Abstract Algebra - Cyclic groups

[basketm19]

(This is my first post on PF btw - I posted on this another thread, but I'm not sure if I was supposed to)

I was doing some practice problems for my exam next week and I could not figure this out.

Homework Statement

Suppose a is a group element such that |a^28| = 10 and |a^22| = 20. Determine |a|.

Homework Equations

Let a be element of order n in group and let k be a positive integer. Then <a^k> = <a^gcd(n,k)> and |a^k| = n/gcd(n,k).

The Attempt at a Solution

10 = n/gcd(n, 28); 20 = n/gcd(n, 22)

Setting n equal to each other, 10gcd(n, 28) = 20gcd(n,22)

gcd(n, 28) = 2gcd(n, 22)

The possible values for n are 4, 8, 12, 16, 20, 24, ... , so on.

Not sure where to go from here.

 

[mighty2000]

Hello! Welcome to the Physics Forums. It looks like you have made some good progress on this problem. Let's continue working through it together.

First, let's use the information given to find the greatest common divisor (gcd) of 28 and 22. We can do this by listing the factors of both numbers and finding the largest one they have in common.

Factors of 28: 1, 2, 4, 7, 14, 28
Factors of 22: 1, 2, 11, 22

The largest number they have in common is 2, so gcd(28, 22) = 2.

Now, we can use the equation |a^k| = n/gcd(n,k) to find the possible values for n. Since we know that |a^28| = 10 and |a^22| = 20, we can substitute these values into the equation to get:

|a^28| = n/gcd(n,28) = 10
|a^22| = n/gcd(n,22) = 20

We can rearrange these equations to get:

n = 10 gcd(n,28)
n = 20 gcd(n,22)

Since we know that gcd(n,28) = 2 and gcd(n,22) = 2, we can substitute these values into the equations to get:

n = 10(2) = 20
n = 20(2) = 40

So, the possible values for n are 20 and 40. Now, we can use the equation |a^k| = n/gcd(n,k) to find the order of a. We know that gcd(n,28) = 2, so we can substitute this into the equation to get:

|a| = n/gcd(n,28) = 20/2 = 10

Therefore, |a| = 10. I hope this helps! Let me know if you have any other questions. Good luck on your exam next week!