Direct Product of Sylow p-Subgroups is Sylow p-Subgroup

Theorem

Let $G_1$ and $G_2$ be groups.

Let $H_1$ and $H_2$ be subgroups of $G_1$ and $G_2$ respectively.

Let $H_1$ be a Sylow $p$-subgroup of $G_1$.

Let $H_2$ be a Sylow $p$-subgroup of $G_2$.

Then $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.

Proof

By definition of Sylow $p$-subgroup:

$\order {H_1} = p^r$, where $p^r$ is the highest power of $p$ which is a divisor of $\order {G_1}$.

$\order {H_2} = p^s$, where $p^s$ is the highest power of $p$ which is a divisor of $\order {G_2}$.

We have that:

$\order {H_1 \times H_2} = p^{r + s}$

We also have that $p^{r + s}$ is the highest power of $p$ which is a divisor of $\order {G_1 \times G_2}$.

Hence it follows by definition that $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Exercise $2$