Problem 5.1.4 (Sylow subgroups of a direct product)

Question

Let $A$ and $B$ be finite groups and let $p$ be a prime. Prove that any Sylow $p$-subgroup of $A 	imes B$ is of the form $P 	imes Q$, where $P \in Syl_p(A)$ and $Q \in Syl_p(B)$. Prove that $n_p(A 	imes B) = n_p(A) n_p(B)$. Generalize both of these results to a direct product of any finite number of finite groups (so that the number of Sylow $p$-subgroups of a direct product is the product of the numbers of Sylow $p$-subgroups of the factors).

Richard Ganaye · Accepted Answer

\begin{proof} Write
$$|A| = p^\alpha r, \ |B| = p^\beta s,\qquad 	ext{where } p 
mid r,\ p 
mid s.$$
Then
\begin{align}
|A 	imes B| = |A||B| = p^{\alpha + \beta} m, \qquad 	ext{where } m = rs, 	ext{ and } p 
mid m.
\end{align}

If  $P \in Syl_p(A)$ and $Q \in Syl_p(B)$, then $|P| = p^\alpha$ and $|Q| = p^\beta$, thus $|P 	imes Q| = p^{\alpha + \beta}$. Therefore, by (1), $P 	imes Q$ is a Sylow $p$-subgroup of $A 	imes B$.

\bigskip
Conversely, suppose that $H$ is a Sylow $p$-subgroup of $A	imes B$. Then $|H| = p^{\alpha + \beta}$.

Put 
\begin{align*}
	ilde{A} &= \{(a , 1) \mid a \in A \} \simeq A,\
	ilde{B} &= \{(1,b) \mid b \in AB\} \simeq B.
\end{align*}
so that $	ilde{A}$ and $	ilde{B}$ are subgroups of $A 	imes B$ isomorphic to $A$ and $B$.

By Sylow's Theorem, there is some Sylow $p$-subgroup $P_0$ of $A$, and some Sylow $p$-subgroup $Q_0$ of $B$, satisfying $|P_0| = p^\alpha$ and $|Q_0| = p^\beta$. Then $	ilde{P}_0 = P_0 	imes \{1\} \in Syl_p(	ilde{A})$ and $	ilde{Q}_0 =  \{1\} 	imes Q_0\in Syl_p(	ilde{B})$, so $	ilde{P}_0 \leq 	ilde{A}$ and $	ilde{Q}_0 \leq 	ilde{B}$ are $p$-subgroups of $G = A	imes B$.

By the third part of Sylow's Theorem, there exist $g = (g_1, g_2)$ and $h = (h_1, h_2)$ such that
$$	ilde{P}_0 \leq gHg^{-1},\qquad 	ilde{Q}_0 \leq hH h^{-1}.$$
We define 
\begin{align*}
	ilde{P} = g^{-1} 	ilde{P}_0 g,
\qquad 	ilde{Q} = h^{-1} 	ilde{Q}_0 h,
\end{align*}
so that $	ilde{P} \leq H$ and $	ilde{Q} \leq H$. Note that $g^{-1} 	ilde{P}_0 g = (g_1^{-1}, g_2^{-1})(P_0 	imes \{1\}) (g_1,g_2) = g_1^{-1} P_0 g_1 	imes \{1\} \leq 	ilde{A}$, thus $	ilde{P} = P 	imes \{1\}$, where $P = g_1^{-1} P_0 g_1$ is a Sylow $p$-subgroup of $A$, and similarly $	ilde{Q} =\{1\} 	imes Q$, where $Q = h_2^{-1} Q_0 h_2$ is a Sylow $p$-subgroup of $B$.

Since $	ilde{P} \leq H$ and $	ilde{Q} \leq H$, then $	ilde{P} 	ilde{Q} \leq H$. Moreover $ P 	imes Q  = 	ilde{P} 	ilde{Q} $, since $(p,q)= (p,1) (1,q)$. Therefore $|	ilde{P} 	ilde{Q}| = |P| |Q| = p^{\alpha + \beta} = |H|$, thus $H = 	ilde{P} 	ilde{Q} = P 	imes Q$.

Any Sylow $p$-subgroup $H$ of $A 	imes B$ is of the unique form $P 	imes Q$, where $P \in Syl_p(A)$ and $Q \in Syl_p(B)$, and conversely.

In other words, the map
$$
\left\{
\begin{array}{ccl}
Syl_p(A) 	imes Syl_p(B) &	o& Syl_p(A	imes B)\
(P, Q) & \mapsto & P	imes Q
\end{array}
ight.
$$
is bijective. Hence $|Syl_p(A	imes B)| = |Syl_p(A)| \cdot | Syl_p(B)|$, so
\begin{align}
n_p(A 	imes B) = n_p(A) n_p(B).
\end{align}

Suppose that this result is true for the product of $n-1$ subgroups, and consider $G = A_1	imes A_2 	imes \cdots 	imes A_n$. If we apply the preceding result to $A = A _1	imes A_2 	imes \cdots 	imes A_{n-1}$ and $B = A_n$, then
\begin{align*}
n_p(A_1	imes A_2 	imes \cdots 	imes A_n) &= n_p(A 	imes B)\
&= n_p(A) n_p(B)\\
&= n_p( A _1	imes A_2 	imes \cdots 	imes A_{n-1}) n_p(A_n)\
&= n_p(A_1) n_p(A_2)\cdots n_p(A_{n-1}) n_p(A_n),
\end{align*}
by the induction hypothesis.

The induction is done, which proves that for all positive integers $n\geq 2$,
$$n_p(A_1	imes A_2 	imes \cdots 	imes A_n) = n_p(A_1) n_p(A_2)\cdots n_p(A_n).$$
\end{proof}

Problem 5.1.4 (Sylow subgroups of a direct product)

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Problem 5.1.4 (Sylow subgroups of a direct product)

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