# Normal Subgroup of Order 25 in Group of Order 100

## Theorem

Let $G$ be a group of order $100$.

Then $G$ has a normal subgroup of order $25$.

## Proof

Let $r$ be the number of Sylow $p$-subgroup of order $5^2 = 25$

The First Sylow Theorem guarantees existence, so $r \ge 1$.

From the Fourth Sylow Theorem:

$r \equiv 1 \pmod 5$

That is:

$r \in \set {1, 6, 11, 16, \ldots}$

From the Fifth Sylow Theorem:

$r \divides \dfrac {100} {25} = 4$

from which it follows that $r = 1$.

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

$\blacksquare$