# Abstract Algebra: Groups

• April 6th 2008, 02:47 AM
Jhevon
Abstract Algebra: Groups
Hello All,

I need help on this one. I'm sure it's simple, but I'm coming to the end off an all-nighter and so my brain has shut down.

Problem: If $k$ and $n$ are positive integers and $k$ is a divisor of $n$, then $\mathbb{Z}_n / \left< [k] \right> \approx \mathbb{Z}_k$.

Thanks,
Jhevon
• April 6th 2008, 07:46 AM
Rebesques
Maybe the statement has an error? If $n=dk$, then I think

$
\mathbb{Z}_n / \left< [k] \right> \approx \mathbb{Z}_d
$

...Just a thought.
• April 6th 2008, 07:48 AM
Jhevon
Quote:

Originally Posted by Rebesques
Maybe the statement has an error? If $n=dk$, then I think

$
\mathbb{Z}_n / \left< [k] \right> \approx \mathbb{Z}_d
$

...Just a thought.

i double checked the textbook and i copied it right. so the error may be with the book and not me, but i won't count on that, i've never been so lucky.

i think a counter-example to your claim would be $\mathbb{Z}_n = \mathbb{Z}_6$ and $\left< [k] \right> = \left< [3] \right>$ ...or is being awake for 38 hours getting to me?
• April 6th 2008, 08:00 AM
Rebesques
Just go to sleep and we 'll figure this out later :p
• April 6th 2008, 08:09 AM
Rebesques
Yeap, I am being dumb. If $n=dk$, then the group generated by $[k]$ has $n/gcd(n,k)=d$ elements, and so $
\mathbb{Z}_n / \left< [k] \right>$
has $n/d=k$ elements.
• April 6th 2008, 08:11 AM
Jhevon
Quote:

Originally Posted by Rebesques
Just go to sleep and we 'll figure this out later :p

hehe, one of the reasons i did not sleep is that i wanted to get this before then :p

are you familiar with the "natural homomorphism" and the "Fundamental homomorphism theorem"? I'm thinking that we have to apply those somehow, but i am having trouble making it all fit
• April 6th 2008, 08:17 AM
ThePerfectHacker
Define $\mu: \mathbb{Z}_n \mapsto \mathbb{Z}_k$ as $\mu ([x]_n ) = [x]_k$. Show this map is well-defined, note, this is were you are going to use the fact that $k|n$. Now argue that this is a homomorphism. This mapping is onto, and so $\mu [ \mathbb{Z}_n ] = \mathbb{Z}_k$. The kernel of the homomorphism will be the set (group) $\left< [k]_n \right>$, I want to mention here another notation, we can write this also as $k\mathbb{Z}_n$, which looks a little neater. And so by fundamental homomorphism theorem $\mathbb{Z}_n/k\mathbb{Z}_n \simeq \mathbb{Z}_k$.
• April 6th 2008, 08:25 AM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
Define $\mu: \mathbb{Z}_n \mapsto \mathbb{Z}_k$ as $\mu ([x]_n ) = [x]_k$. Show this map is well-defined, note, this is were you are going to use the fact that $k|n$. Now argue that this is a homomorphism. This mapping is onto, and so $\mu [ \mathbb{Z}_n ] = \mathbb{Z}_k$. The kernel of the homomorphism will be the set (group) $\left< [k]_n \right>$, I want to mention here another notation, we can write this also as $k\mathbb{Z}_n$, which looks a little neater. And so by fundamental homomorphism theorem $\mathbb{Z}_n/k\mathbb{Z}_n \simeq \mathbb{Z}_k$.

i just want to clarify something before i go storming off to do this. we did a theorem that says a homomorphism is one-to-one if and only if the kernel is the set with only the identity. that does not seem to be happening here, but we need one-to-one for the isomorphism
• April 6th 2008, 08:38 AM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
i just want to clarify something before i go storming off to do this. we did a theorem that says a homomorphism is one-to-one if and only if the kernel is the set with only the identity. that does not seem to be happening here, but we need one-to-one for the isomorphism

We need one-to-one to prove that $\mathbb{Z}_n\simeq \mathbb{Z}_k$. But we are not proving that, we are proving that, $\mathbb{Z}_n/k\mathbb{Z}_n\simeq \mathbb{Z}_k$.

The fundamental homormohpsim theorem says that if you got a homomorphism $\phi: G\mapsto G'$ then $G/H \simeq \phi [G]$ where $H = \ker \phi$. It does not say anything about being one-to-one or onto.
• April 6th 2008, 08:52 AM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
We need one-to-one to prove that $\mathbb{Z}_n\simeq \mathbb{Z}_k$. But we are not proving that, we are proving that, $\mathbb{Z}_n/k\mathbb{Z}_n\simeq \mathbb{Z}_k$.

The fundamental homormohpsim theorem says that if you got a homomorphism $\phi: G\mapsto G'$ then $G/H \simeq \phi [G]$ where $H = \ker \phi$. It does not say anything about being one-to-one or onto.

Ok, thanks. does this work?

Define $\theta : \mathbb{Z}_n \mapsto \mathbb{Z}_k$ by $\theta ( [a]_n) = [a]_k$ for all $[a]_n \in \mathbb{Z}_n$. This function is well-defined if and only if $k \mid n$ by homework problem 21.6 (you don’t expect me to re-invent the wheel, do you?).

We claim that this defines a homomorphism. For if $[a]_n,[b]_n \in \mathbb{Z}_n$, then $\theta ([a]_n \oplus [b]_n) = \theta ([a + b]_n) = [a + b]_k = [a]_k \oplus [b]_k = \theta ([a]_n) \theta ([b]_n)$.

This mapping is onto, since for any $[a]_k \in \mathbb{Z}_k$ we can find a $[a]_n \in \mathbb{Z}_n$ such that $\theta ([a]_n) = [a]_k$ because $k \mid n$ (should I explain this part more?). So that we have $\theta [ \mathbb{Z}_n] = \mathbb{Z}_k$. Clearly the kernel of this homomorphism is $\left< [k]_n \right> = K$.

Thus we can define a mapping, $\phi : \mathbb{Z}_n /K \mapsto \mathbb{Z}_k$ by $\phi (Ka) = \theta (a)$ for each $Ka \in \mathbb{Z}_n /K$.

This mapping is an isomorphism of $\mathbb{Z}_n / K$ onto $\mathbb{Z}_k$ by the Fundamental Homomorphism Theorem. So that $\mathbb{Z}_n /K \approx \mathbb{Z}_k$

$QED$
• April 6th 2008, 09:32 AM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
This mapping is onto, since for any $[a]_k \in \mathbb{Z}_k$ we can find a $[a]_n \in \mathbb{Z}_n$ such that $\theta ([a]_n) = [a]_k$ because $k \mid n$ (should I explain this part more?)

You are overkilling it. If $[a]_k \in \mathbb{Z}_k$ then $[a]_n \in \mathbb{Z}_n$ and $\theta ([a]_n) = [a]_k$.
• April 6th 2008, 07:02 PM
ThePerfectHacker
This can also be proved using the third isomorphism theorem. Note that $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$. And $k\mathbb{Z}_n = k\mathbb{Z}/n\mathbb{Z}$*.
Thus, $\mathbb{Z}_n / k\mathbb{Z}_n = (\mathbb{Z}/n\mathbb{Z})/(k\mathbb{Z}/n\mathbb{Z}) \simeq \mathbb{Z}/k\mathbb{Z} = \mathbb{Z}_k$.

*)Where $m\mathbb{Z} = \{ 0,\pm m,\pm 2 m, ... \}$.
• April 6th 2008, 07:09 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
This can also be proved using the third isomorphism theorem. Note that $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$. And $k\mathbb{Z}_n = k\mathbb{Z}/n\mathbb{Z}$*.
Thus, $\mathbb{Z}_n / k\mathbb{Z}_n = (\mathbb{Z}/n\mathbb{Z})/(k\mathbb{Z}/n\mathbb{Z}) \simeq \mathbb{Z}/k\mathbb{Z} = \mathbb{Z}_k$.

*)Where $m\mathbb{Z} = \{ 0,\pm m,\pm 2 m, ... \}$.

that is very interesting. it seems you can treat it just like a quotient and actually "cancel" the $n \mathbb{Z}$. very nice.

thanks for the link as well. i'll read through it when i have the time, or maybe we'll get to this in class and i can use it as supplementary material
• April 6th 2008, 07:28 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
that is very interesting. it seems you can treat it just like a quotient and actually "cancel" the $n \mathbb{Z}$. very nice.

Exactly. This is also interesting because it is a quotient group of a quotient group. So you form the cosets. And using the cosets you form a cosets of cosets.

This is Mine 92:):)th Post!!!
• April 6th 2008, 07:52 PM
ThePerfectHacker
It happens to be a good problem to work on if you want to work on your homomorphism.

Third Theorem: If $N\triangleleft H\triangleleft G$ then $H/N\triangleleft G/N$ and $(G/N)/(H/N) \simeq G/H$.

Try to prove it. Hint: Define $\theta: G/N\mapsto G/H$ as $\theta (xN) = xH$.