PsychonautQQ said:
(Ka^5)^n = 1? or perhaps (Ka^5)^n = K?
##(Ka^5)^n## is equal to the identity element of ##G/K##, which is ##K##.
Ka^5 = {a^5, a^17}
(Ka^5)^12 = {a^120, a^204} = {a^12,1} = K
so... (Ka^5)^12 = K
12 is the order.
Yes, this is correct. As you said in your first post, the only possibilities for ##o(Ka^5)## are 1,2,3,4,6 or 12. So to rule out the numbers smaller than 12, just verify that ##(Ka^5)^m \neq K## if ##m## is 1,2,3,4, or 6. You can do this easily by recognizing that ##(Ka^5)^m = Ka^{5m}##, and that ##Ka^{5m} = K## if and only if ##a^{5m} \in K##, if and only if ##5m## is an integer multiple of 12.
How come sometimes order means number of elements in a group??
Yes, the word "order" is used in a couple of different but related ways. Usually it should be clear from the context which one is meant:
- Order of a group ##G## = the number of elements contained in ##G##. Usually we use the notation ##|G|##.
- Order of an element ##g## = the number of elements in ##\langle g\rangle## (the subgroup generated by ##g##) = the smallest positive ##n## for which ##g^n = 1##. Usually we use the notation ##o(g)## or ##|\langle g\rangle|##. It is better not to write ##|g|## because it creates an ambiguity if ##g## is a coset: should ##|g|## mean the number of elements in the coset, or the order of the cyclic group generated by the coset?
Ka^5 has the same number of elements as K since you can go from one to the other with a bijection right?
Correct, ##Ka^5## is a coset of ##K##, and all cosets of ##K## have the same size, by Lagrange's theorem. But the problem is asking for the order of the
element ##Ka^5## in the group ##G/K##. In other words, what is the smallest positive ##n## such that ##(Ka^5)^n = K##? Just as you answered above.
or does Ka^5 even have elements? does it have more elements than K?
Sure, it has elements, the same number as ##K## has. If ##K = \{k_1, k_2, \ldots, k_n\}## then ##Ka^5 = \{k_1a^5, k_2a^5, \ldots k_na^5\}##.
