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I have the following question that I am stumped on:

Let $a,b\in\mathbb{R}$. Show that there exists a group isomorphism such that $\mathbb{R}/a\mathbb{Z}\simeq\mathbb{R}/b\mathbb{Z}$, where $\mathbb{R},a\mathbb{Z}$, and $b\mathbb{Z}$ are considered as groups under addition.

How would one prove this? Do I need to use the first isomorphism theorem and construct a mapping that way? Or is that completely off the mark? Any help is appreciated!

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As others have noted above, you need to assume further that $a,b \neq 0$.

Let's define a map $\phi: \Bbb R \to \Bbb R/(b\Bbb Z)$, by:

$\phi(x) = \dfrac{b}{a}(x) + b\Bbb Z$.

It should be clear this is an additive homomorphsim, since:

$\phi(x + y) = \dfrac{b}{a}(x + y) + b\Bbb Z = \left(\dfrac{b}{a}x + \dfrac{b}{a}y\right) + b\Bbb Z$

$= \left(\dfrac{b}{a}x + b\Bbb Z\right) + \left(\dfrac{b}{a} + b\Bbb Z\right) = \phi(x) + \phi(y)$.

Note if $k \in \Bbb Z$, that $\phi(ak) = \dfrac{b}{a}(ak) + b\Bbb Z = bk + b\Bbb Z$,

and since $bk \in b\Bbb Z$, then $bk + b\Bbb Z = b\Bbb Z = 0_{\Bbb R/(b\Bbb Z)}$.

Thus $a\Bbb Z \subseteq \text{ker }\phi$.

On the other hand, if $x \in \text{ker }\phi$, we have:

$\dfrac{b}{a}x + b\Bbb Z = b\Bbb Z$, that is: $\dfrac{b}{a}x \in b\Bbb Z$, so:

$\dfrac{b}{a}x = bk$, for some $k \in \Bbb Z$, and since $b \neq 0$, by cancellation we have:

$\dfrac{x}{a} = k \in \Bbb Z$. Thus $x = ak \in a\Bbb Z$, so $\text{ker }\phi \subseteq a\Bbb Z$, and the two sets are equal.

Finally, by the First Isomorphism Theorem, we have $\Bbb R/(b\Bbb Z) \cong \Bbb R/(\text{ker }\phi) = \Bbb R/(a\Bbb Z)$.

People who have been doing this for years pretty much see it as "obvious" that since the map:

$x \mapsto \dfrac{b}{a}x + b\Bbb Z$

annihilates $a\Bbb Z$, it descends to a well-defined homomorphism $\overline{\phi}$ on the quotient $\Bbb R/(a\Bbb Z)$.

Proving $\overline{\phi}$ is injective is the "hard" part, which is why the previous two answers start instead with the isomorphism $f: \Bbb R \to \Bbb R$ given by:

$x \mapsto \dfrac{b}{a}(x)$ which is more clearly bijective from the get-go.

If $\pi_{b\Bbb Z}: \Bbb R \to \Bbb R/(b\Bbb Z)$ is the canonical quotient homomorphism, then:

$\phi = \pi_{b\Bbb Z} \circ f$, which hopefully sheds some light on the relationship between the two approaches.

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It is not true if $a=0, b\neq 0$. If $a,b\neq 0$ let $f:R\rightarrow R$ defined by $f(x)={a\over b}x, f(x+b)=f(x)+a$. This $f$ induces an isomorphism $\bar f:R/bZ\rightarrow R/aZ$.

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  • $\begingroup$ I'm still a bit confused. Does this use the first isomorphism theorem? Or am I missing something obvious? $\endgroup$ – Sir_Math_Cat Feb 19 '17 at 7:02
  • $\begingroup$ If $\bar x\in R/bZ, x,y $ above $\bar x$, $x-y=nb$, $n\in Z$ $f(x)=f(y)+na$, so $f(x)$ and $f(y)$ defines the same element in $R/aZ$ so $\bar f$ is well-defined. $\endgroup$ – Tsemo Aristide Feb 19 '17 at 7:06
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The result won't be true unless you suppose $a,b \ne 0$. In that case, note that $f \colon x \mapsto (b/a)x$ is an automorphism of $\mathbf{R}$ and $f(a\mathbf{Z}) = b\mathbf{Z}$. So the isomorphism holds by transport of structure via $f$.

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  • $\begingroup$ How does one explicitly define the maps? $\endgroup$ – Sir_Math_Cat Feb 19 '17 at 7:07
  • $\begingroup$ The class of $x$ in $\mathbf{R}/a\mathbf{Z}$ corresponds to the class of $f(x)$ in $\mathbf{R}/b\mathbf{Z}$. This works because $f$ is really an isomorphism between two copies of $\mathbf{R}$. And $a$ in the first copy corresponds to $b$ in the second. $\endgroup$ – user49640 Feb 19 '17 at 7:10
  • $\begingroup$ So I would define something like $\phi:\mathbb{R}/a\mathbb{Z}\to\mathbb{R}/b\mathbb{Z}$ by $x\in\mathbb{R}/a\mathbb{Z}\mapsto (b/a)x\in\mathbb{R}/b\mathbb{Z}$? $\endgroup$ – Sir_Math_Cat Feb 19 '17 at 7:13
  • $\begingroup$ Not $x$, but the class of $x$. $\endgroup$ – user49640 Feb 19 '17 at 7:13
  • $\begingroup$ But under that mapping applied to a class, that defines an isomorphism? $\endgroup$ – Sir_Math_Cat Feb 19 '17 at 7:16

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