I am trying to prove that a group of order $105$ has a subgroup of order $21$. I know it can be done using Sylow theorems, I was just wondering if the proof below could be another way of doing that.
$3$ is prime and $3$ divides $105$, so $G$ has an element of order $3$, say $x$ (By Cauchy's). Similarly, let $y$ be an element of order $7$. Then $xy$ is in $G$, and since $\gcd(3,7)=1$ order of $xy$ is $21$. Then the group $\langle xy \rangle$ generated by $xy$ has an order $21$ and is a subgroup of $G$.
I am a little concerned about my proof, since it implies that group of order $21$ is cyclic $\Rightarrow$ abelian. Is there a problem with that? I know if $p$ doesn't divide $q-1$ then it is true, but in this case $3$ divides $7-1$.