Group of Order 105 has Normal Cyclic Subgroup of Index 3

Theorem
Let $G$ be a group of order $105$.

Then $G$ has a normal cyclic subgroup $N$ such that:
 * $\index G N = 3$

where $\index G N$ denotes the index of $N$ in $G$.

Proof
Let $G$ be a group of order $105$ whose identity is $e$.

From Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup, $G$ has either:
 * exactly one normal Sylow $5$-subgroup

or:
 * exactly one normal Sylow $7$-subgroup.

Suppose $G$ has exactly one normal Sylow $5$-subgroup, which we will denote $P$.

Then $G / P$ is the quotient group of $G$ by $P$ and is of order $21$.

From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G / P$ has exactly one normal Sylow $7$-subgroup which we denote $N / P$.

From the Correspondence Theorem, $N$ is a normal subgroup of $G$ of order $7 \times 5 = 35$.

It follows from Group of Order 35 is Cyclic Group that $N$ is cyclic.

Now suppose that $G$ has exactly one normal Sylow $7$-subgroup, which we will denote $Q$.

Then $G / Q$ is the quotient group of $G$ by $Q$ and is of order $15$.

From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G / Q$ has exactly one normal Sylow $5$-subgroup which we denote $M / Q$.

From the Correspondence Theorem, $m$ is a normal subgroup of $G$ of order $5 \times 7 = 35$.

It follows from Group of Order 35 is Cyclic Group that $M$ is cyclic.

Both cases have been covered, and the existence of a normal cyclic subgroup of order $35$ has been proved.

The result follows by definition of the index of a subgroup.