I want to prove that 'Let $G$ be a group containing normal subgroups of orders $3$ and $5$, respectively. Prove $G$ contains an element of order $15$'.

How can I use this fact for proof: 'Let $r$ and $s$ be relatively prime integers. A cyclic group of order $rs$ is isomorphic to the product of a cyclic group of order $r$ and a cyclic group of order $s$.' ?

I first thought that I get the proof obviously, but then I saw 'normal' in question, and stopped.

Also, the cyclic subgroup generated by that order $15$ element has to be normal in $G$?


Here is a roadmap:

  • Prove that $H_3 H_5 = H_5 H_3$

  • Prove that $H_3 H_5$ is a normal subgroup

  • Prove that $H_3 H_5 \cong H_3 \times H_5 \cong C_3 \times C_5 \cong C_{15}$

  • $\begingroup$ What happens when normality is taken away? We need normality in proof of $H_3 H_5 = H_5 H_3$? $\endgroup$ – Silent Apr 12 '18 at 10:47
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    $\begingroup$ The result is no longer true. For example $S_5$ has elements of order $3$ and $5$ but no element of order $15$. $\endgroup$ – Derek Holt Apr 12 '18 at 10:48
  • $\begingroup$ @DerekHolt Thank you. When normality assumed, can $C_{15}$ be non-normal in $G$? Also we need normality in $H_3 H_5 = H_5 H_3$, right? $\endgroup$ – Silent Apr 12 '18 at 10:50
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    $\begingroup$ Yes, if $M$ and $N$ are normal subgroups of any group $G$, then so is $MN = NM$. (Since you have accepted the answer, you must understand everything already!) $\endgroup$ – Derek Holt Apr 12 '18 at 11:44
  • $\begingroup$ @DerekHolt Do we really need HK to be normal in G , is it not enough for HK to be just a subgroup of G ? and since H and K has order 3 and 5 they contain an element of order 3 and 5 , x , y , resp, so the order of that element xy is 15. $\endgroup$ – Kasmir Khaan Jan 10 at 8:31

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