Finding the order of a factor group

[PsychonautQQ]

Homework Statement

G is a cyclic group generated by a, G = <a>. |a| = 24, let K=<a^12>.

Q: In G/K, find the order of the element Ka^5

Work:
K=<a^12> = <1,a^12> --> |K| = 2
|G/K| = |G| / |K| = 24 / 2 = 12,

so |Ka^5| = 1,2,3,4,6 or 12.

now I'm lost ;-/

 

[jbunniii]

If we let $n$ denote the order of $Ka^5$, then what can you say about $(Ka^5)^n$?

 

[PsychonautQQ]

(Ka^5)^n = 1? or perhaps (Ka^5)^n = 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.

How come sometimes order means number of elements in a group?? Ka^5 has the same number of elements as K since you can go from one to the other with a bijection right? or does Ka^5 even have elements? does it have more elements than K?

 

[jbunniii]

(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 $Ka^5 = \{k_1a^5, k_2a^5, \ldots k_na^5\}$.