Sylow Theorems/Examples/Sylow 2-Subgroups in Group of Order 48

Example of Use of Sylow Theorems

In a group of order $48$, there are either $1$ or $3$ Sylow $2$-subgroups.

Proof

Let $G$ be a group of order $48$.

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:

$48 = 3 \times 2^4$

and so the order of $H$ is $16$.

Thus:

\(\ds \index G H\)

\(=\)

\(\ds \dfrac {48} {16}\)

\(\ds \)

\(=\)

\(\ds 3\)

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 $1 \ \text{(b)}$