Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
2  
@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. –  B. S. Jun 22 '11 at 18:30
    
@Arturo: Thank you for your help. –  stacy Jun 22 '11 at 19:07

1 Answer 1

up vote 6 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.

share|improve this answer
    
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 Bezout...in 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.