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Currently working on a Group Theory question:

Let $G$ be a cyclic group of order 100 and let $a \in G$ denote a generator. Find two subgroups of $G$ which contain both $a^{20}$ and $a^{55}$. Are there any other subgroups of $G$ containing these two elements?

So my attempt at this question is taking the given. The order of $G$ is 100 and I know that $a^m = n/gcd(m,n)$. But I'm unsure as to how this helps. Any hints on how to proceed are appreciated.

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Hints: fill in details.

Observe that

$$ord(a^{20})=5\;,\;\;ord(a^{55})=20\;\implies\;\text{any subgroup containing this two elements will}$$

have at least order $\;20\;$ ....but also order $\;20\;$ at most (why? Lagrange theorem), and since any finite cyclic group has one unique proper subgroup of any order dividing the group's order then there is only one such subgroup

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  • $\begingroup$ I think that there are at least two: yours and $G$ itself. $\endgroup$ – Antoine Oct 26 '16 at 18:44
  • $\begingroup$ @Antoine Thanks, edited...though I believe the intention was "proper subgroup". $\endgroup$ – DonAntonio Oct 26 '16 at 18:45
  • $\begingroup$ @DonAntonio I have not touched Lagrange Theorem yet. But to be clear, the only two subgroups that contain both of these elements is a subgroup or order 20 and G itself. I think G is an answer in this case since there was no indication of "proper subgroup" $\endgroup$ – El Spiffy Oct 26 '16 at 19:06
  • $\begingroup$ @ElSpiffy Both groups, $\;G\;$ itself and the subgroup generated by $\;a^{55}\;$ , are indeed the answer. $\endgroup$ – DonAntonio Oct 26 '16 at 19:51

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