How Many Left Cosets of <a^4> in <a> Are There?

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<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>.
 
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Muzza said:
<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 theorem that states the order of a^n?
 
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).
 
Muzza said:
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..
 

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