Group of Order 40 has Normal Subgroup of Order 5

Theorem

Let $G$ be of order $40$.

Then $G$ has a normal subgroup of order $5$.

Proof

We have that:

$40 = 2^3 \times 5$

From the First Sylow Theorem, $G$ has at least one Sylow $5$-subgroup, which is of order $5$.

Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.

From the Fourth Sylow Theorem:

$n_5 \equiv 1 \pmod 5$

and from the Fifth Sylow Theorem:

$n_5 \divides 8$

where $\divides$ denotes divisibility.

It follows that $n_5 = 1$.

From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $5$-subgroup is normal.

$\blacksquare$