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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.

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    $\begingroup$ 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
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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}

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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