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**Danneedshelp** Q: Suppose that $\displaystyle a$ has order 15. Find all of the left cosets of

$\displaystyle < a^{5} >$ in $\displaystyle < a >$.

Proof. First, lets list the elements of $\displaystyle < a >$

$\displaystyle < a >=\{e, a, a^{2}, a^{3} , a^{4} , a^{5} , a^{6} , a^{7} , a^{8} , a^{9} , a^{10} , a^{11} , a^{12} , a^{13} , a^{14}\}$

The cosets would be:

$\displaystyle < a^{5} >, a < a^{5} >, a^{2} < a^{5} >, a^{3} < a^{5} >, a^{4} < a^{5} > $

I don't understand where the choices $\displaystyle e, a, a^{2}, a^{3}, a^{4}$ came from. I see that there will be 5 cosets, since $\displaystyle |<a^{5}>|||<a>|=\frac{15}{3}=5$, but I don't understand the choices of the cosets.

As I understand it, there will be more cosets, but cosets parition the group into distinct sets. So, once you have found all the cosets whos union is the orginal group, you are done. Correct?