# Homework Help: Prove (Q/Z)/H is isomorphic to Q/Z

1. Sep 30, 2015

### jackmell

1. The problem statement, all variables and given/known data

Let $H$ be a finite subgroup of the group $G = \mathbb{Q}/\mathbb{Z}$. Prove $G/H$ is isomorphic to $G$.

2. Relevant equations

My plan is to model the proof after a simple example but I would need to show there is a bijection between $\frac{\mathbb{Q}/\mathbb{Z}}{ H}$ and $\mathbb{Q}/\mathbb{Z}$ which I'm not sure how to do.

Consider first an easier example: Let $G=\mathbb{Z}$ and $H=2\mathbb{Z}$. Then even though we intuitively sense $H$ is smaller than $G$, $G\cong H$ since the map $\phi: G\to H$ with $\phi(n)=2n$ is a bijective homomorphism: $\phi(0)=0, \phi(n+m)=\phi(n)+\phi(m)$. And this is the same case with mod'ing-out any finite subgroup of $\mathbb{Q}/\mathbb{Z}$: The factor group $\displaystyle\frac{\mathbb{Q}/\mathbb{Z}}{\mathbb{H}}$ will still contain a countable infinite number of elements and therefore has the same cardinality as $\mathbb{Q}/\mathbb{Z}$ (however, not sure that's sufficient to demonstrate the two groups are isomorphic).

3. The attempt at a solution

Let:
\begin{align} \mathbb{Q}/\mathbb{Z}&=\{\mathbb{Z},\frac{p_1}{q_1}+\mathbb{Z},\frac{p_2}{q_2}+\mathbb{Z},\cdots\}: 0\leq \frac{p_i}{q_i}<1\\ \mathbb{H}&=\{\mathbb{Z},\frac{r_1}{s_1}+\mathbb{Z},\frac{r_2}{s_2}+\mathbb{Z},\cdots,\frac{r_n}{s_n}+\mathbb{Z}\} : \end{align}
such that $H$ is a finite subgroup of $\mathbb{Q}/\mathbb{Z}$. We can then form the factor group:
$\frac{\mathbb{Q}/\mathbb{Z}}{H}=\left\{\mathbb{Z},\frac{a_1}{b_1}+\mathbb{Z},\frac{a_2}{b_2}+\mathbb{Z},\cdots\right\}: \frac{a_i}{b_i}\notin \left\{\frac{r_i}{s_i}\right\}$
Claim $\displaystyle\frac{\mathbb{Q}/\mathbb{Z}}{H}$ has the same cardinality as $\mathbb{Q}/\mathbb{Z}$.

Construct the following mapping:
$\phi:\displaystyle\frac{\mathbb{Q}/\mathbb{Z}}{H}\to\mathbb{Q}/\mathbb{Z}$
with
$\phi(g+H)=g+\mathbb{Z}$
Now,
\begin{align} \displaystyle g+H&=\left(\frac{p_i}{q_i}+\mathbb{Z}\right)+\left(\frac{r_i}{s_1}+\mathbb{Z}\right)=\left(\frac{p_i}{q_i}+\frac{r_i}{s_i}\right)+\mathbb{Z} \\ g+Z&=\frac{p_i}{q_i}+\mathbb{Z} \end{align}
and $\left(\frac{p_i}{q_i}+\frac{r_i}{s_i}\right)+\mathbb{Z}=\frac{p_i}{q_i}+\mathbb{Z}$ when $\displaystyle\frac{r_i}{s_i}\in \mathbb{Z}$ for $\displaystyle 0\leq \frac{r_i}{s_i}< 1$ which is never the case so that $\frac{\mathbb{Q}/ \mathbb{Z}}{H}$ and $\mathbb{Q}/\mathbb{Z}$ have the same cardinality.

This makes intuitive sense to me but I'm not sure how to prove this mapping is indeed a bijection which I think is required to prove there is an isomorphism between the two groups.

Jack

Edit: Correct typos

Last edited: Sep 30, 2015
2. Sep 30, 2015

### andrewkirk

No, that's not sufficient. Having the same cardinality means there's a bijection. The bijection need not necessarily preserve the algebraic structure.
That doesn't look right. The items on both sides of the equals are elements of different groups. Can you clarify what you were trying to write here?

3. Sep 30, 2015

### andrewkirk

I just thought about this a bit, and realized that, for the subset $H$ of $\mathbb{Q}/\mathbb{Z}$ to be a subgroup (ie closed under addition) and finite, it must have a special property that means the subgroup can be uniquely characterized by a positive integer (ie the set of all such finite subgroups is in bijection with $\mathbb{N}$). Can you work out what that property is?

Once you have that property, it's easy to see what the algebra-preserving bijection (isomorphism) will be.

4. Oct 1, 2015

### jackmell

Ok thanks for that Andrew.

I'm not really clear what the elements of $H$ are. Can you clarify the following for me:

(1) If

$\displaystyle G=\mathbb{Q}/\mathbb{Z}=\left\{\frac{p}{q}+\mathbb{Z} : 0\leq p/q < 1\right\}=\left\{\mathbb{Z}\;,\frac{p_1}{q_1}+\mathbb{Z},\;\frac{p_2}{q_2}+\mathbb{Z}\;,\cdots\right\}: \frac{p_i}{q_i}<1$

Then any finite subgroup $H$ of $G$ will have the form:

$H =\displaystyle \left\{ \mathbb{Z},\;\frac{a_1}{b_1}+\mathbb{Z},\; \frac{a_2}{b_2}+\mathbb{Z},\cdots, \frac{a_n}{b_n}+\mathbb{Z}\right\}$

so that when we mod-out $H$ from $G$, I assume we obtain the factor group:

$K=\displaystyle\frac{\mathbb{Q}/\mathbb{Z}}{H}=\left\{H\;,\frac{r_1}{s_1}+H\;,\frac{r_2}{s_2}+H\;,\frac{r_3}{s_3}+H\cdots\right\}$

such that the set $\displaystyle\big\{\frac{r_i}{s_i}\big\}\notin \big\{\frac{a_i}{b_i}\big\}$.

This, assuming of course the factor group is indeed unbounded. I'm not quite confident about what exactly an element $\displaystyle\frac{r_i}{s_i}+H$ of $K$? Is it:

$\displaystyle\frac{r_i}{s_i}+H=\left\{\frac{r_i}{s_i}+\mathbb{Z},\left(\frac{r_i}{s_i}+\frac{a_1}{b_1}\right)+\mathbb{Z},\left(\frac{r_i}{s_i}+\frac{a_2}{b_2}\right)+\mathbb{Z},\left(\frac{r_i}{s_i}+\frac{a_3}{b_3}\right)+\mathbb{Z},\cdots \right\}$

Now regarding the requirements of $H$ being a finite group: Certainly it must meet the group axioms:

(i) The identity of $G$ has to be an element of $H$. So we allow $\mathbb{Z}\in H$.

(ii) If $h_1\in H$ then $h_1^{-1}\in H$. so let $\frac{a_i}{b_i}+\mathbb{Z}\in H$. Then we must have $-\left(\frac{a_i}{b_i}+\mathbb{Z}\right)=\left(-\frac{a_i}{b_i}+\mathbb{Z}\right)\in H$.

(iii) If $h_1,h_2\in H$ then their sum (since additive groups) must be in $H$ and this is a little confusing for me: Say:

\begin{align} h_1&=\left(\frac{a_1}{b_1}+\mathbb{Z}\right)\\ h_2&=\left(\frac{a_2}{b_2}+\mathbb{Z}\right) \end{align}

Then $\displaystyle \left(\frac{a_1}{b_1}+\frac{a_2}{b_2}\right)+\mathbb{Z}\in H$. Or:
$\displaystyle \left(\frac{a_1}{b_1}+\frac{a_2}{b_2}\right)+\mathbb{Z}=\frac{a_k}{b_k}+\mathbb{Z}$
for some $\displaystyle \frac{a_k}{b_k}+\mathbb{Z}\in H$ and this is only true when:
$\left(\frac{a_1}{b_1}+\frac{a_2}{b_2}-\frac{a_k}{b_k}\right)\in \mathbb{Z}$
that is, that sum must be an integer.

Do I have this much correct in the analysis?

Last edited: Oct 1, 2015
5. Oct 1, 2015

### andrewkirk

Look at it like this. H is a subgroup so it must contain zero. It is finite and all its members are fractions. Assume WLOG that every fraction is expressed in irreducible form - ie the numerator and denominator have no common factor. Since the set is finite, there is a least common multiple of the denominators. Call that D. If we express the fractions in H with denominator D, we have a bunch of numerators that are a subset of the numbers in {0,1,...,D-1}. Can we use the fact that H is a subgroup to work out what that subset must be?

6. Oct 1, 2015

### jackmell

Afraid I'm not quite following you Andrew. Let me however, first post what I think I would need to do if we prove the factor group $\frac{\mathbb{Q}/\mathbb{Z}}{H}$ is unbounded for any finite group $H$:

Let me just flat-out assume I have a mapping:

$\phi:\frac{\mathbb{Q}/\mathbb{Z}}{H}\to \mathbb{Q}/\mathbb{Z}$
and I want to prove it's a bijective homomorphism for any finite subgroup $H$ of $\mathbb{Q}/\mathbb{Z}$, and for the moment, let's assume $\frac{\mathbb{Q}/\mathbb{Z}}{H}$ is unbounded so that if I get this part, then show it's unbounded, we'll have the problem solved.

So I'll first prove it's well-defined. That means if $\frac{r_a}{s_a}+H=\frac{r_b}{s_b}+H$ for $\frac{r_i}{s_i}+H\in \frac{\mathbb{Q}/\mathbb{Z}}{H}$ then necessarily, we must have $\phi(\frac{r_a}{s_a}+H)=\phi(\frac{r_b}{s_b}+H)$ or $\frac{r_a}{s_a}+\mathbb{Z}=\frac{r_b}{s_b}+\mathbb{Z}$ and this is only true if $\frac{r_a}{s_a}-\frac{r_b}{s_b}\in \mathbb{Z}$ but by definition $0\leq \frac{r_i}{s_i}<1$ and so $\frac{r_a}{s_a}-\frac{r_b}{s_b}=0$ that is $\frac{r_a}{s_a}=\frac{r_b}{s_b}$. And $\phi$ is certainly a (additive) group homomorphism which I won't Latex that out here to save space and time. So far then I have a well-defined homomorphism between two groups of the same cardinality.

Now let's try for injectivity: Assume the contrary, that is, $\phi(\frac{r_a}{s_a}+H)=\phi(\frac{r_b}{s_b}+H)$. Then that would mean $\frac{r_a}{s_a}+\mathbb{Z}=\frac{r_b}{s_b}+\mathbb{Z}$ and again we have $\frac{r_a}{s_a}=\frac{r_b}{s_b}$ so then $\phi$ is one-to-one.

Let's see if it's onto as well: Let $\frac{r_a}{s_a}+\mathbb{Z}\in \mathbb{Q}/\mathbb{Z}$. Then I need to show there exists some $\frac{r_i}{s_i}+H\in \frac{\mathbb{Q}/\mathbb{Z}}{H}$ such that $\phi(\frac{r_i}{s_i}+H)=\frac{r_a}{s_a}+\mathbb{Z}$. That seems easy: let $\frac{r_i}{s_i}=\frac{r_a}{s_a}$.

Thus, if the factor group is unbounded, then
$\frac{\mathbb{Q}/\mathbb{Z}}{H}\cong \mathbb{Q}/\mathbb{Z}$

Not quite sure how rigorous that is but it's a start. So now, all I would need to show is that the factor group is unbounded.

Could you or others too help me show this?

Ok thanks,
Jack

7. Oct 1, 2015

### andrewkirk

Jack, I'm afraid the approach you are taking looks very complicated. I don't think it needs to be that complicated.

Also, the homomorphism $\phi$ you are using, which is $\phi(q+H)=q+\mathbb{Z}$, is not a bijection, so that approach won't lead to a successful solution.

I think the difficulty, and the complication, arises from the fact that in your formulation you have used neither the finiteness of H, nor the fact that all its elements are proper fractions (or, more strictly speaking, each element is $q+\mathbb{Z}$ where $q$ is a proper fraction). It is those two properties that guarantee a solution, and that can provide a path towards it.

I suggest you go over my post #5. If there is something in it you do not understand, feel free to ask. If you can get through it, and work out what the numerators of the fractions in H must be, when expressed with a common denominator of D, then I am confident you'll be able to see your path towards a solution.

8. Oct 1, 2015

### andrewkirk

PS, it might help to work with an example. Say we know that H contains $\frac{2}{3}+\mathbb{Z},.\frac{4}{5}+\mathbb{Z}$ and $\frac{1}{2}+\mathbb{Z}$ What is the smallest subgroup of $\mathbb{Q}/\mathbb{Z}$ containing those elements?

9. Oct 1, 2015

### jackmell

Thanks. Ok, part of my problem maybe is that I'm having problems constructing sub-groups of $\mathbb{Q}/\mathbb{Z}$. In your example above, I would think the smallest subgroup would be:
$\left\{\mathbb{Z},\frac{1}{30}+\mathbb{Z},\frac{2}{30}+\mathbb{Z},\cdots,\frac{29}{30}+\mathbb{Z}\right\}$

Pretty sure that's an additive subgroup containing those three elements. So that in the general case, the smallest subgroup containing all of a finite $H$ would of course be likewise for LCM of the denominators=D:

$R=\left\{\mathbb{Z},\frac{1}{D}+\mathbb{Z},\frac{2}{D}+\mathbb{Z},\cdots,\frac{D-1}{D}+\mathbb{Z}\right\}$
Then $H\leq R$ (pretty sure but need to double check) so perhaps I should be looking at the subgroup $R$ that's easier to work with instead of $H$?

10. Oct 1, 2015

### andrewkirk

OK, now if for each element $q+H$ of $(\mathbb{Q}/\mathbb{Z})/H$, where $H$ is that set you just described in terms of $D$, you choose the smallest possible non-negative representative element $q$, what will be the range of all such $q$s? Compare that to the range of smallest representative $q$s for the sets $q+\mathbb{Z}$. Does that comparison suggest an isomorphism, given that you need to map one range onto the other?

11. Oct 1, 2015

### jackmell

Not sure here but let me take in in pieces. First:

Ok, so if I understand this correctly, any finite $H$ would be contained in a subgroup $R=\left\{\mathbb{Z},\frac{1}{D}+\mathbb{Z},\cdots,\frac{D-1}{D}+\mathbb{Z}\right\}$. Then the smallest representative of $h+H$ would be a member not contained in $H$ so $\frac{1}{D+1}+H\in \frac{\mathbb{Q}/\mathbb{Z}}{H}$. But then we would also have
$\frac{1}{D+2}+H\in \frac{\mathbb{Q}/\mathbb{Z}}{H}$ and in fact $\frac{1}{D+n}+H\in \frac{\mathbb{Q}/\mathbb{Z}}{H}$ for $n\in\mathbb{N}$.

12. Oct 1, 2015

### andrewkirk

Not just 'contained in'. H will be identical to some such R. So we can write $H=H(D)=\left\{\mathbb{Z},\frac{1}{D}+\mathbb{Z},\cdots,\frac{D-1}{D}+\mathbb{Z}\right\}$.

I don't think that'll help find the range of such smallest representatives.
Consider this: $q+H(D)=q-\frac{1}{D}+H(D)$, since $\frac{1}{D}+\mathbb{Z}\in H$.

So if we subtract $\frac{1}{D}$ repeatedly from $q$ until we can't do it any more without going negative, in what range must the revised $q$ lie?

13. Oct 1, 2015

### jackmell

Ok, since $0\leq q<1$, then if we repeatedly subtract $1/D$, then the revised $q$ is in the range of $[0,1/D)$.