# Abstract Algebra - Cyclic groups

1. Oct 26, 2011

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

I was doing some practice problems for my exam next week and I could not figure this out. (This is my first post on PF btw)

2. Relevant 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).

3. Attempt at 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.

2. Oct 26, 2011

### Deveno

gcd(440,280) = 40.

this means we can find integers a,b with

440a + 280b = 40, that is

11a + 7b = 1. a = 2, b = -3 will work.

thus a^40 = (a^(880))(a^-(840))

= (a^(440))^2(a^(280))^(-3)

= ((a^22)^20)^2((a^28)^10)^(-3)

= e^2(e^-3) = ee = e.

that means |a| divides 40, which gives us just 1,2,4,5,8,10,20 and 40 as possibilities.

since a^28 ≠ e, we can rule out 1,2, and 4 as possibilities.

suppose a^5 = e. then a^28 would have order 5,

since (a^28)^5 = (a^5)^28. similarly, we can rule out

every divisor of 10 and 20, leaving just 8 and 40.

so suppose |a| = 8.

then (a^28)^2 = a^56 = (a^8)^7 = e, but |a^28| = 10.

what's left?

3. Oct 26, 2011

Thanks for the response. But I don't understand why you are starting out with gcd(440,280) = 40.

4. Oct 26, 2011

### micromass

Staff Emeritus
We know that the order of $a^{28}$ is 10. So $a^{280}=e$. Similarly, we know that $a^{440}=e$.

For each integers x and y, it follows that $a^{280x+440y}=e$. We wish to minimize 280x+440y (since this will be a small number such that the order divides this number). The minimal number is exactly gcd(280,440). This is why he started by this.