Prove that CnXCm isomorphic to Cgcd(m,n)XClcm(m,n)

  • MHB
  • Thread starter Alex224
  • Start date
In summary, Alex224 is trying to solve a problem involving cyclic groups, but is stuck. He has come across two basic theorems which he will need to use to prove the problem.
  • #1
Alex224
3
0
Hi, I have a problem in my homework that I am stuck with.
Let there be two natural numbers n,m.
let there be d = greatest common divisor of m and n - gcd(m,n)
and l = least common multiple of m and n - lcm(m,n)
I need to prove that CnXCm isomprphic to ClXCd (Cm Cn Cl Cd are all cyclic groups)

I have tried to see what happens if I look at m and n as products of prime numbers but I am kind of stuck around that idea without knowing where to take it.
I also think I should use the fact that gcd(m,n)*lcm(m,n) = m*n but also, can't figure out where to take it where to take it.
Another thing, if the gcd(m,n) = 1 I know that CmXCn is cyclic and the order of it is mn. I thought maybe this fact could help me somehow (m*n = l*d), but I don't know how, because it is possible that gcd(m,n) > 1

can somebody push me towards the right path to solution?
thanks!
 
Physics news on Phys.org
  • #2
Hi Alex224,

Are you still interested in a solution to this problem? I have something worked out and want to make sure you're still interested before doing a write up.
 
  • #3
GJA said:
Hi Alex224,

Are you still interested in a solution to this problem? I have something worked out and want to make sure you're still interested before doing a write up.

yes I am interested. thank you!
 
  • #4
I'm not sure what book you're using or what theorems you my have come across in your studies, so I will quote the two we'll be needing here. Regardless, these are basic theorems for cyclic groups and will very likely be in any textbook you come across.

Theorem 1

If $G$ is a finite cyclic group with order $n$, then $G\cong \mathbb{Z}_{n}.$

Theorem 2

Let $n$ be a positive integer which has prime decomposition $n=p_{1}^{i_{1}}p_{2}^{i_{2}}\cdots p_{k}^{i_{k}}$, where $p_{1}<p_{2}<\ldots < p_{k}.$ Then $$\mathbb{Z}_{n}\cong\mathbb{Z}_{p_{1}^{i_{1}}}\times\mathbb{Z}_{p_{2}^{i_{2}}}\times\cdots\times\mathbb{Z}_{p_{k}^{i_{k}}}.$$

Exercise Proof Sketch

According to Theorem 1, it suffices to prove the exercise using $\mathbb{Z}_{m}$ and $\mathbb{Z}_{n}.$ Let $m$ and $n$ have the following prime decompositions: $m = p_{1}^{i_{1}}\cdots p_{k}^{i_{k}}$ and $n = q_{1}^{j_{1}}\cdots q_{s}^{j_{s}}.$

Case 1: $m$ and $n$ are relatively prime.

This is a short exercise using Theorem 2 above that I will let you try first.

Case 2: $m$ and $n$ are not relatively prime.

Since $m$ and $n$ are not relatively prime, they share prime factors in their decompositions. For the sake of concreteness, suppose they share $r$ common primes, where $1\leq r\leq \min\{k,s\}$. Since multiplication is commutative, we may assume without loss of generality that $p_{1} = q_{1},\ldots, p_{2} = q_{2},\ldots,$ and $p_{r} = q_{r}.$

Now, note that gcd$(m,n) = p_{1}^{*_{1}}\cdots p_{r}^{*_{r}},$ where $*_{t} = \min\{i_{t}, j_{t}\}$ for $1\leq t\leq r,$ and lcm$(m,n) = p_{1}^{**_{1}}\cdots p_{r}^{**_{r}}p_{r+1}^{i_{r+1}}\cdots p_{k}^{i_{k}}q_{r+1}^{j_{r+1}}\cdots q_{s}^{j_{s}},$ where $**_{t}=\max\{i_{t},j_{t}\}$ for $1\leq t\leq r.$

According to Theorem 2, it follows that

$$\mathbb{Z}_{m}\times\mathbb{Z}_{n}\cong \underbrace{\mathbb{Z}_{p_{1}^{i_{1}}}\times\cdots\mathbb{Z}_{p_{r}^{i_{r}}}\times\mathbb{Z}_{p_{r+1}^{i_{r+1}}}\times\cdots\times\mathbb{Z}_{p_{k}^{i_{k}}}}_{\mathbb{Z}_{m}}\times\underbrace{\mathbb{Z}_{p_{1}^{j_{1}}}\times\cdots\mathbb{Z}_{p_{r}^{j_{r}}}\times\mathbb{Z}_{q_{r+1}^{j_{r+1}}}\times\cdots\times\mathbb{Z}_{q_{s}^{j_{s}}}}_{\mathbb{Z}_{n}}\qquad (1).$$ From this point we are nearly done. I will leave it to you to try and use the facts about gcd$(m,n)$ and lcm$(m,n)$ above to rewrite equation (1) using the $p^{*}$ and $p^{**}$ terms. Once you have that, you can apply Theorem 2 above once more to obtain the result you want.

Please let me know if you have any further questions after making an honest attempt of your own. Good luck!
 
Last edited:

1. What does the equation "CnXCm isomorphic to Cgcd(m,n)XClcm(m,n)" mean?

The equation means that the Cartesian product of two cyclic groups with orders m and n is isomorphic to the Cartesian product of two cyclic groups with orders gcd(m,n) and lcm(m,n).

2. How is isomorphism defined in this context?

In this context, isomorphism refers to a bijective homomorphism between two groups. This means that the two groups have the same structure and can be mapped onto each other in a way that preserves the group operation.

3. What is the significance of the greatest common divisor (gcd) and least common multiple (lcm) in this equation?

The gcd and lcm represent important properties of the orders of the cyclic groups involved. The gcd represents the largest number that divides both m and n, while the lcm represents the smallest number that is divisible by both m and n. These properties are used to determine the orders of the cyclic groups in the Cartesian product.

4. Can you provide an example to illustrate this equation?

For example, let's consider the cyclic groups C4 and C6. The order of C4 is 4 and the order of C6 is 6. The gcd of 4 and 6 is 2 and the lcm is 12. Therefore, according to the equation, C4XC6 is isomorphic to C2XC12.

5. How is this equation relevant in mathematics or other fields?

This equation is relevant in abstract algebra, specifically in group theory. It is used to study the properties and relationships between cyclic groups. It also has applications in computer science, particularly in coding theory and cryptography.

Similar threads

  • Linear and Abstract Algebra
Replies
8
Views
892
  • Math POTW for University Students
Replies
1
Views
530
  • Calculus and Beyond Homework Help
Replies
2
Views
953
  • Linear and Abstract Algebra
Replies
4
Views
937
  • Precalculus Mathematics Homework Help
Replies
3
Views
798
  • Linear and Abstract Algebra
Replies
1
Views
728
  • Linear and Abstract Algebra
Replies
11
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
892
Back
Top