# For $x,y \in \mathbb{Q}\setminus\{ 0\}$, prove that $\mathbb{Q}/\langle x \rangle \simeq \mathbb{Q}/ \langle y \rangle$.

For $x$, $y$ nonzero rationals, I need to prove that $\mathbb{Q}/\langle x \rangle \simeq \mathbb{Q}/ \langle y \rangle$.

To do this, I was thinking about using the Third Isomorphism Theorem:

Let $M, N \trianglelefteq G$ and $M \leq N$. Then, $(G/M)/(N/M)\simeq G/N$.

However, this would only make sense if $\langle y \rangle$ was a normal subgroup of $\mathbb{Q}$. I know it is cyclic, and hence abelian, but is it also normal?

Then, I could let $M = e_{G}$ (the identity element in $G$), but then $\langle x \rangle = N/M = N/e = N = \langle y \rangle$, which doesn't make any sense.

So, suffice it to say, I am at a loss as to how to prove this result! :(

• There are so many things here you really need to take another look at. Abelian subgroups need not be normal (subgroups of abelian groups are normal might be what you had in mind). Cyclic groups are however abelian. Oct 17 '16 at 12:28
• @TobiasKildetoft you're right, I just realized (lack of sleep will do that to a person). So, I just edited that part of the question. I wonder then, if the 3rd Isomorphism Theorem is not the way to go. And if not, do you have any suggestions?
– user100463
Oct 17 '16 at 12:33
• @JessyCat: In a commutative group, every subgroup is normal. Since $\Bbb Q$ is commutative, it follows that you need not worry about normality in this problem. Oct 17 '16 at 12:56
• @AlexM. it still doesn't help me though, because if I apply the third isomorophism theorem, I still end up with $\langle x \rangle = \langle y \rangle$, which is weird.
– user100463
Oct 17 '16 at 13:05
• Do you know the homomorphisms $\mathbb{Q}\to\mathbb{Q}$? Can you find one that maps $\langle x\rangle$ to $\langle y\rangle$? Can you see it induces a homomorphism $\mathbb{Q}/\langle x\rangle\to\mathbb{Q}/\langle y\rangle$? Is it an isomorphism? Oct 17 '16 at 13:21

I would suggest you look no further than the first isomorphism theorem.

Take some nonzero $a \in \mathbb{Q}$ and convince yourself of the fact that if we show that $\mathbb{Q}/\langle a \rangle \cong \mathbb{Q}/\langle 1 \rangle$ we are done. Define the map $$f: \mathbb{Q} \to \mathbb{Q}/\langle 1 \rangle,\quad x \mapsto \frac{1}{a} \cdot x$$ and do the following:

• Show that $f$ is an homomorphism.
• Show that $f$ is surjective.
• Show that $\ker{(f)} = \langle a \rangle$.
• Apply the first isomorphism theorem to find $\mathbb{Q}/\ker{(f)} \cong \text{im}{(f)}$.
• This requires the OP to know how $\Bbb Q / \langle 1 \rangle$ looks like, wich is essentially the solution of the problem. In other words, how does the OP know that $\frac x a$ is an element of $\Bbb Q / \langle 1 \rangle$? Your proof is therefore circular. Oct 17 '16 at 13:00
• @Pjotr5 is the reason why if we show $\mathbb{Q}\langle a \rangle \simeq \mathbb{Q}\langle 1 \rangle$ because we can obtain any $\langle b \rangle$ (where $b \in \mathbb{Q}$) from $\langle 1 \rangle$ by multiplying all the elements of $\langle 1 \rangle$ by $b$?
– user100463
Oct 17 '16 at 13:01
• @AlexM. do you have a better solution?
– user100463
Oct 17 '16 at 13:02
• @AlexM. I don't agree. I haven't given a proof but rather I've given a recipe that can be made into a rigorous proof. Since $\frac{x}{a}$ is an element of $\mathbb{Q}$, it becomes an element of $\mathbb{Q}/\langle 1\rangle$ after reduction. What reduction exactly means is for the OP to figure out. Oct 17 '16 at 13:04
• @JessyCat Exactly. For example $\frac{1}{2}$ and $\frac{3}{2}$ are different elements of $\mathbb{Q}$, but the same element in $\mathbb{Q}/\langle 1 \rangle$. When I write $\frac{x}{a} \in \mathbb{Q}/\langle 1 \rangle$ I abuse the notation a little bit, I should actually write the element $\frac{x}{a} + \langle 1 \rangle$ of $\mathbb{Q}/\langle 1 \rangle$. Oct 17 '16 at 13:16