NBHM 2020 PART C Question 27 Solution (Structure of the group $\frac{\Bbb Q}{n \Bbb Z}$)

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.
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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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4 comments:

  1. Can u plz elaborate H(m) in option 3

    ReplyDelete
  2. How H(m) is infinite, can't understand

    ReplyDelete
    Replies
    1. There is no condition on $p$. Therefore $H(m)$ is infinite. is it fine now?

      Delete
  3. There 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

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