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Assume that $G= \langle a \rangle$ is a cyclic group, and $H$ is a subgroup of $G$. Assume $H= \langle a^k \rangle$ , where $k$ is the smallest positive integer such that $a^k \in H$. Note that $G/H$ is cyclic, since $G$ is cyclic. Assume that $[G:H]=t$. Prove that $t=k$.

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closed as off-topic by Thomas, Davide Giraudo, Dominic Michaelis, user26857, user61527 Sep 20 '13 at 7:00

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    $\begingroup$ I kind of understand how to prove it, but I have difficulty wright it down formally. My thoughts: G/H = <Ha>, Since [G:H]=t, Thus |G/H| = t. So then I have a difficulty to formulate that k=t, because H, Ha, Ha^1,..., Ha^(k-1) are distinct and are all the cosets. $\endgroup$ – Nadia C Sep 18 '13 at 17:53
  • $\begingroup$ The condition $H\ne\emptyset$ is superfluous because every subgroup contains the neutral element. $\endgroup$ – Matemáticos Chibchas Sep 18 '13 at 18:13
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There are a few ways to do this. One way (if the group is finite) is to show directly that $|H|=\frac{n}{k}$ where $n = |G|$ and then use Lagrange's theorem: $|G|=[G:H]|H|$. (You can show that $|H|$ is the smallest positive integer $m$ such that $(a^k)^m=e$ because $H$ is cyclic, then use this to deduce that $|H|=\frac{n}{k}$.)

Edit: possibly an easier way to do this is to use the fact that $G/H$ is cyclic, so it must have a generator. Clearly $aH$ generates as $(aH)^i=a^iH$ so the size of the group is just the order of $aH$. But this is the smallest positive integer $m$ such that $a^mH=H$ i.e. $a^m \in H$. This is, by definition, $k$.

Another way to do it is to use the fact $|G/H|=[G:H]$ as you seem to be suggesting and then explicitly compute what $G/H$ is. In your comment, you successfully identify the quotient group, but if you want to do so formally:

1) Identify the possible elements of $G/H$: Notice that since all elements in $G$ are of the form $a^i$ for some integer $0\leq i\leq n$, all possible cosets are of the form $a^iH$.

2) Show when two of these elements are the same: You can do this by saying that $a^iH=a^jH$ is equivalent to $a^{i-j}\in H$. Thus $a^iH=a^jH$ iff $k | i-j$.

3)Finally, you can now say that any of the (at most $n$) possible cosets from part 1 can be represented by one of the cosets you describe in your comment (by part 2) and that all of these are distinct (by part 2 again) so you have successfully identified $G/H$ which now clearly has $k$ elements.

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  • $\begingroup$ Can you, please, just wright down the proof how you would do it for a teacher. Cause you just outlined it. $\endgroup$ – Nadia C Sep 18 '13 at 18:45
  • $\begingroup$ @NadiaC Sorry, I don't like giving word-for-word answers to questions. There should be enough here for you to write out something you could show to a teacher. Try following the steps outlined and cutting out anything that feels too obvious and also try and make them more concise. $\endgroup$ – Tom Oldfield Sep 18 '13 at 19:29
  • $\begingroup$ I was asking how to formulate it formally. I get the outline. My difficulty is to communicate it precisely. $\endgroup$ – Nadia C Sep 18 '13 at 20:42
  • $\begingroup$ Your first way does not work if $G = \Bbb{Z}$. $\endgroup$ – ronno Sep 19 '13 at 9:56
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    $\begingroup$ @NadiaC The ability to communicate precisely can be improved far more efficiently by practice, rather than seeing examples. My answer shows what points need to be made and roughly how they relate to each other, I think that you'd be far better off trying to make things precise on your own. You will see many examples of clear, concise proofs in lectures and books, so I think that you should take this opportunity to practice. $\endgroup$ – Tom Oldfield Sep 19 '13 at 13:40

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