# Structure theorem for finitely generated Abelian groups: proof

In this question about the structure theorem for finitely generated Abelian groups, a nice proof was given in an answer by Arturo Magidin. However, as noted by Lorban in a comment, a crucial step lacking motivation seems to be made in the proof. I could not fill in the gap. I'll give a minimal context here:

If $G$ is a free Abelian group of rank $n$ and $H$ is a subgroup, define the set $S \subset \mathbb{Z}$ by the following rule: $d \in S$ when there exists a basis $\{ b_1, b_2, \dots, b_n \}$ of $G$ and integers $k_2, k_3, \dots, k_n$ such that $d b_1 + k_2 b_2 + k_3 b_3 + \dots + k_n b_n \in H$.

Prove that $S$ is closed under summation.

I do not see any natural way to prove this result (in particular, any way to directly prove that $S$ is closed under addition rather than first proving that $S$ must be the set of multiples of some particular integer). However, note that Arturo Magidin's answer does not require the full strength of this result. Instead of asserting that $S$ is closed under summation and taking $d_1$ to be its positive generator, you can just define $d_1$ to be the least positive element of $S$.

The argument then uses the assumption that $d_1$ generates $S$ as a group in three places. In each place, you can give a direct argument that $d_1$ divides the particular elements of $S$ in question.

First, you need to know that if $b_1,\dots,b_n$ is a basis and $h_1 = d_1b_1+k_2b_2+\cdots+k_nb_n\in H$, then $d_1$ divides each $k_i$. To prove this, suppose without loss of generality that $d_1$ does not divide $k_2$, and write $k_2=qd_1+r$ where $0<r<d_1$. Let $b_1'=b_1+qb_2$. Then $b_2,b_1',b_3,\dots,b_n$ is also a basis, and $h_1=rb_2+d_1b_1'+k_3b_3+\dots+k_nb_n$. Thus $r\in S$. But since $0<r<d_1$, this contradicts the minimality of $d_1$.

Second, you need to know that if $a_1,b_2,\dots,b_n$ is a basis, $h_1=d_1a_1\in H$, and $h=j_1a_1+j_2b_2+\cdots+j_nb_n\in H$, then $d_1$ divides $j_1$. To prove this, suppose $d_1$ does not divide $j_1$ and write $j_1=qd_1+r$ for $0<r<d_1$. Then note that $h-qh_1=ra_1+j_2b_2+\dots+j_nb_n$ is an element of $H$, so $r\in S$. This contradicts the minimality of $d_1$.

Finally, you need to know that if $a_1,a_2,\dots,a_n$ is a basis such that $h_1=d_1a_1$ and $h_2=d_2a_2$ are both in $H$, then $d_1$ divides $d_2$. This is again a similar argument: if not, write $d_2=qd_1+r$ with $0<r<d_1$ and let $a_1'=a_1+qa_2$. Then $a_2,a_1',a_3,\dots,a_n$ is a basis, and $h_2-qh_1=ra_2-qda_1'\in H$. Thus $r\in S$, contradicting minimality of $d_1$.

(Once you have then used this to patch up Arturo Magidin's answer, the fact that $S$ is a subgroup follows from what he has proven, since his argument shows that actually every element of $H$ is divisible by $d_1$. It follows that every element of $S$ is divisible by $d_1$, and conversely it is clear that every multiple of $d_1$ is in $S$, so $S=(d_1)$ and in particular $S$ is a subgroup of $\mathbb{Z}$.)

• Thank you, this is perfect. – Latrace Mar 3 '17 at 22:07