Problem 5.4.10 (A finite abelian group is the direct product of its Sylow subgroups)

Question

Prove that a finite abelian group is the direct product of its Sylow subgroups.

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $G$ is an abelian  group with subgroups $H$ and $K$ such that $n = |H|$ and $m = |K|$ are relatively prime. Then
\begin{itemize}
\item $H \unlhd G$ and $K \unlhd G$ since $G$ is abelian, and
\item $H \cap K = \{1\}$ because $|H \cap K|$ divides $n$ and $m$, where $\mathrm{g.c.d.}(m,n) = 1$, so $|H \cap K | = 1$.
\end{itemize}
By Theorem 9, 
$$HK \simeq H 	imes K.$$
Let $S_1, S_2, \ldots, S_r$ be the Sylow subgroups of $G$, corresponding to the distinct primes $p_1,p_2,\ldots,p_r$, where $|G| = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r}$. Put $H = S_1S_2\ldots S_{r-1}$ and $K = S_r$. Then $|H| = p_1p_2\cdots p_{r-1}$ and $|K| = p_r$ are relatively prime, so $HK \simeq H 	imes K$, i.e.,

$$S_1S_2 \cdots S_r \simeq (S_1S_2\ldots S_{r-1}) 	imes S_r.$$
By induction, we obtain
\begin{align*}
S_1S_2\cdots S_r \simeq S_1 	imes S_2 	imes \cdots 	imes S_r.
\end{align*}
Then  
$$|S_1S_2\cdots S_r| = |S_1 	imes S_2 	imes \cdots 	imes S_r| = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_r^{\alpha_r} = |G|,$$
where $S_1S_2\cdots S_r \leq G$, thus $G = S_1S_2\cdots S_r$, and so
$$G \simeq S_1 	imes S_2 	imes \cdots 	imes S_r.$$
A finite abelian group is the direct product of its Sylow subgroups.
\end{proof}

Problem 5.4.10 (A finite abelian group is the direct product of its Sylow subgroups)

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Problem 5.4.10 (A finite abelian group is the direct product of its Sylow subgroups)

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