Quotient (Factor) Groups of Z

**mpetnuch** · November 4th 2007, 05:36 PM

I know that $Z/nZ \cong Z_n$ but would $(Z \times Z)/<(n,n)> \cong Z_n \times Z_n$ ?

P.S. Sorry for the bad formatting, trying to learn the LaTeX formatting for this site :-)

**ThePerfectHacker** · November 4th 2007, 06:01 PM

Originally Posted by mpetnuch

I know that $Z/nZ \cong Z_n$ but would $(Z \times Z)/<(n,n)> \cong Z_n \times Z_n$ ?

P.S. Sorry for the bad formatting, trying to learn the LaTeX formatting for this site :-)

Define,
$\phi: \mathbb{Z}\times \mathbb{Z} \mapsto \mathbb{Z}_n\times \mathbb{Z}_n$ as $\phi: (x,y) = \phi([x]_n,[y]_n)$ where $[x]_n$ and $[y]_n$ are the congruence classes of $x$ and $y$ respectively. This is a homomorpism. And clearly $\phi$ is an onto map.
Thus, by the fundamental homorphism theorem,
$\mathbb{Z}_n\times \mathbb{Z}_n\simeq (\mathbb{Z}\times \mathbb{Z})/\ker (\phi)$ but $\ker \phi = \{ kn,jn \} = \left< n \right> \times \left< n \right>$ .
So you probabily meant to write,
$(\mathbb{Z}\times \mathbb{Z})/(\left< n\right>\times \left< n \right>)$ .

Are you in New York?

**mpetnuch** · November 4th 2007, 06:08 PM

Yes I live in New York? Why or how did you guess?

**ThePerfectHacker** · November 4th 2007, 06:12 PM

Originally Posted by mpetnuch

Yes I live in New York? Why or how did you guess?

City College?

**mpetnuch** · November 4th 2007, 06:21 PM

Originally Posted by ThePerfectHacker

So you probabily meant to write,
$(\mathbb{Z}\times \mathbb{Z})/(\left< n\right>\times \left< n \right>$

This question arose from a question in my text that asked:

Show that $(\mathbb{Z}\times \mathbb{Z})/\left<(2, 2)\right>)$ is not a cyclic group. I was assuming that the quotient group was isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$ , i.e., the Klein 4-group, which is not cyclic (prob. #21 pg 176 in Beachy's Abstract Algebra)

By, $\left<(n, n)\right>$ I meant the cyclic subgroup generated by (n,n), i.e. { ..., (-2n, -2n), (-n,n), (0,0), (n,n), (2n,2n), ...}

So I think we agree, but just different notation right?

**mpetnuch** · November 4th 2007, 06:25 PM

Originally Posted by ThePerfectHacker

City College?

Yeah. LOL. Just started there this semester. Prepping for my alegbra exam on Tuesday? Are you in the class/another grad student/prof?

**ThePerfectHacker** · November 4th 2007, 06:43 PM

Originally Posted by mpetnuch

This question arose from a question in my text that asked:

Show that $(\mathbb{Z}\times \mathbb{Z})/\left<(2, 2)\right>)$ is not a cyclic group. I was assuming that the quotient group was isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$ , i.e., the Klein 4-group, which is not cyclic.

By, $\left<(n, n)\right>$ I meant the cyclic subgroup generated by (n,n), i.e. { ..., (-2n, -2n), (-n,n), (0,0), (n,n), (2n,2n), ...}

So I think we agree, but just different notation right?

This problem appears in Beachy+Blair problem 21 on section 3.8. It is very possible you go to CCNY and are in the abstract algebra course. However, it was not assignent as a homework problem (I am not in the class but I do know which homework is assigned).

No you are wrong. There is a large difference between $\left< n \right> \times \left< n \right>$ and $\left< (n,n) \right>$ . One is a Cartesian product and the other is not.

We can prove that (like you asked that),
$(\mathbb{Z}\times \mathbb{Z})/(\left< n \right> \times \left< n \right> ) \simeq \mathbb{Z}_n \times \mathbb{Z}_n$ for $n>1$ .
First, $\left< n \right> \times \left< n \right>$ is a subgroup of $\mathbb{Z}\times \mathbb{Z}$ so it is normal because subgroups of an abelian group are normal.

Okay now the subgroup $\left< n \right> \times \left< n \right>$ is the set $\{ ... , (-n,-n), (0,0), (n,n), ... \}$ . The cosets have the form $(x,y) + \left< n \right> \times \left< n \right> = \{..., (x-n,y-n),(x,y),(x+n,y+n),...\}$ . Note that is $x_1\equiv x_2 (\bmod n)$ then $(x_1,y)+\left< n \right> \times \left< n \right> = (x_2,y)+\left< n \right> \times \left< n \right>$ .* So we can express uniquely everything as $(a,b) + \left< n \right> \times \left< n \right>$ where $0\leq a,b\leq n-1$ . Thus, there are a total of $n\times n = n^2$ elements in this coset. You can prove isomorphism with $\mathbb{Z}_n\times \mathbb{Z}_n$ by defining $\phi: (a,b)+\left< n \right> \times \left< n \right> \mapsto \mathbb{Z}_n\times \mathbb{Z}_n$ and $\phi(a,b) = ([a]_n,[b]_n)$ . So your conjecture is true.

So are you in CCNY, everything tells me that you are? Are you afraid that I am going to kill you or something like that.

*)Important side note: This is why this group modulo $\left< (1,1) \right>$ is infinite (problem 20) while $\left< (2,2) \right>$ is actually finite because $x_1=x_2$ for no matter what $x_1,x_2$ you chose because everything is congruent to mod 1. So the cosets behave completely differently if you know what I mean.

**ThePerfectHacker** · November 4th 2007, 06:46 PM

Originally Posted by mpetnuch

Yeah. LOL. Just started there this semester. Prepping for my alegbra exam on Tuesday? Are you in the class/another grad student/prof?

Really. That is awesome, I am a student at CCNY also. But I am not taking Algebra. I happen to know the automorphism of the symettric group was assigned as an extra credit problem (it really is an interesting problem). I am a (undergraduate) student, who likes to think I am a professor.

Maybe we have seen eachother and never realized it. I am taking Multi-variable Advanced Calculus and Probability Theory.

You have an exam on Tuesday? He did not do a review session as he usually does.

Did you find this site by pure coincidence?

**mpetnuch** · November 4th 2007, 06:48 PM

Guess you missed my above post. I am in CCNY. And I hope you don't kill me. Cause there are some really 'ghetto' people that go there.

**mpetnuch** · November 4th 2007, 06:54 PM

Originally Posted by ThePerfectHacker

Did you find this site by pure coincidence?

I actually found this site maybe 2 years ago when I was at Cooper Union. I don't remember exactly how I stumbled upon it, but I had a problem in my Analysis on Manifolds class that I could not do - and my professor didn't give me the best explaination either. So I wound up surfing the web (as I do frequently since so many math problems are solved here - usually with good explainations) hoping to find some help.

I am a first year graduate student at CCNY. I am taking topology, real analysis, algebra (as you figured out), and statistics II.

It seems that you are really advanced for an undergrad. I was a chemical engineering major undergrad. I had a decent math background coming to CCNY, but algebra is giving me the most headaches, although it also gives me the biggest satisfaction when I solve a problem as well.

**mpetnuch** · November 4th 2007, 07:01 PM

Originally Posted by ThePerfectHacker

We can prove that (like you asked that),
$(\mathbb{Z}\times \mathbb{Z})/(\left< n \right> \times \left< n \right> ) \simeq \mathbb{Z}_n \times \mathbb{Z}_n$ for $n>1$ .
First, $\left< n \right> \times \left< n \right>$ is a subgroup of $\mathbb{Z}\times \mathbb{Z}$ so it is normal because subgroups of an abelian group are normal.

You meant $\left< (n,n) \right>$ ?

**ThePerfectHacker** · November 4th 2007, 07:14 PM

Originally Posted by mpetnuch

You meant $\left< (n,n) \right>$ ?

Yes, I made a mistake.

I might not be able to help anymore, I just taking a pill against cold, it supposed to make you go to sleep. But since I hate swallowing pills I just cut of the plastic and drank the liquid.

(I know why they keep it in the plastic).

**mpetnuch** · November 4th 2007, 07:17 PM

No problem, thanks for all the help tonight. Maybe I will see you around CCNY!

**ThePerfectHacker** · November 5th 2007, 08:54 AM

Post #7 was a mistake. Let me try again.

I said that $\mathbb{Z}\times \mathbb{Z}/ (\left< n\right> \times \left< n \right> ) \simeq \mathbb{Z}_n\times \mathbb{Z}_n$ which is absolute garbage. Sorry about that.

Okay here is how it works.
Note that, $\left< n\right> \times \left< n \right> = \{...,(-n,-n), (0,0),(n,n),...\}$ .
Now if $x,y$ are integers then $(x,y) + \left< n\right> \times \left< n \right>$ determines the coset $\{ ... , (x-n,y-n),(x,y),(x+n,y+n),...\}$ . Note that there is an invariance that exists among all its ordered pairs, namely, the difference from the first entry and the second entry is fixed. This means $(x_1,y) + \left< n\right> \times \left< n \right>$ and $(x_2,y)+\left< n\right> \times \left< n \right>$ can only be the same coset if $x_1 - y = x_2 - y$ (because of this invariance) so $x_1 = x_2$ . Thus, two numbers being congruent to each other mod $n$ is not sufficient for them to be the same coset. Now turning the question around we ask if $(x_1,y_1)+\left< n\right> \times \left< n \right>$ and $(x_2,y_2)+\left< n\right> \times \left< n \right>$ such that $x_1-y_1 = x_2 - y_2$ do they determine the same coset? The easiet way to answer this question is to look at a specific example say $n=3$ note $(1,2)+\left< 3\right> \times \left< 3 \right>$ and $(2,3)+\left< n\right> \times \left< n \right>$ have the same difference of their coordinates but these cosets are not the same, namely, the first one has $...,-2,1,2,...$ among its first member while the second one $...,-1,2,4,...$ amond its second member. However if $x_1-y_1 = x_2 - y_2$ and $x_1 \equiv x_2 (\bmod n)$ and $y_1\equiv y_2 (\bmod n)$ then this is sufficient for them to be the exactly the same coset. This condition is also necessary, thus, we have completely described all the cosets.

So we can think of this cosets as the following. Let $G$ be a group of $(a,b)$ where $a,b\in \mathbb{Z}$ . Define an equivalence relation on $G$ as two elements are equivalent if the first and second members are congruent, and the differences are the same. Define the group operation as basic component addition. This will be a group. The special case is when $n=1$ because everything is congruent to eachother so the only condition is that the difference between the elements is the same to determine the same equivalence class.

So I do not see any "nice" way of describing what this group is when $n>1$ , it certainly is not finite.

**mpetnuch** · November 5th 2007, 11:44 AM

I think I got a "nice" representation. I haven't had time to check everything properly, but your equilvance relation idea sparked a few ideas in my head. I think it is that case that:

$\mathbb{Z}\times \mathbb{Z}/ \left< (n, n) \right> \simeq \mathbb{Z}_n\times \mathbb{Z}$

Hopefully, I will get a chance later to flush everything out and I will post it there.

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