Abstract Algebra - Cyclic groups

[basketm19]

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. 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).

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.

 

[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?

 

[basketm19]

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

 

[micromass]

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

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.

 

[blue_raver22]

Dear student,

Thank you for sharing your question on cyclic groups in abstract algebra. Your attempt at solving the problem is a good start. Let's continue with your solution and see if we can find the value of |a|.

From your work, we have determined that the possible values for n are 4, 8, 12, 16, 20, 24, ... , so on. However, we can narrow down the possible values of n further by considering the fact that the order of an element in a cyclic group must divide the order of the group. In this case, the group element a has order n, but what is the order of the group?

We can use the fact that |a^28| = 10 to find the order of the group. Since |a^28| = 10, this means that the order of the group must be a multiple of 10. In other words, the order of the group is either 10, 20, 30, 40, ... , so on. This narrows down the possible values of n to 20, 40, 60, 80, ... , so on.

Now, let's plug in these possible values of n into our equations:

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

For n = 20, we have 20/gcd(20, 28) = 10 and 20/gcd(20, 22) = 20, which are both true.

Therefore, we can conclude that |a| = n = 20 is the only possible value that satisfies both equations.

I hope this helps you in your understanding of cyclic groups. Keep up the good work in your studies and good luck on your exam!