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The duplicate question doesn't really answer my problem.

Given $G$ with order of 15. $G_3$ and $G_5$ are the only sub-groups of $G$ with order of 3 and 5 respectively. (In other words, there could be other sub-groups with different order but none with order 3 that's not $G_3$).

I want to prove that $G$ is Cyclic group.

My try:

To prove that $G$ is Cyclic group I was searching for $a$ in $G$ such that: $a^{|G|}=a^{15}=e$

3 and 5 are prime numbers so $G_3$ and $G_5$ are both Cyclic groups. ie, there are $b$ in $G_3$ and $c$ in $G_5$ such that: $b^3=e, c^5=e$.

if $b=c$ then we finished, since our $a$ would be $bc$ but that's not always correct to assume this equality...

Any help?

Algo
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  • This is the standard way to prove that every group of order $15$ is cyclic. See the duplicates how to finish it. We have $G=G_3\times G_5\cong C_{15}$. – Dietrich Burde Nov 28 '21 at 15:55
  • @DietrichBurde sorry that didn't help. I don't understand what this multiplication means and why it solves my problem. – Algo Nov 28 '21 at 16:09
  • Not every product of two primes is like 15. For example, there is a non-Abelian group of order 21, whose subgroups ( apart from itself, of course) are cyclic (clearly, as 3 and 7 are primes). – Hagen von Eitzen Nov 28 '21 at 16:34
  • Algo, it solves your problem, because it shows that every group of order $15$ is the direct product of its unique subgroups $G_3$ and $G_5$. So it must be cyclic by the Chinese remainder theorem. – Dietrich Burde Nov 28 '21 at 17:14
  • @HagenvonEitzen how to prove =? plus why once I prove this I finish? I don't understand.... – Algo Nov 28 '21 at 17:56

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