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,

• You haven't explained why $xy$ has order $21$, and in fact it need not. – Chris Eagle Dec 27 '11 at 21:56
• I think you're assuming that $x$ and $y$ commute. How else can you be confident of the order of $xy$? – Dylan Moreland Dec 27 '11 at 21:59
• Note that your proof would imply that $S_3$ is cyclic. Indeed $S_3$ has a permutation $x$ of order $2$, and another one $y$ of order $3$....The mistake was explained above ... – N. S. Dec 27 '11 at 22:03
• OK, I got it :) I don't know if x and y commute, so order of xy need not be 21. Thanks a lot. – Jennie Dec 27 '11 at 22:04
• A finite group whose order $n$ is not divisible by the square of any prime (or more generally has no chief factor whose order is divisible by the square of any prime) has subgroups of order $d$ for every divisor $d$ of $n$, but those subgroups don't have to be cyclic (though in the square-free case they are all built from at most two cyclic subgroups, so you might want to look more carefully at the possible groups of order 21 to see the full story isn't much more complicated). – Jack Schmidt Dec 28 '11 at 4:21

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).

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 $21$.