Isomorphism of Quotient Subgroups

Question

I have a problem on the assignment and I've got stuck at one point.

Let $\Bbb{R}$ be the group of real numbers under addition, and let $x\Bbb{Z} ⊂ \Bbb{R}$ be the subgroup which consists of all integral multiples of a given real number $x$, i.e., $x\Bbb{Z} = \{mx | m ∈ Z\}$. Prove that for any non-zero real numbers $x$, $y$, the quotient groups $\Bbb{R}/x\Bbb{Z}$ and $\Bbb{R}/y\Bbb{Z}$ are isomorphic.

First I noticed that subgroups $x\Bbb{Z}$, $y\Bbb{Z}$ form cosets of $\Bbb{R}$ which implies that they are normal subgroups. Then I know that if I can find a isomorphism (bijection) $\phi: x\Bbb{Z} \to y\Bbb{Z}$ then the quotient groups are isomorphic too. The problem is that I cannot find such a bijection.

Answer 1

$\mathbb R / y \mathbb Z$ is a scaled version of $\mathbb R / xZ$, by the scale factor $y/x$. This suggests defining $\phi : \mathbb R / xZ \to \mathbb R / yZ$ by $$\phi(r + xZ) = r(y/x) + yZ$$ and verifying that $\phi$ is an isomorphism.

First check that this is well-defined: if $r + xZ = s + xZ$, then $r - s \in xZ$, so $r(y/x) - s(y/x) \in yZ$, and therefore $r(y/x) + yZ = s(y/x) + yZ$. In other words, $\phi(r + xZ) = \phi(s + xZ)$. So $\phi$ is well-defined.

Now check that $\phi$ is a homomorphism. We have $$\begin{aligned} \phi((r + xZ) + (s + xZ)) &= \phi(r + s + xZ) \\ &= (r + s)(y/x) + yZ \\ &= r(y/x) + yZ + s(y/x) + yZ \\ &= \phi(r + xZ) + \phi(s + xZ) \\ \end{aligned}$$

Now check that $\phi$ is injective. Suppose that $\phi(r + xZ) = 0 + yZ = yZ$. This means that $r(y/x) + yZ = yZ$, so $r(y/x) \in yZ$, which means that $r/x$ is an integer, and therefore $r \in xZ$, so $r + xZ = 0 + xZ$. So $\phi$ is indeed injective.

Surjectivity is clear: the element $s + yZ$ is equal to $\phi(s(x/y) + xZ)$.

Answer 2

How about $\phi : x\mathbb{Z} \to y\mathbb{Z}$ given by $\phi(mx) = my$? I presume neither $x$ nor $y$ are zero.

Answer 3

If $x$ and $y$ are both non-zero, then there is a non-zero $r \in \Bbb{R}$ such that $y=rx$. Define the map $\varphi: \Bbb{R}/x\Bbb{Z}\rightarrow \Bbb{R}/y\Bbb{Z}$ for each $a \in \mathbb{R}$ by $\varphi(a+x\Bbb{Z})=ar+y\Bbb{Z}$. If $a+x\Bbb{Z}=a'+x\Bbb{Z}$ in $\mathbb{R}/x\mathbb{Z}$, then $a-a' = nx$ for some $n \in \Bbb{Z}$. Consequently $ar-a'r=(a-a')r= nxr=ny$. Thus, this is a well-defined map.

$\varphi$ is surjective, for given $b+y\Bbb{Z}$, $\varphi(\frac{b}{r}+x\mathbb{Z}) = \frac{b}{r}r+y\mathbb{Z} = b+\Bbb{Z}$. If $\varphi(a+x\Bbb{Z})=\varphi(a'+x\Bbb{Z})$, then $ar-a'r=my$ for some $m \in \mathbb{Z}$. Then, $(a-a')r=my=mrx$ and since $r\not=0$, we must have $a-a'=mx$, hence $\varphi$ is also injective and consequently a bijection.

Now, $\varphi((a+x\mathbb{Z})+(a'+\Bbb{Z}))=\varphi((a+a')+x\Bbb{Z}=(a+a')r+y\Bbb{Z} = (ar+y\Bbb{Z}) + (a'r+\Bbb{Z}) = \varphi(a+x\Bbb{Z})+\varphi(a'+x\Bbb{Z}).$