# Generating Sets for Subgroups of $(\Bbb Z^n,+)$.

The question Finite Generated Abelian Torsion Free Group is a Free Abelian Group led me to conjecture and prove an interesting thing about generating sets for $\Bbb Z^n$ and certain subgroups.

If $x=(x_1,\dots,x_n)\in\Bbb Z^n$ we will write $$\gcd(x)=\gcd(x_1,\dots,x_n).$$

Exercise. If $v_1,\dots,v_n$ generate $\Bbb Z^n$ then $\gcd(v_1)=\dots=\gcd(v_n)=1$.

Conversely,

Theorem. Suppose $m\in\Bbb Z^n$ and $\gcd(m)=1$. There exists a set $v_1,\dots,v_n$ of generators for $\Bbb Z^n$ with $v_1=m$.

Actually a day ago I was just conjecturing that, and was going to ask how to prove it. But I have a nice little proof, so the best I can do to twist this post into the form of a question is this:

Question. Whose theorem is that? Or does it have a name or what? I suppose I might also ask whether the proof below is more or less standard.

The theorem is a special case of a result about certain subgroups:

Definition. The group $G\subset\Bbb Z^n$ is a $\Bbb Q$-group if there exists a linear subspace $V\subset\Bbb Q^n$ with $G=V\cap\Bbb Z^n$. In this case we will write $\dim(G)=\dim(V)$.

At least I'm going to call them $\Bbb Q$-groups until someone tells me what they're really called. Yes, I gather the dimension of a $\Bbb Q$-group is the rank of same; but we don't know yet that it's a free abelian group. (An easy exercise shows that the dimension of a $\Bbb Q$-group is well-defined.)

Note that if $G$ is a $\Bbb Q$-group and $G\ne\{0\}$ then there exists $v\in G$ with $\gcd(v)=1$.

If $x,y\in\Bbb Z^n$ we will write $$x\cdot y=\sum_{j=1}^n x_jy_j.$$

Lemma. Say $G\subset\Bbb Z^n$ is a $\Bbb Q$-group and $\dim(G)=k\ge1$. Let $v\in G$ be such that $\gcd(v)=1$. Then $$G=\langle v\rangle\oplus H,$$where $H$ is a $\Bbb Q$-group with $\dim(H)=k-1$.

Proof. There exists $\mu\in\Bbb Z^n$ with $$\mu\cdot v=1.$$If $x\in G$ then $$x=\left((x\cdot\mu)v\right)+\left(x-(x\cdot\mu)v\right),$$and$$\left(x-(x\cdot\mu)v\right)\cdot\mu=0.$$If $V$ is the subspace of $\Bbb Q^n$ with $G=V\cap\Bbb Z^n$ let $$W=\{y\in V\,:\,\mu\cdot y=0\}.$$Since $v\in V\setminus W$ it follows that $\dim(W)=k-1$. Let $$H=W\cap\Bbb Z^n=\{y\in G\,:\,\mu\cdot y=0\}.$$We've shown that $G=\langle v\rangle+H$, and it's clear that $\langle v\rangle\cap H=\{0\}$, since $\mu\cdot v\ne0$.

And now by induction

Corollary. Suppose $G$ is a $\Bbb Q$-group, $\dim(G)=k>0$. If $m\in G$ and $\gcd(m)=1$ then there exists a set of generators $v_1,\dots,v_k$ for $G$ with $v_1=m$.

In particular $G$ is a free abelian group of rank $k$ (it's clear that $v_1,\dots,v_k$ are $\Bbb Q$-linearly independent).

• math.stackexchange.com/questions/1360805/… – user26857 Aug 17 '15 at 22:57
• How to show if $v_1,\dots,v_n$ generate $\mathbf{Z}^n$, then $\gcd(v_1)=\cdots=\gcd(v_n)=1$? Could you give some hints? – Xiang Yu Nov 20 '15 at 16:40
• First, $v_1,\dots,v_n$ must be linearly independent in $\Bbb Q^n$. Say $gcd(v_1)=2$. Then $v_1=2w_1$ for some $w_1\in\Bbb Z^n$. Now there exist integers $k_j$ such that $w_1=2k_1w_1+k_2v_2+\dots+k_nv_n$. Since $1\ne2k_1$ this contradicts the linear independence of $w_1,v_2,\dots,v_n$. – David C. Ullrich Nov 20 '15 at 20:08