Group of order $105$ has a subgroup of order $21$

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 $7-1$.

Thanks,

Solution 1:

Hints for an elementary approach:

Step #1: Show that if $P_3$ is a Sylow 3-subgroup, then $5\mid |N_G(P_3)|$. Thus $N_G(P_3)$ has a subgroup $H$ of order 15. A standard exercise shows that $H$ is cyclic.

Step #2: The number of Sylow 5-subgroups is either 1 or 21. By Step #1 at least one, hence all, of the Sylow 5-subgroups have a normalizer of size at least ??? Therefore the number of Sylow 5-subgroups is ???

Step #3: Identify all the automorphisms of $C_5$ of order that is a factor of seven. Conclude that if a $P_5$ is normalized by an element of order 7, then there exist elements of orders 5 and 7 that commute.

Step #4: Using steps #2 and #3 conclude that the normalizer of at least one, hence all, Sylow 7-subgroups is at least 35. Show that the group $G$ has a normal Sylow 7-subgroup $P_7$.

Step #5: Show that the quotient group $G/P_7$ has a subgroup of order 3. Apply the correspondence theorem (as in Step #1).

Solution 2:

Another approach is that there are $1$ or $15$ Sylow $7$-subgroups. If there were $15,$ then there would be $90$ elements of order $7.$ Hence there could only be one Sylow $5$-subgroup which is normal. There can't be $7$ Sylow $3$-subgroups in tht case either, otherwise we would have $14$ elements of order $3$, and no room for any elements of order $5,$ which there must be. Hence if there were $15$ Sylow $7$-subgroups, there would be a normal cyclic subgroup of order $15.$ In that case, a Sylow $7$-subgroup would centralize a Sylow $3$-subgroup since a group of order $3$ admits no automorphism of order $7$.Similarly a Sylow $7$-subgroup centralizes a Sylow $5$-subgroup, leading to the contradiction that the Sylow $7$-subgroup is normal. If there is one Sylow $7$-subgroup, it is normal, and is normalized by a Sylow $3$-subgroup, yielding a subgroup of order $21$.