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Let $G$ be free abelian of rank $n$ and $H \subseteq G$ a subgroup also of rank $n$. It is known that $G/H$ is finite, in fact a direct sum of at most $n$ cyclic groups. Thus we can write $$G/H = \langle x_1,\ldots,x_n | d_1x_1 = \cdots =d_nx_n = 0\rangle,$$ where the $d_i$'s are the orders of those cyclic groups. From here why does it follow that there is a basis $\beta_1,\ldots,\beta_n$ of $G$ such that $d_1\beta_1,\ldots ,d_n\beta_n$ is a basis for $H$?


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Suppose that $G/H\simeq \mathbb Z_{d_1}\oplus\cdots\oplus\mathbb Z_{d_n}$. (Note that some of $d_i$'s can be equal to $1$.)

Set $D=\operatorname{diag}(d_1,\dots,d_n)$. This matrix is equivalent to a matrix $C=\operatorname{diag}(c_1,\dots,c_n)$ (called the Smith Normal Form of $D$) with $c_1\mid\cdots\mid c_n$, that is, there are $U,V\in\mathrm{GL}_n(\mathbb Z)$ such that $UDV=C$. One knows that there exists a basis $g_1,\dots,g_n$ in $G$ such that $c_1g_1,\dots,c_ng_n$ is a basis of $H$. Note that $(c_1g_1,\dots,c_ng_n)^T=C(g_1,\dots,g_n)^T$. Since $C=UDV$ we get that $UDV(g_1,\dots,g_n)^T$ is a basis of $H$. But $U$ is invertible, so $DV(g_1,\dots,g_n)^T$ is also a basis of $H$. On the other side, since $V$ is invertible it follows that $(x_1,\dots,x_n)^T:=V(g_1,\dots,g_n)^T$ is a basis of $G$. Finally we get that $D(x_1,\dots,x_n)^T$ is a basis of $H$, i.e. $d_1x_1,\dots,d_nx_n$ is a basis of $H$.

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By the fundamental theorem of finitely generated abelian groups, a priori don't we have $d_1|d_2|\ldots | d_n$? – user38268 Jan 5 '13 at 10:39
@BenjaLim The fundamental theorem is one thing, and an arbitrary decomposition as a direct sum of cyclic groups is other thing. – user26857 Jan 5 '13 at 10:41
when you say "One knows that there exists a basis $g_1,\ldots$ ... is a basis for $H$", that is basically following the argument in my comment to rschwieb's answer below yes? – user38268 Jan 5 '13 at 10:42
Of course it is! This argument is the key for proving the fundamental theorem and can be found in many textbooks. – user26857 Jan 5 '13 at 11:01

Hint: Since $H$ and $G$ are free $\Bbb Z$ modules of the same rank, there is an isomorphism between them which you can express in whatever basis you like. The statement you have supplied "there exists a basis... such that..." amounts to changing bases so that the transformation matrix becomes diagonal.

With your knowledge of how the invariant numbers work, I think this is what you need for your question.

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Suppose that $x_1,\ldots,x_n$ is a basis for $G$ and $y_1,\ldots,y_n$ is a basis for $H$. Then we can say that $y_i = \sum a_{ij} x_j$. Now let $A = (a_{ij})$. Then the Smith Normal Form shows that there are invertible matrices $B,C$ so that $BAC = D$ where $D$ is diagonal. Now let $y = (y_1,\ldots,y_n)^T$, $x = (x_1,\ldots,x_n)^T$. Then we have $$\begin{eqnarray*} BAx &=& DC^{-1}x \\ &=& By. \end{eqnarray*}$$ Now because $B$ is invertible, $By$ will also be a basis for $H$. $C^{-1}x$ is also a basis for $G$ and hence the problem is proven. However, how do we know that the elements on – user38268 Jan 3 '13 at 15:52
the diagonal of $D$ are precisely $d_1,\ldots,d_n$? – user38268 Jan 3 '13 at 15:53
@BenjaLim My memory was telling me this was incorporated in the proof about SNF, and that the connection to the invariants (which I'm assuming is what the $d_i$ stand for) was laid out there. – rschwieb Jan 3 '13 at 15:59
The $d_i$ in the OP's question above are the orders of those cyclic groups. – user38268 Jan 3 '13 at 16:01
@BenjaLim I think the issue here is that the OP was not specific enough about how the $d_i$ were chosen and what constraints are on them. It seems that the invariant factors were intended. By quoting the OP at me again, for some reason, it seems you might find this unreasonable? Without those constraints, I definitely agree that the problem is not clear. – rschwieb Jan 3 '13 at 16:12

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