Let $G = <a>$ be a cyclic group of order 15, and $H = <a^{10}>$. List all the left cosets of $H$ in $G$.
I know that $<a> = \{a^0, a^1,…,a^{15}\}$. The problem doesn't say that $H$ is a cyclic subgroup so I'm if I assume that $H = <a^{10}> = \{a^{10}\}$.
A left co set would be say $x \in G, xH = \{xh: h \in H\}$ So then all of the left cosets would be $\{a^{15}\}, \{a^{16}\},…,\{a^{25}\}.$
I don't think this is correct but my book doesn't really give any details aside from the definition of a cyclic group. Any help in solving this or pointing me in the right direction would be appreciated.
Best Answer
I gave a similar solution here.
For brevity, I'll just give a general idea: first, you should determine the exact contents of H. You can do this just by repeatedly exponentiating a10 until you get e. You should see that H={a10, a5, e}.
Then, you go through elements of G and find what coset they produce. You already know that a10, a5, and e are in < a10>. You should look at elements that you haven't yet seen in a coset, and see what you get. For example, a*< a10>={a11, a6, a}. Just keep repeating until every element has been put in a coset. Note that cosets partition the group, so you should see each element once and only once, and that each coset should be the same size- that of the subgroup, so here 3- so you should get 5 cosets total since there are 15 elements in the whole group.