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I find it hard to understand a part of a proof on torsion free abelian groups of rank 1. Let $A$ and $B$ be torsion free groups of rank 1 and of the same type. Let $a'$ and $b'$ arbitrary non zero elements from $A$ and $B$, respectively. Then the height of $a'$ and $b'$ $H(a')=(k_1,...,k_n,...)$ and $H(b')=(k'_1,...,k'_n,...)$ are equivalent. Let $n_1,...,n_s$ be those indices $n$ for which $k_n$ is different from $k'_n$ . Since $k_{n_i}$ and $k'_{n_i}$ are non negative integers, we can solve the equations $p_{n_1}^{k_{n_1}}... p_{n_s}^{k_{n_s}}x=a'$ and $p_{n_1}^{k'_{n_1}}... p_{n_s}^{k'_{n_s}}y=b'$ and denote their solutions by $a\in A$ and $b\in B$, respectively. Clearly $H(a)=H(b)$ since they are obtained from $H(a')$ and $H(b')$ by replacing the $k_n$ $(k'_n)$ of index $n_1,n_2,...,n_s$ by $0$. It follows that an equation $mx=ta$ $(t,m$ non zero integers) is solvable in $A$ if and only if the equation $my=tb$ admits a solution in $B$. Why is it true? I think it is related to the fact that in a torsion free abelian group of rank 1 two non zero elements are dependent but I think it is not enough. And why in torsion free groups may such equations have at most one solution? Someone can help me? Thanks

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Your nomenclature is a bit of a mess. Is $(k_n)_i$ supposed to be $k_{n_i}$? What is $((p_n)_1)$? Why the up-raised parentheses, ${}^{(}(k_n)_1$? – Arturo Magidin Jun 22 '11 at 18:03
yes the first you said is right and the second is the same of the first. In The third case they are powers of $(p_n)_i$ but I don't know how to write double index and I don't know the reason why the power doesn't work. I wrote all between $ .Can you help me? – stacy Jun 22 '11 at 18:18
@stacy: To write a double index, you must put it in curly brackets: k_{n_i} produces $k_{n_i}$. For exponents, likewise, use curly brackets, not parentheses. i.e., p_{n_i}^{k_{n_i}} to get $p_{n_i}^{k_{n_i}}$. It's curly brackets that are used for (invisible) grouping in $\LaTeX$, not parentheses. – Arturo Magidin Jun 22 '11 at 18:28
@stacy: Examine two subgroups of $\mathbb Q$ as $A$ and $B$. I think this is a theorem of Baer. – Babak S. Jun 22 '11 at 18:30
@Arturo: Thank you for your help. – stacy Jun 22 '11 at 19:07
up vote 7 down vote accepted

This is probably overkill, but anyway:

First, note that if $G$ is abelian and torsion free, then an equation of the form $nx = a$ with $n\gt 0$ has at most one solution. For if $g$ and $h$ are both solutions, then $ng = a = nh$, hence $0 = a-a = ng-nh = n(g-h)$, and since $n\gt 0$, that means that $g-h$ is a torsion element. But since $G$ is torsion free, this means $g-h = 0$, hence $g=h$. Thus, the equation has at most one solution.

Now, for the rest and to give context... To recall the definitions:

Theorem. If $G$ is a torsion free abelian group, then any two maximal independent sets in $G$ have the same cardinality. If a maximal independent subset of $G$ has $r$ elements, then $G$ is an additive subgroup of an $r$-dimensional vector space over $\mathbb{Q}$.

Proof. We can embedd $G$ into a divisible group; then moding out by the torsion of the divisible group, we obtain an embedding of $G$ into a torsion-free divisible group $D$, which has a natural structure of a $\mathbb{Q}$-vector space (since $nx=y$ has a unique solution for any $n\gt 0$). Now let $X$ be a maximal independent subset of $G$. Then the subspace generated by $X$, viewing $X$ as a subset of the $\mathbb{Q}$-vector space $D$, contains $G$ (since for any $g\in G$ there exists $n\in \mathbb{Z}$, $n\neq 0$, such that $ng$ is in the subgroup generated by $X$); so $G$ can be embedded in a vector space of dimension $|X|$). But that means that $X$ is a basis for the subspace generated by $G$, so the cardinality of $X$ is completely determined by $G$. QED

The rank of a torsion-free abelian group $G$ is the number of elements in a maximal independent subset.

If $G$ is a torsion-free abelian group, $x\in G$, and $p$ is a prime, then the $p$-height of $x$, $h_p(x)$, is the highest power of $p$ "dividing $x$ in $G$"; that is, the largest $n$ such that $p^ng = x$ has a solution in $G$; if the equation is solvable for all $n$, then $h_p(x) = \infty$.

If $x\neq 0$ is an element of a torsion-free group $G$, then the height sequence of $x$ is the sequence $$h(x) = (h_2(x), h_3(x), h_5(x),\ldots,h_p(x),\ldots).$$

A characteristic is a sequence of nonnegative integers and the symbol $\infty$. Two characteristics are equivalent if and only if they have $\infty$ in the same coordinates and they differ in at most a finite number of other coordinates. An equivalence class of characteristics is called a type.

Lemma. If $G$ is a torsion free group of rank $1$, and $x,y\in G$ are both nonzero, then their height sequences are equivalent.

Proof. If $y=nx$, with $n=p_1^{e_1}\cdots p_t^{e_t}$, then trivially, $h_p(y)\geq h_p(x)$ in all cases, with equality for certain if $h_p(x)=\infty$ for all $p$.

For $p\neq p_i$, $i=1,\ldots,t$, suppose that $p^kg = y = nx$. Then $\gcd(p,n)=1$, so there exists $r$ and $s$ such that $p^kr + sn =1$. Hence $$x = (p^kr + sn)x = p^k(rx) + s(nx) = p^k(rx) + sy = p^k(rx)+s(p^kg) = p^k(rx+sg),$$ so $h_p(x)\geq h_p(y)$.

Finally, $h_{p_i}(y) = e_i + h_{p_i}(x)$ (essentially the same argument), with the convention that $e_i+\infty=\infty$. Thus, the two sequences are equivalent.

In the general case, there are nonzero integers $m$ and $n$ such that $my=nx$. From the previous part, we know the height sequences for $y$ and $my$ are equivalent; the height sequences for $x$ and $nx$ are equivalent; and therefore the height sequences for $y$ and $x$ are equivalent. QED

In light of this, if $G$ is a torsion free abelian group of rank $1$, then we can define the type of $G$ to be the equivalence class of the height sequence of any nonzero element of $G$.

Theorem. Let $G$ and $G'$ be torsion-free abelian groups of rank $1$. Then $G\cong G'$ if and only if the type of $G$ is equal to the type of $G'$.

Proof. If $\phi\colon G\to G'$ is any homomorphism, and $x\in G$ is not in the kernel, then $h_p(x)\leq h_p(\phi(x))$ for all primes $p$; if $\phi$ is invertible, then we get the reverse inequality by applying $\phi^{-1}$, so $G\cong G'$ implies the types are equal.

Conversely, if $x\in G$ and $x'\in G'$, then $x$ and $x'$ have equivalent height sequences. Let $P$ be the finite (and possibly empty) set of primes for which $h_p(x)\lt h_p(x')$, let $Q$ be the finite (and possibly empty) set of primes for which $h_p(x)\gt h_p(x')$. Let $$m = \prod_{p\in P} p^{h_p(x')-h_p(x)},\qquad n=\prod_{p\in Q} p^{h_p(x)-h_p(x')}.$$ Then $mx$ and $nx'$ have the same height sequence (using calculations as we did above). Thus, we can find elements of $G$ and $G'$ that have identical height sequences.

Since both $G$ and $G'$ are torsion free of rank $1$, we can view them as subgroups of $\mathbb{Q}$, and we have elements $y\in G$ and $y'\in G'$ that have the same height sequence, both nonzero. Write $y = \frac{a}{b}$ and $y'=\frac{a'}{b'}$ (viewing them as rational numbers). Then $\frac{b}{a}G$ is isomorphic to $\frac{b'}{a'}G'$, both contain $1$, and $1$ has the same height sequence in both.

It now follows that $\frac{b}{a}G = \frac{b'}{a'}G'$: for any $p$ for which $h_p(1)=\infty$, the groups contain the corresponding Prüfer group; if $h_p(1)=e_p\lt\infty$, then it contains $\frac{1}{p^{e_p}}$ and no reciprocal of any higher power of $p$; and that is it. QED

Since the height sequence of $ta$ is completely determined by the height sequence of $a$ and the prime factorization of $t$, your question comes down to proving the following:

If $G$ is a torsion free abelian group of rank $1$, $a$ and $b$ are two nonzero elements, and the height sequences of $a$ and $b$ are identical, then $nx = a$ has a solution if and only if $nx=b$ has a solution.

This follows from:

Proposition Let $G$ be a torsion free abelian group of rank $1$, let $a\in G$ be a nonzero element, let $h(a) = (h_2(a),h_3(a),\ldots,)$ be the height sequence for $a$, and let $n= p_1^{e_1}\cdots p_k^{e_k}$ be a positive integer. Then $nx=a$ has a solution if and only if $e_i\leq h_{p_i}(a)$ for each $i$.

(I'll let you prove this one on your own; one possibility for the "if" part is to do induction on $k$.)

The highlighted question now has an immediate answer, since whether or not $nx=y$ has a solution depends only on the height sequence of $y$ and the prime factorization of $n$, and since $a$ and $b$ have identical height sequences it follows that $nx=a$ and $nx=b$ either both have solutions, or neither has solutions; and if it has solutions, it has at most one since the group is torsion free.

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Thank you so much for your help and for in depth explanation:) very useful! As regards the last proof I tried and I think it works with induction and my book it was only stated without proof. So in conclusion the only important information about the two equation are $n$ , $h(a)$ and $h(b)$. In this case are equal, hence the theory about their solutions works:) It 'was very important to understand the issue of factorization. thanks! – stacy Jun 23 '11 at 8:57

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