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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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 '19 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.