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.

  • 3
    $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) $\endgroup$ – Patrick Stevens May 20 '18 at 17:22
  • $\begingroup$ I think you are overlooking a fact that $\langle a \rangle $ is cyclic and therefore abelian. Thus every subgroup is normal. $\endgroup$ – hardmath May 20 '18 at 17:24
  • $\begingroup$ Patrick Stevens - I don't know what you're talking about. The question has been posed on this forum exactly as it was posed to me. There's no issue with what's displayed. $\endgroup$ – Mhan7 May 20 '18 at 17:55
  • $\begingroup$ hardmath - I understand that point, but how do we apply the index to identifying the elements of the subgroups? $\endgroup$ – Mhan7 May 20 '18 at 17:55
  • $\begingroup$ @Mhan7: There's a clear difference; see the history. $\endgroup$ – hmakholm left over Monica May 20 '18 at 19:20


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}

| cite | improve this answer | |
  • $\begingroup$ But I'm not looking for order, am I? I'm looking for all of the elements of the left cosets. $\endgroup$ – Mhan7 May 20 '18 at 18:15
  • 1
    $\begingroup$ The number of left cosets is the number of elements of the quotient. Then you can use Lagrange's theorem. $\endgroup$ – Bernard May 20 '18 at 18:32
  • $\begingroup$ Right, but once I have that "index", now what? I know there are 5 left cosets, and that there are 3 elements in each coset. Now... about those 3 elements in the index? They are generators for the remainders of the cosets. Please explain... $\endgroup$ – Mhan7 May 20 '18 at 18:41
  • 1
    $\begingroup$ The three items are just the elements of the subgroup generated by $a^5$. What else can I say? $\endgroup$ – Bernard May 20 '18 at 19:11
  • 1
    $\begingroup$ I'm not sure I fully understand. The values of $\langle a^5\rangle$ are just the distinct powers of $a^5$. As soon as you obtain a multiple of $15$, the power of $a^5$ is equal to $1$, so we're back at the starting point. Maybe an arithmetic explanation will be more illuminating: $\dfrac{n}{\gcd(k,n)}$ is the least integer $\ell$ such that $\ell\, k=\operatorname{lcm}(n,k)$. $\endgroup$ – Bernard May 20 '18 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.