Sylow Theorems/Examples/Sylow 7-Subgroups in Group of Order 28

Example of Use of Sylow Theorems

In a group of order $28$, there exists exactly $1$ Sylow $7$-subgroup.

This Sylow $7$-subgroup is normal in $G$.

Proof

Let $G$ be a group of order $28$.

From the Fourth Sylow Theorem, the number of Sylow $7$-subgroups is congruent to $1$ modulo $7$.

From the Fifth Sylow Theorem, the number of Sylow $7$-subgroups is a divisor of $\dfrac {28} 7 = 4$.

Thus there is exactly $1$ Sylow $7$-subgroup of $G$.

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

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(a)}$