Is there a formula for cosets?

[semidevil]

so let's say |a| = 30. How many left cosets of <a^4> in <a> are there?

ok, so |a| = 30. and think I need to find the order of <a^4> also. I thought the order of it is

<a^4> = e, a^4, a^8, a^12, a^16, a^20, a^24, a^28. so it has order of 8. but my book said the order of it is 15...

is there a formula they used to figure this out?

 

[Muzza]

<a^4> can't have an order of 8 - that would violate Lagrange's theorem.

You've simply forgotten some elements, for example, (a^4)^8 = a^(32) = a^2 is in <a^4>.

 

[semidevil]

<a^4> can't have an order of 8 - that would violate Lagrange's theorem.

You simply forgotten some elements, for example, (a^4)^8 = a^(32) = a^2 is in <a^4>.

ok. I still don't really know and can't find the answer. Is there a formula or theorm that states the order of a^n?

 

[Muzza]

Why can't you find the answer? Just continue like you did, but instead of stopping listing elements when you reach (a^4)^7, continue. It will eventuelly start to repeat.

It's relatively easy to see that the order of a^4 is 15, since (a^4)^k = e <=> a^(4k) = e is equivalent to "|a| divides 4k", i.e. "30 divides 4k" (and the smallest solution to that is k = 15).

 

[semidevil]

Why can't you find the answer? Just continue like you did, but instead of stopping listing elements when you reach (a^4)^7, continue. It will eventuelly start to repeat.

It's relatively easy to see that the order of a^4 is 15, since (a^4)^k = e <=> a^(4k) = e is equivalent to "|a| divides 4k", i.e. "30 divides 4k" (and the smallest solution to that is k = 15).

ah...ok, I get what you mean now...thanx..