Suppose that $a$ has order $15$. Find all of the left cosets of $\langle a^5\rangle $ in $\langle a\rangle$ .
Ok, so I know by Lagrange's Theorem, that the order of the subgroup divides the order of the group. Therefore, the index of the cosets must be $3$.
However… How do I apply this index to the subgroups? I know the final answer, I just want the breakdown, in a meaningful way. The text I use did not really elaborate on this with the notation given. I think I may just be confused for that reason.
Best Answer
Hint:
If the order of an element $a$ is $n$, the order of $a^k$ is $\;\dfrac n{\gcd(k,n)}$.
Some explicit details:
The cosets are : \begin{align} &\langle\mkern2mu a^5\mkern1mu\rangle=\{\, 1,a^5, a^{10}\,\},&&a\,\langle\mkern2mu a^5\mkern1mu\rangle=\{\, a,a^6, a^{11}\,\},&&a^2\langle\mkern2mu a^5\mkern1mu\rangle=\{\, a^2,a^7, a^{12}\,\},\\[1ex] &a^3\langle\mkern2mu a^5\mkern1mu\rangle=\{\, a^3,a^8, a^{13}\,\}, &&a^4\langle\mkern2mu a^5\mkern1mu\rangle=\{\, a^4,a^9, a^{14}\,\}. \end{align}