# is A/B is cyclic?

Let $A$ be the additive group $\mathbb Z\oplus \mathbb Z$ and $B$ be the subgroup $\{(5m+7n, 2m+4n)| m,n\in \mathbb Z\}$ is $A/B$ is cyclic?

Can we write that subgroup $B$ has the form $\mathbb Z\oplus \mathbb 2Z$ since $\gcd(5,7)=1$ and $\gcd(2,4)=2$? Is it true can we write B as this form?

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Is it really true that gcd(5,7) = 35? –  Ben Feb 25 '12 at 2:46
Are you sure $\gcd(5,7)=35$? Perhaps you mean $\lcm(5,7)=35$, in which case you should say $\lcm(2,4)=4$? –  Alex Becker Feb 25 '12 at 2:46
totally sorry corrected it –  the code Feb 25 '12 at 2:49
Notice by taking $(m,n) = (4,-2)$ that $(6,0) \in A$, and taking $(m,n) = (-7,5)$ shows $(0,6) \in A$. So $6\mathbb{Z}^2 \subset A$ and $\mathbb{Z}^2 / A$ has at most $6^2$ elements. –  Joel Cohen Feb 25 '12 at 2:55

$B$ is actually isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$. The general solution to the system

$$\begin{cases} 5m+7n=a\\ m+2n=b \end{cases}$$

is

$$\begin{cases}m=\frac13(2a-7b)\\\\ n=\frac13(5b-a)\;. \end{cases}$$

Thus, $\langle a,2b\rangle\in B$ if and only if $3\mid 2a-7b$ and $3\mid 5b-a$, i.e., if and only if $a\equiv 5b\pmod 3$ and $2a\equiv 7b\pmod 3$. These reduce to $a\equiv 2b\pmod 3$ and $b\equiv 2a\pmod 3$, respectively, which are equivalent to each other and to $a+b\equiv 0\pmod 3$. In short,

\begin{align*}B&=\{\langle a,2b\rangle\in\mathbb{Z}\oplus\mathbb{Z}:3\mid a+b\}\\ &=\{\langle a,b\rangle\in\mathbb{Z}\oplus\mathbb{Z}:a\equiv b\!\!\!\!\pmod 3\text{ and }2\mid b\}\;. \end{align*}

Thus, $\langle a,b\rangle\in B$ iff there are $m,n\in\mathbb{Z}$ such that $b=2n$ and $a=3m+2n$, and conversely, any $m,n\in\mathbb{Z}$ uniquely determine a pair $\langle 3m+2n,2n\rangle\in B$. Thus, $$B=\{m\langle 3,0\rangle+m\langle 2,2\rangle:m,n\in\mathbb{Z}\}\;,$$

and the map

$$h:\mathbb{Z}\oplus\mathbb{Z}\to B:\langle m,n\rangle\mapsto\langle 3m+2n,2n\rangle$$

is an isomorphism.

Now let $\langle m,n\rangle\in A$; then exactly one of the following pairs belongs to $B$:

$$\begin{cases}\langle m,n\rangle,\langle m,n-2\rangle,\text{ or }\langle m,n-4\rangle,&\text{if }n\text{ is even}\\ \langle m,n-1\rangle,\langle m,n-3\rangle,\text{ or }\langle m,n-5\rangle,&\text{if }n\text{ is odd}\;. \end{cases}$$

Equivalently, every element of $A$ belongs to $\langle 0,k\rangle+B$ for some $k\in\{0,1,2,3,4,5\}$, and from that you should easily be able to answer the question in the title.

It might help to look at a graphical display of part of $A$; members of $B$ are colored red or blue, the two blue elements being the generators for $h$.

$$\begin{array}{c} \langle -3,4\rangle&\color{red}{\langle -2,4\rangle}&\langle -1,4\rangle&\langle 0,4\rangle&\color{red}{\langle 1,4\rangle}&\langle 2,4\rangle&\langle 3,4\rangle&\color{red}{\langle 4,4\rangle}&\langle 5,4\rangle\\ \langle -3,3\rangle&\langle -2,3\rangle&\langle -1,3\rangle&\langle 0,3\rangle&\langle 1,3\rangle&\langle 2,3\rangle&\langle 3,3\rangle&\langle 4,3\rangle&\langle 5,3\rangle\\ \langle -3,2\rangle&\langle -2,2\rangle&\color{red}{\langle -1,2\rangle}&\langle 0,2\rangle&\langle 1,2\rangle&\color{blue}{\langle 2,2\rangle}&\langle 3,2\rangle&\langle 4,2\rangle&\color{red}{\langle 5,2\rangle}\\ \langle -3,1\rangle&\langle -2,1\rangle&\langle -1,1\rangle&\langle 0,1\rangle&\langle 1,1\rangle&\langle 2,1\rangle&\langle 3,1\rangle&\langle 4,1\rangle&\langle 5,1\rangle\\ \color{red}{\langle -3,0\rangle}&\langle -2,0\rangle&\langle -1,0\rangle&\color{red}{\langle 0,0\rangle}&\langle 1,0\rangle&\langle 2,0\rangle&\color{blue}{\langle 3,0\rangle}&\langle 4,0\rangle&\langle 5,0\rangle\\ \langle -3,-1\rangle&\langle -2,-1\rangle&\langle -1,-1\rangle&\langle 0,-1\rangle&\langle 1,-1\rangle&\langle 2,-1\rangle&\langle 3,-1\rangle&\langle 4,-1\rangle&\langle 5,-1\rangle\\ \langle -3,-2\rangle&\color{red}{\langle -2,-2\rangle}&\langle -1,-2\rangle&\langle 0,-2\rangle&\color{red}{\langle 1,-2\rangle}&\langle 2,-2\rangle&\langle 3,-2\rangle&\color{red}{\langle 4,-2\rangle}&\langle 5,-2\rangle\\ \langle -3,-3\rangle&\langle -2,-3\rangle&\langle -1,-3\rangle&\langle 0,-3\rangle&\langle 1,-3\rangle&\langle 2,-3\rangle&\langle 3,-3\rangle&\langle 4,-3\rangle&\langle 5,-3\rangle\\ \langle -3,-4\rangle&\langle -2,-4\rangle&\color{red}{\langle -1,-4\rangle}&\langle 0,-4\rangle&\langle 1,-4\rangle&\color{red}{\langle 2,-4\rangle}&\langle 3,-4\rangle&\langle 4,-4\rangle&\color{red}{\langle 5,-4\rangle}\\ \langle -3,-5\rangle&\langle -2,-5\rangle&\langle -1,-5\rangle&\langle 0,-5\rangle&\langle 1,-5\rangle&\langle 2,-5\rangle&\langle 3,-5\rangle&\langle 4,-5\rangle&\langle 5,-5\rangle\\ \color{red}{\langle -3,-6\rangle}&\langle -2,-6\rangle&\langle -1,-6\rangle&\color{red}{\langle 0,-6\rangle}&\langle 1,-6\rangle&\langle 2,-6\rangle&\color{red}{\langle 3,-6\rangle}&\langle 4,-6\rangle&\langle 5,-6\rangle\\ \end{array}$$

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$$\det\pmatrix{5&7\cr2&4\cr}=6$$ so $A/B$ has order 6. It's abelian, since $A$ is abelian, so what do you know about abelian groups of order 6?

$B$ is an infinite group, so $B$ is certainly not ${\bf Z}_1\oplus{\bf Z}_2$. As an exercise, prove $A/B$ has order 6 directly, by finding representatives of 6 different cosets of $B$ in $A$.

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Since $B$ is a noncyclic subgroup of a free abelian group of rank $2$, it is free abelian of rank $2$; that is, $B\cong\mathbb{Z}\oplus\mathbb{Z}$.

Perhaps you are asking: does there exist a basis $\mathbf{a}=(a_1,a_2)$, $\mathbf{b}=(b_1,b_2)$ of $A$ such that $B=\langle \mathbf{a},2\mathbf{b}\rangle$?

The answer is "no"; because $A/B$ has order $6$, but any subgroup such as the one you describe would have index $2$.

(Of course, having order $6$ tells you that the quotient will be cyclic, but I suspect that cyclicity was not really the heart of your question).

You are looking at the subgroup of $\mathbb{Z}\oplus\mathbb{Z}$ generated by $(5,2)$ and $(7.4)$; however, you cannot simply look at the gcds of the coordinates separately: for instance, while we can indeed express $1$ as a linear combination of $5$ and $7$ (so that we can find some element in $B$ of the form $(1,y)$), we don't know ahead of time whether we can get $(1,0)$ just from the fact that the projection onto the first coordinate is everything; and while we can obtain elements of $B$ of the form $(x,2)$, again we do not know whether we can get $x=0$, just form the fact that the projection onto the second coordinate is $2\mathbb{Z}$. In fact, we cannot: if $m(5,2) + n(7,4) = (0,2)$, then we would have that $m=7k$, $n=-5k$, and then the second coordinate would be $14k - 20k = -6k$, a multiple of $6$, hence it cannot be equal to $2$.

You are, in fact, looking at the wrong thing. You can't handle the coordinates independently, as you attempted to do. What you want to do is look at the Smith normal form. I'll do the necessary transformations here in an ad hoc manner, but the Smith normal form is the algorithmic way to get these results.

We can obtain $(2,2) = (7,4)-(5,2)$. Note that $(2,2)$ and $(5,2)$ generate $B$. Then we can replace $(5,2)$ with $(5,2)-(2,2) = (3,0)$; so $B$ is generated by $(3,0)$ and $(2,2)$.

From $(3,0)$ and $(2,2)$, we can replace $(3,0)$ with $(3,0)-(2,2)=(1,-2)$. So $B$ is generated by $(1,-2)$ and $(2,2)$. Then we can replace $(2,2)$ with $(2,2)-2(1,-2) = (0,6)$; so $B$ is generated by $(1,-2)$ and $(0,6)$. That gives the Smith normal form: take $\mathbf{a}=(1,-2)$ and $\mathbf{b}=(0,1)$ as the basis for $A$, and then the basis for $B$ is $\mathbf{a}$ and $6\mathbf{b}$.

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