# Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup

## Theorem

Let $G$ be a group of order $56$.

Then $G$ has either:

exactly one normal Sylow $2$-subgroup

or:

exactly one normal Sylow $7$-subgroup.

## Proof

Let $G$ be a group of order $56$ whose identity is $e$.

We have that:

$56 = 2^3 \times 7$

From the First Sylow Theorem, $G$ has at least one Sylow $2$-subgroup and Sylow $7$-subgroup.

Let:

$n_2$ denote the number of Sylow $2$-subgroups of $G$
$n_7$ denote the number of Sylow $7$-subgroups of $G$.
if $n_2 = 1$ then the unique Sylow $2$-subgroup is normal

and:

if $n_7 = 1$ then the unique Sylow $7$-subgroup is normal.

It remains to be shown that either $n_2 = 1$ or $n_7 = 1$.

By the Fourth Sylow Theorem:

$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd)

and from the Fifth Sylow Theorem:

$n_2 \divides 56$

where $\divides$ denotes divisibility.

It follows that $n_2 \in \set {1, 7}$.

By the Fourth Sylow Theorem:

$n_7 \equiv 1 \pmod 7$

and from the Fifth Sylow Theorem:

$n_7 \divides 56$

where $\divides$ denotes divisibility.

It follows that $n_7 \in {1, 8}$.

If either $n_2 = 1$ or $n_7 = 1$ the proof is finished.

Suppose $n_7 = 8$.

As all of these Sylow $7$-subgroups are prime, the intersection of any $2$ of them is $\set e$.

Thus, these $8$ Sylow $7$-subgroups contribute $6 \times 8 = 48$ distinct elements to $G$.

This leaves $7$ elements still to be accounted for.

A Sylow $2$-subgroup is of order $2^3 = 8$.

Thus all $7$ of these remaining elements must all be in that one Sylow $2$-subgroup

So if $n_7 \ne 1$, then $n_2 = 1$.

The result follows.

$\blacksquare$