Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12

Example of Use of Sylow Theorems

In a group of order $12$, there are either $1$ or $4$ Sylow $3$-subgroups.

Proof

Let $G$ be a group of order $12$.

Let $n_3$ be the number of Sylow $3$-subgroups in $G$.

From the Fourth Sylow Theorem, $n_3$ is congruent to $1$ modulo $3$, that is, in $\set {1, 4, 7, \ldots}$

Let $H$ be a Sylow $3$-subgroup of $G$.

We have that:

$12 = 4 \times 3$

and so the order of $H$ is $3$.

Thus:

\(\ds \index G H\)

\(=\)

\(\ds \dfrac {12} {3}\)

\(\ds \)

\(=\)

\(\ds 4\)

From the Fifth Sylow Theorem:

$n_3 \divides 4$

where $\divides$ denotes divisibility.

Thus there may be $1$ or $4$ Sylow $3$-subgroups of $G$.

$\blacksquare$

Sources

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