Z/n)k(Z/n)k for some nn and kk