1. ## Quotient Groups

Hi,

So I have a problem that I have been looking at for 4 days now.

Let F be the quotient group G/Z (Z is integers) where G= { ap^b in Q| a, b in Z} where Q is the rational numbers and Z is the integers.

Let let x,y be in F and o(x)=< o(y) (order of x is less or equal to order of y) if and only if x is in <y> (cyclic group generated by y).

Going <== (backwards direction) is fine. I completely understand this direction. Going ==> is where I'm having the trouble.

==> Let x,y in F. Then x=ap^b + Z and y= cp^d + Z. I know the o(x)=< o(y). My professor says that by re-writing ap^b (because it is rational), I can find the exact order of what x is (the smallest number I can multiply x by to get an integer). I tried using actual numbers, but I still cannot get what he was talking about. Also, even if i get what the exact order of x is, I have no idea how to use that to get that x is in <y>.
I cannot seem to understand how using the order will tell me that there exists so t in Z such that x=ty. (which would put x in <y>).

Any help will be great. Thank you.

2. Originally Posted by Benmath
Hi,

So I have a problem that I have been looking at for 4 days now.

Let F be the quotient group G/Z (Z is integers) where G= { ap^b in Q| a, b in Z} where Q is the rational numbers and Z is the integers.

Let let x,y be in F and o(x)=< o(y) (order of x is less or equal to order of y) if and only if x is in <y> (cyclic group generated by y).

Going <== (backwards direction) is fine. I completely understand this direction. Going ==> is where I'm having the trouble.

==> Let x,y in F. Then x=ap^b + Z and y= cp^d + Z. I know the o(x)=< o(y). My professor says that by re-writing ap^b (because it is rational), I can find the exact order of what x is (the smallest number I can multiply x by to get an integer). I tried using actual numbers, but I still cannot get what he was talking about. Also, even if i get what the exact order of x is, I have no idea how to use that to get that x is in <y>.
I cannot seem to understand how using the order will tell me that there exists so t in Z such that x=ty. (which would put x in <y>).

Any help will be great. Thank you.

Ok, so $t\in G=\{ap^b\, \mid\, a\,,\,b\in\mathbb{Z}\}$ is an integer iff $b\geq 0$ (I'm guessing p is a fixed prime number...), so we can say that if $x=ap^b+\mathbb{Z}\in G\slash\mathbb{Z}$ , then:

$o(x)=\left\{\begin{array}{cc}0\;\;\;\;\,,\,\,if\,\ ,b\geq 0\\\!\!\!p^{-b}\,\,\,\,\,\,,\,if\,\,b<0\end{array}\right.$ , so if we also have $y=cp^d+\mathbb{Z}\in G\slash\mathbb{Z}$ and assuming $x\neq \overline{1}=$the quotient group's unit (i.e., $b<0$), then

$ap^b+\mathbb{Z}\in\,\Longleftrigh tarrow\,ap^b+\mathbb{Z}=mcp^d+\mathbb{Z}\,,\,\,m\i n \mathbb{Z}\Longleftrightarrow\,ap^b-mcp^d\in \mathbb{Z}$ (note that it must be $d<0$ as well! (why??).

Now, if $-b\geq-d$ , then we'd get that $p^b(a-mcp^{d-b})\in \mathbb{Z}\Longrightarrow\,p^b\in \mathbb{Z}$ , as $a-mcp^{d-b}\in \mathbb{Z}\,\,since\,\,d-b\geq 0$ , but this can't be since $p^b<1\,\,as\,\,b<0$...!

So it must be $-b\leq -d\Longrightarrow\,o(x)=p^{-b}\leq p^{-d}=o(y)$. Q.E.D.

Tonio

3. I see how this proves the reverse direction, but I am unsure of how it proves the forward direction. I assumed that o(x)<=o(y) then I know that means d>=b then d-b>=0, but I don't know how to go forward after this.

4. Originally Posted by Benmath
I see how this proves the reverse direction, but I am unsure of how it proves the forward direction. I assumed that o(x)<=o(y) then I know that means d>=b then d-b>=0, but I don't know how to go forward after this.

But that direction you wrote you already did it: "Going <== (backwards direction) is fine. I completely understand this direction" !

Tonio

5. I think I posted something incorrectly. What I understand is the direction where I assume that x is in <y>, then go from there.

The direction I don't understand is assuming that o(x)<o(y) then showing that means x is in <y>. Sorry for the confusion.

6. So I know for the forward direction to say:

Let x,y be in F and let o(x) =< o(y). Then x=ap^b + Z and y=cp^d + Z. Then o(x)=p^-b when b<0 and o(y)=p^-d when d<0. Then I know that p^-b =< p^-d, I know in the end I need to get x=ty for some t in Z, but I get stuck in the middle of the proof.

7. Originally Posted by Benmath
So I know for the forward direction to say:

Let x,y be in F and let o(x) =< o(y). Then x=ap^b + Z and y=cp^d + Z. Then o(x)=p^-b when b<0 and o(y)=p^-d when d<0. Then I know that p^-b =< p^-d, I know in the end I need to get x=ty for some t in Z, but I get stuck in the middle of the proof.

I really thought a lot on this problem last day and I think I've gotten to some nice place:

== (1) let us agree to write elements of the quotient group as follows: $ap^{-b}+\mathbb{Z}\,,\,\,a\in\mathbb{Z}\,,\,b\in\mathbb {N}\cup\{0\}\,,\,p$ a fixed prime

== (2) if $r=s+p^t\,,\,r\,,\,s\,,\,k\in\mathbb{Z}$ , then it's easy to check (task for ya) that $rp^{-t}+\mathbb{Z}=sp^{-t}+\mathbb{Z}$ , and from this it follows that

== (3) every element in $\left$ is uniquely determined by its residue modulo $p^t$

== (4) let us agree that for an element $rp^{-t}+\mathbb{Z}\,\in\,G\slash\mathbb{Z}$ we have that $(p,r)=1$ , since if $r=hp^f\,,\,\,then\,\,rp^{-t}+\mathbb{Z}=hp^{-t+f}+\mathbb{Z}$ and we'll want to be sure what exponent of p we're working with.

So let then $x=ap^{-b}+\mathbb{Z}\,,\,\,y=cp^{-d}+\mathbb{Z}\,\in\,G\slash\mathbb{Z}$ , with $p^b\leq p^d\Longleftrightarrow\,b\leq d\Longleftrightarrow\,p^{-b}\geq p^{-d}$ , then:

Lemma: $ap^{-b}+\mathbb{Z}=mp^{d-b}cp^{-d}+\mathbb{Z}\,,\,m\,\in\,\mathbb{Z}$ has a solution.

Proof.- The above has a solution iff $ap^{-b}-mcp^{-b}\,\in\,\mathbb{Z}\Longrightarrow\,a-mc=kp^b\,,\,for\,\,some\,\,k\,\in\,\mathbb{Z}$ $\Longrightarrow\,mc=a-kp^b$ , and this last equations ALWAYS has a solution since both $a\,,\,c\,\in\,\left(\mathbb{Z}\slash p^b\mathbb{Z}\right)^{*}=$ the units modulo $p^b$ ! Q.E.D.

Example: Take $x=2\cdot 7^{-3}+\mathbb{Z}\,,\,y=11\cdot 7^{-6}+\mathbb{Z}\,\Longrightarrow\,\,with\,\,\,a=2\,, \,c=11\,,\,b=3\,,\,d=6$ $\,and\,\,as\,\,o(x)=7^3<7^6=o(y)\,,\,we\,\,get\,\, that\,\,x\,\in\,$, and copying from the proof above:
both 2 and 11 are coprime with 7, and $11\cdot 156=1\!\!\!\pmod {7^3}\,,\,11^{-1}=156$ , all this, of course, in $\left(\mathbb{Z}\slash 7^3\mathbb{Z}\right)^{*}$ $\Longrightarrow\,m\cdot 11=2\!\!\!\pmod {7^3}\,\Longrightarrow\,m=2\cdot 11^{-1}=2\cdot 156=312\!\!\!\pmod {7^3}$ , and thus definining $m=312\cdot 7^3$ we get

$2\cdot 7^{-3}+\mathbb{Z}=(312\cdot 7^3)11\cdot 7^{-6}+\mathbb{Z}$ , as you can easily check now with a calculator.

Tonio

8. Thank you so much for all your time and effort. It really is a very interesting problem.

9. Originally Posted by Benmath
Thank you so much for all your time and effort. It really is a very interesting problem.

Indeed it was and the bastard gave me some tough moments...but it was a pleasure all in all.

Tonio