Group of Order 42 has Normal Subgroup of Order 7

Theorem

Let $G$ be of order $42$.

Then $G$ has a normal subgroup of order $7$.

Proof

We have that:

$42 = 2 \times 3 \times 7$

From the First Sylow Theorem, $G$ has at least one Sylow $7$-subgroup, which is of order $7$.

Let $n_7$ denote the number of Sylow $7$-subgroups of $G$.

From the Fourth Sylow Theorem:

$n_7 \equiv 1 \pmod 7$

and from the Fifth Sylow Theorem:

$n_7 \divides 6$

where $\divides$ denotes divisibility.

It follows that $n_7 = 1$.

From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $7$-subgroup is normal.

$\blacksquare$