1. ## Modulo

I have two questions about modulo n:

1). In my notes I have $\mathbb{Z}_n$ quoted everywhere! My big problem is: how do I know if this is addition modulo n or multiplication modulo n?

2). My second question is that I have this:

$\mathbb{Z}_n \oplus \mathbb{Z}_m$.

What does this mean? I know it's the direct sum, so does it mean this:

For example, $\mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2 \} \oplus \{0,1,2,3 \}=\{0,2,4,3\}$?

(I just went through and added the 1st number of each set to the 1st number of the other (and so on)). My big problem was that I didn't have anything to add to the 3 =S

If someone can clarify these it would help a lot!

2. If you're talking about group theory, you probably mean addition modulo n.

3. Thanks!

I don't suppose you also know how to do the direct product as well? I'm getting stuck on this proof i'm trying to do!!

4. Originally Posted by Showcase_22

1). In my notes I have $\mathbb{Z}_n$ quoted everywhere! My big problem is: how do I know if this is addition modulo n or multiplication modulo n?
$\mathbb{Z}_n$ is just a number system at this point, there is no operation. You can apply either addition or multiplication to it.

Originally Posted by Showcase_22

2). My second question is that I have this:

$\mathbb{Z}_n \oplus \mathbb{Z}_m$.

What does this mean? I know it's the direct sum, so does it mean this:

For example, $\mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2 \} \oplus \{0,1,2,3 \}=\{0,2,4,3\}$?

(I just went through and added the 1st number of each set to the 1st number of the other (and so on)). My big problem was that I didn't have anything to add to the 3 =S

If someone can clarify these it would help a lot!
Not 100% sure here but this is what I would suggest.

I would take the first element of the first set and add it to each element of the second, then take the second element of the first set and add it to every element of the second and so on.

$\mathbb{Z}_3 \oplus \mathbb{Z}_4$ will be the set of distinct solutions.

5. The question says that $\mathbb{Z}_n+\mathbb{Z}_m=\mathbb{Z}_{mn}$ if n and m are relatively prime.

So I used the example $\mathbb{Z}_3 \oplus \mathbb{Z}_4$ since 3 and 4 are coprime (and hence relatively prime).

$\mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2\} \oplus \{0,1,2,3 \}=\{0,1,2,3,1,2,3,4,2,3,4,5 \}=\{0,1,2,3,4,5 \} \neq \mathbb{Z}_{12}$

I really can't fathom what this question is going on about =S

6. ## Direct Sum

The symbol

$
\mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2\} \oplus \{0,1,2,3 \}=
$

is often used to denote the set of ordered pairs, with the first element from the first set, and the second from the second set. Namely:

(0,0), (0,1), (0,2), (0,3),
(1,0), (1,1), (1,2), (1,3),
(2,0), (2,1), (2,2), (2,3)

7. Originally Posted by Showcase_22
The question says that $\mathbb{Z}_n+\mathbb{Z}_m=\mathbb{Z}_{mn}$ if n and m are relatively prime.

So I used the example $\mathbb{Z}_3 \oplus \mathbb{Z}_4$ since 3 and 4 are coprime (and hence relatively prime).

$\mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2\} \oplus \{0,1,2,3 \}=\{0,1,2,3,1,2,3,4,2,3,4,5 \}=\{0,1,2,3,4,5 \} \neq \mathbb{Z}_{12}$

I really can't fathom what this question is going on about =S
Firstly. I assume you know what $A\times B$ is? The cartestian product $A\times B=\left\{(x,y): x\in A\text{ }y\in B\right\}$. Well it is fairly simple to show that if $G,G'$ are groups then $G\times G'$ is a group with $(g_1,g_1')(g_2,g_2')=\left(g_1g_2,g_1'g_2'\right)$. When the two groups are abelian it is customary to write $G\oplus G'$ in lieu of $G\times G'$

Secondly, you need to be a tad careful here. Firstly it should be $(m,n)=1\implies \mathbb{Z}_n\oplus\mathbb{Z}_m\cong \mathbb{Z}_{nm}$. The equals sign is clearly false since the LHS is a set of ordered pairs and the RHS is not. But as is indicated the two groups are "isomorphic" (I believe from previous threads that you are familiar with this concept). To see why the above is true here is a brief outline. For $\mathbb{Z}_n\oplus\mathbb{Z}_m$ to be cyclic we need there to be a generator of order $\left|\mathbb{Z}_n\right|\cdot\left|\mathbb{Z}_m\r ight|=mn$. Note that for any $z\in\mathbb{Z}_n\oplus\mathbb{Z}_m$ it is true that $z^{\text{lcm}(m,n)}=e$ (why?). Therefore, to have (and this of course is not a formal proof) a generator of order $mn$ we must have that $\text{lcm}(m,n)=mn\implies (m,n)=1$. So if $(m,n)=1$ then $\mathbb{Z}_n\oplus\mathbb{Z}_n$ is a cyclic group of order $mn$. Also, you should know that any cyclic group of order $\ell$ is isomorphic to $\mathbb{Z}_{\ell}$. From there the rest is trivial.

8. $

(m,n)=1\implies \mathbb{Z}_n\oplus\mathbb{Z}_m\cong \mathbb{Z}_{nm}
$
Sorry, I did mean to write this!

$

z^{\text{lcm}(m,n)}=e
$
What is your notation here? What's e?

9. $e$ is the identity element.