TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A p-primary component has the full p-power order and is the unique subgroup of that order

Statement

Let $G$ be finite abelian and write $∣ G ∣ = p^{a} m$ with $p ∤ m$ . Then $G (p)$ is a subgroup of order $p^{a}$ . It is the unique subgroup of $G$ having that order. In particular, if $p ∤ ∣ G ∣$ , then $G (p) = {e}$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be an abelian group and $p$ a prime. Its $p$ -primary component is $G (p) = {g \in G : g^{p^{k}} = e for some k \in N} .$ Thus the identity is included by $k = 0$ . In additive notation, $G (p) = {g : p^{k} g = 0 for some k \in N}$ . Powers and element orders use def-group-power and def-order-in-a-group. No finiteness or maximality is part of the definition. (The p-primary component of an abelian group).

[L2]

Let $G$ be a finite abelian group and let $p$ be a prime dividing $∣ G ∣$ . Then $G$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem for finite abelian groups).

[L3]

Let $G$ be a finite group and $H \leq G$ . Then $∣ G ∣ = [G : H] ∣ H ∣ .$ Consequently, under the canonical embedding $ι : N \to Z$ , $∣ H ∣$ divides $∣ G ∣$ . (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L4]

Let $N ⊴ G$ . If $[G : N]$ is finite, then the quotient group $G / N$ is finite and $∣ G / N ∣ = [G : N] .$ In particular, if $G$ is finite, then $∣ G / N ∣ = \frac{∣ G ∣}{∣ N ∣} .$ (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

[L5]

Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved. For $N ⊴ G$ , the maps $H \mapsto H / N$ and $K \mapsto π^{- 1} (K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N \leq H \leq G$ and subgroups $K \leq G / N$ ; they preserve normality. (Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

[L6]

Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ . If $H \leq G$ and $N ⊴ G$ , then $H / (H \cap N) ≅ H N / N .$ (Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ ).

[L7]

If $G$ is abelian and $N ⊴ G$ , then $G / N$ is abelian. (Every quotient group of an abelian group is abelian).

[L8]

Let $G$ be a group (def-group) with identity $e$ , let $g, h \in G$ , and let powers be as in def-group-power. For all $m, n \in Z$ : 1. $g^{m + n} = g^{m} g^{n}$ ; 2. $g^{- m} = (g^{m})^{- 1}$ ; 3. $(g^{m})^{n} = g^{m n}$ ; 4. $g^{m} g^{n} = g^{n} g^{m}$ : any two powers of one element commute; 5. if $g h = h g$ then $(g h)^{n} = g^{n} h^{n}$ . Claim 5 is false in general without its hypothesis: in a group in which $g$ and $h$ do not commute the equation can fail already at $n = 2$ , and a witness is recorded on the companion page. Claims 1 and 3 hold in any monoid (def-semigroup-and-monoid) for exponents in $N$ , and so does claim 5 for exponents in $N$ under the same commuting hypothesis; only the extension to negative exponents needs inverses. (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · direct

1.1

If $x^{p^{r}} = y^{p^{s}} = e$ and $t = \max {r, s}$ , commutativity and the power laws give $(x y)^{p^{t}} = e$ and $(x^{- 1})^{p^{t}} = e$ , so $G (p)$ is a subgroup. If a prime $q$ divides $∣ G (p) ∣$ , Cauchy's theorem in $G (p)$ gives an element of order $q$ ; the definition of $G (p)$ forces $q = p$ . Hence $∣ G (p) ∣ = p^{b}$ for some $b \leq a$ .

givenL1L2L3L4L5L6L7L8

2.1

If $b < a$ , then $p$ divides $∣ G / G (p) ∣$ . Cauchy's theorem in this abelian quotient gives a nonidentity coset $x G (p)$ of order $p$ .

step 1.1

3.1

Then $x^{p} \in G (p)$ , so $(x^{p})^{p^{k}} = e$ for some $k$ and therefore $x \in G (p)$ , contradicting the choice of a nonidentity coset. Hence $b = a$ .

step 2.1

4.1

If $H \leq G$ has order $p^{a}$ , Lagrange applied inside $H$ makes every $h \in H$ have $p$ -power order, so $H \subseteq G (p)$ ; equal finite orders give $H = G (p)$ . The case $a = 0$ gives the trivial subgroup.

step 3.1∎

Depends on

Used by

A finite abelian group is the internal direct product of its primary components Theorem

Dependency tree · two levels

46 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

Keith Conrad, Decomposition of Finite Abelian Groups, §§1-4 (standard reference, not scraped)
Richard Elman, Lectures on Abstract Algebra, Ch. 14 (standard reference, not scraped)