# every group of order $105$ has a cyclic normal subgroup of index $3$ ?

Does every group of order $105$ has a cyclic normal subgroup of index $3$ ? (Please don't use Sylow theorems )

• Well since every group of order $35$ is indeed cyclic and subgroups of index "smallest prime divisor of $|G|$" are always normal it would suffice to prove that there is a subgroup of index $3$. Nov 20 '14 at 8:27

You can decompose $105=35\times 3=3\times 5\times 7$. This is a group of order $pqr$ where $p<q<r$ are primes. Note that we cannot have both $n_5>1$ and $n_7>1$ since counting elements gives a contradiction. Now, if we pick $P,Q$ a $5$ Sylow and a $7$ Sylow respectively, at least one is normal, so $PQ$ is a subgroup which is normal, since $3$ is the least prime dividing $105$. It is also cyclic, since $5\nmid 6$. Since $PQ$ is normal, taking $P_1$ to be any other $5$-Sylow and conjugating to get $P$ gives that $P_1$ is a $5$-Sylow of $PQ$, which is abelian, so $P=P_1$. Thus $n_5=1$, and likewise $n_7=1$.