For $n$ a positive integer, let $\frac{\Bbb Q}{n\Bbb Z}$ be the quotient of the group of rational numbers $\Bbb Q$ (under addition) by
the subgroup $n\Bbb Z$. For each of the following statements, state whether it is true or false.

1) Every element of $\frac{\Bbb Q}{n\Bbb Z}$ is of finite order,

2) There are only finitely many elements in $\frac{\Bbb Q}{n\Bbb Z}$ of any given finite order,

3) Every proper subgroup of $\frac{\Bbb Q}{n\Bbb Z}$ is finite,

4)$\frac{\Bbb Q}{2\Bbb Z}$ and $\frac{\Bbb Q}{5\Bbb Z}$ are isomorphic as groups.

The elements of $\frac{\Bbb Q}{ \Bbb Z}$ are cosets of the form $ \Bbb Z + s$ where $s = \frac{p}{q}$ is rational (

Let $x = \Bbb Z + \frac{p}{q}$ be an arbitrary element of $\frac{\Bbb Q}{\Bbb Z}$. Let $m$ be the smallest integer such that

We have, $$x + x = \Bbb Z + \frac{2p}{q}$$ and adding $x$ with itself for $m$ times gives us $$x^{m} = (\Bbb Z + \frac{p}{q})^m = \Bbb Z + \frac{mp}{q} = \Bbb Z + r = \Bbb Z.$$ This shows that $m$ is the smallest power such that $x^m =e$ and so the order of $x = n \Bbb Z + \frac{p}{q}$ is equal to $m$.

Hence, whenever we consider an arbitrary element $\Bbb Z + \frac{p}{q}$ of $\frac{\Bbb Q}{\Bbb Z}$

Proof: Verify that the element $\Bbb Z + \frac{1}{n}$ is a generator. In general the elements $\Bbb Z + \frac{m}{n}$ with (m,n)=1 in $H_n$ forms the $\phi(n)$ generators of this cyclic group.

Note that, as an element of $\frac{\Bbb Q}{\Bbb Z}$, we always assume $(m,n) = 1$ in $\Bbb Z + \frac{m}{n}$.

It is straight forward to verify that $H(m)$ is indeed an infinite subgroup of $\frac{\Bbb Q}{\Bbb Z}$.

1) Every element of $\frac{\Bbb Q}{n\Bbb Z}$ is of finite order,

2) There are only finitely many elements in $\frac{\Bbb Q}{n\Bbb Z}$ of any given finite order,

3) Every proper subgroup of $\frac{\Bbb Q}{n\Bbb Z}$ is finite,

4)$\frac{\Bbb Q}{2\Bbb Z}$ and $\frac{\Bbb Q}{5\Bbb Z}$ are isomorphic as groups.

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**Solution:**The elements of $\frac{\Bbb Q}{ \Bbb Z}$ are cosets of the form $ \Bbb Z + s$ where $s = \frac{p}{q}$ is rational (

**We have gcd(p,q)=1**). We have $ \Bbb Z + s$ is equal to the identity element $ \Bbb Z + 0 = \Bbb Z$ if and only if $s \in \Bbb Z$.**Main Result : Let $n$ be a positive integer then $$\frac{\Bbb Q}{\Bbb Z} \cong \frac{\Bbb Q}{n \Bbb Z}.$$****Proof:**The function which takes $\Bbb Z + x$ to $n \Bbb Z + nx$ is the required isomorphism.**In view of this result, it is enough to verify the given options to the group $\frac{\Bbb Q}{\Bbb Z}$.****option 4**: (**True**) By the above result we have the group $\frac{\Bbb Q}{\Bbb Z}$ is isomorphic to the groups $\frac{\Bbb Q}{2 \Bbb Z}$ and $\frac{\Bbb Q}{5 \Bbb Z}$. By transitivity, we have**$\frac{\Bbb Q}{2 \Bbb Z}$ is isomorphic to $\frac{\Bbb Q}{5 \Bbb Z}$.**

**option 1**: (**True**) We claim that every element of $\frac{\Bbb Q}{\Bbb Z}$ has finite order.Let $x = \Bbb Z + \frac{p}{q}$ be an arbitrary element of $\frac{\Bbb Q}{\Bbb Z}$. Let $m$ be the smallest integer such that

**$mp = q r$**for some $r \in \Bbb N$ (i.e, $q$ divides $mp$). Such a $m$ exists because $m = q$ is one possibility such that $q \mid mp$ and so we can consider the minimal one among the $m$ satisfying this condition by the well ordering principle of $\Bbb N$.We have, $$x + x = \Bbb Z + \frac{2p}{q}$$ and adding $x$ with itself for $m$ times gives us $$x^{m} = (\Bbb Z + \frac{p}{q})^m = \Bbb Z + \frac{mp}{q} = \Bbb Z + r = \Bbb Z.$$ This shows that $m$ is the smallest power such that $x^m =e$ and so the order of $x = n \Bbb Z + \frac{p}{q}$ is equal to $m$.

**Hence every element in $\frac{\Bbb Q}{n \Bbb Z}$ is of finite order.**

**option 2:(True)****Observation 1**: Let $x = \Bbb Z + \frac{p}{q}$ be an arbitrary element of $\frac{\Bbb Q}{\Bbb Z}$. Let $p = qm + r$ ($0 \le r <q $). Since $gcd(p,q)=1$ we have $1<r<q$ and $\frac{p}{q} = m + \frac{r}{q}$. Now, $$\Bbb Z + \frac{p}{q} = \Bbb Z + (m+\frac{r}{q}) = \Bbb Z + \frac{r}{q}.$$Hence, whenever we consider an arbitrary element $\Bbb Z + \frac{p}{q}$ of $\frac{\Bbb Q}{\Bbb Z}$

**we can always assume that $1<p<q$**.**Observation 2**: Let $n$ be a positive integer. The set $H_n = \{\Bbb Z + \frac{m}{n} : 0 \le m \le n\}$ (here we are not assuming (m,n)=1) is a cyclic subgroup of $\frac{\Bbb Q}{\Bbb Z}$ order $n$.Proof: Verify that the element $\Bbb Z + \frac{1}{n}$ is a generator. In general the elements $\Bbb Z + \frac{m}{n}$ with (m,n)=1 in $H_n$ forms the $\phi(n)$ generators of this cyclic group.

Note that, as an element of $\frac{\Bbb Q}{\Bbb Z}$, we always assume $(m,n) = 1$ in $\Bbb Z + \frac{m}{n}$.

**Hence $\Bbb Z + \frac{m}{n}$ of $\frac{\Bbb Q}{\Bbb Z}$ has order $n$.****This shows that the set of elements of order $n$ in $\frac{\Bbb Q}{\Bbb Z}$ are given by $\Bbb Z + \frac{p}{n}$ with $1 \le p \le n$, $(p,n) = 1$ and there are $\phi(n)$ of them.**

**Hence there are only finitely many elements of order $n$ in $\frac{\Bbb Q}{\Bbb Q}$ for every positive integer $n$.****option 3:**(**False**) Let $m$ be a positive integer. Define $$H(m) = \{\Bbb Z + \frac{p}{q} : q \text{ divides } m\}.$$It is straight forward to verify that $H(m)$ is indeed an infinite subgroup of $\frac{\Bbb Q}{\Bbb Z}$.

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Can u plz elaborate H(m) in option 3

ReplyDeleteHow H(m) is infinite, can't understand

ReplyDeleteThere is no condition on $p$. Therefore $H(m)$ is infinite. is it fine now?

DeleteThere is no condition on p but due to condition on q as it divides m it must be less than or equal to m and p is also an integer ,which is also must be less than m bcz in Z+p/q if p/q >1 then p/q=1+r/q that imply Z+p/q =Z+r/q ,there no of elements in Hm become finite

ReplyDelete