let order of a=24. find generator of <a to the power 21> intersect <a to the power 10>?

Question

suppose that order of a = 24 ..find a generator for intersect .... pleas explain well to me because i do not understand this topic well

Answer 1

I gather you have a group, and a is an element of the group with order 24. I.e. a^24=1 (where I just call the identity "1"), and so do a^48, a^96, and so on. But no other powers of a equal 1.

Since 7 and 24 are relatively prime, the group generated by a^7 is exactly the same as the group generated by a. That is, if you list a, a^2, a^3, ..., a^24 =1 and b, b^2, b^3, ..., b^24 =1, where b = a^7, you get exactly the same list, just in a different order. (The last/24th terms are the same, but most of the others are different.)

I.e., the group generated by a^7 is exactly the same as the group generated by a. And similarly the group generated by a^21 is exactly the same as the group generated by a^3.

Similarly, the group generated by a^10 is exactly the same as the group generated by a^2.

Proving to yourself all of what I said above isn't exactly necessary to solving the problem, but it certainly will make understanding things a lot easier.

Now, let's tackle the problem head-on. To be in the group generated by a^21 means exactly to equal a^n, where n is congruent to a multiple of 21, modulo 24. (Because 24 is the order of a.) Factor 21 into 3 x 7, where 3 divides 24 and 7 is relatively prime to 24. n is congruent to a multiple of 21 modulo 24 if and only if n is congruent to a multiple of 3 modulo 24.

Similarly, a^n is in the group generated by a^10 if and only if n is congruent to a multiple of 10 and hence also to a multiple of 2 modulo 24. But n is congruent to multiples of 2 and 3 modulo 24 if and only if n is congruent to 6 modulo 24.

So the intersection is the order-4 group {a^6, a^12, a^18, a^24=1}.