Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12
< Sylow Theorems | Examples
Jump to navigation
Jump to search
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 $1 \ \text{(e)}$