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!


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.


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 $\bar f:R/bZ\rightarrow R/aZ$.

  • $\begingroup$ 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

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 $f$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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