group containing normal subgroups of orders $3$ and $5$ contains element of order $15$

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 $G$?

• Prove that $H_3 H_5 = H_5 H_3$
• Prove that $H_3 H_5$ is a normal subgroup
• Prove that $H_3 H_5 \cong H_3 \times H_5 \cong C_3 \times C_5 \cong C_{15}$
• What happens when normality is taken away? We need normality in proof of $H_3 H_5 = H_5 H_3$? – Silent Apr 12 '18 at 10:47
• The result is no longer true. For example $S_5$ has elements of order $3$ and $5$ but no element of order $15$. – Derek Holt Apr 12 '18 at 10:48
• @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? – Silent Apr 12 '18 at 10:50
• 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!) – Derek Holt Apr 12 '18 at 11:44