Group of Order 42 has Normal Subgroup of Order 7
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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.
$\blacksquare$