Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 12
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Example of Use of Sylow Theorems
In a group of order $12$, there are either $1$ or $3$ Sylow $2$-subgroups.
Proof
Let $G$ be a group of order $12$.
Let $n_2$ be the number of Sylow $2$-subgroups in $G$.
From the Fourth Sylow Theorem, $n_2$ is congruent to $1$ modulo $2$, that is, odd.
Let $H$ be a Sylow $2$-subgroup of $G$.
We have that:
- $12 = 3 \times 2^2$
and so the order of $H$ is $4$.
Thus:
\(\ds \index G H\) | \(=\) | \(\ds \dfrac {12} {4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3\) |
From the Fifth Sylow Theorem:
- $n_2 \divides 3$
where $\divides$ denotes divisibility.
Thus there may be $1$ or $3$ Sylow $2$-subgroups of $G$.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Exercise $1 \ \text{(d)}$