Free abelian group

Subgroup of a free abelian group

Suppose $𝐺$ is a free abelian group of rank $𝑛$ and $𝐻 \leq ℤ 𝐺$ . Then: group

$𝐻$ is free of rank $𝑚 \leq 𝑛$ ;
There exists a basis ${𝛼 𝑖} 𝑛 𝑖 = 1$ for $𝐺$ and integers ${𝑐 𝑖} 𝑚 𝑖 = 1$ such that ${𝑐 𝑖 𝛼 𝑖} 𝑚 𝑖 = 1$ forms a basis for $𝐻$ ;
The Lagrange index $| 𝐺 / 𝐻 |$ is finite iff $𝑚 = 𝑛$ .

Moreover, if $𝑚 = 𝑛$ we have a change of basis $𝐴 \in M 𝑛, 𝑛 (ℤ)$ from a basis ${𝛼 𝑖} 𝑛 𝑖 = 1$ of $𝐺$ to ${𝛽 𝑖} 𝑛 𝑖 = 1$ of $𝐻$ such that

⎡ ⎢ ⎢ ⎣ 𝛽 1 ⋮ 𝛽 𝑛 ⎤ ⎥ ⎥ ⎦ = 𝐴 ⎡ ⎢ ⎢ ⎣ 𝛼 1 ⋮ 𝛼 𝑛 ⎤ ⎥ ⎥ ⎦ .

Then $| 𝐺 / 𝐻 | = | d e t 𝐴 |$ .¹

Proof

proof

develop | en | SemBr

Footnotes

2022. Algebraic number theory course notes, ¶A.11, p. 144 ↩

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Backlinks

Absolute norm of an ideal of the ring of integers of a number field
Discriminant of a number field
Free abelian group

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