Exercise 5.1.2 (Decomposition of direct products)

Question

Let $G_1,G_2,\ldots,G_n$ be groups and let $G = G_1	imes G_2	imes \cdots 	imes G_n$. Let $I$ be a proper, nonempty subset of $\{1,2,\ldots,n\}$ and let $J = \{1,\ldots,n\} - I$. Define $G_I$ be the set of elements of $G$ that have the identity of $G_j$ in position $j$ for all $j \in J$.
\begin{enumerate}
\item[\bf(a)] Prove that $G_I$ is isomorphic to the direct product of the group $G_i$, $i \in I$.
\item[\bf(b)] Prove that $G_I$ is a normal subgroup of $G$ and $G/G_I\simeq G_J$.
\item[\bf(c)] Prove that $G \simeq G_I 	imes G_J$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\begin{proof} 
\item[(a)]
We define
\begin{align*}
&I = \{i_1,i_2,\ldots i_k\} \subset [\![1,n]\!],& i_1<i_2<\cdots<i_k,\
&J = \{j_1,j_2,\ldots,j_l\} = [\![1,n]\!] - I,& j_1<j_2<\ldots< j_l.
\end{align*}
Let $\varphi$ and $\psi$ be the maps defined by
$$
\varphi
\left\{
\begin{array}{ccl}
G_I & 	o &G_{i_1}	imes G_{i_2}	imes \cdots  	imes G_{i_k} \
(g_1,g_2,\ldots,g_n)  & \mapsto & (g_{i_1}, g_{i_2}, \ldots,g_{i_k}) 
\end{array}
ight.
$$
and
$$ \psi
\left\{
\begin{array}{ccl}
G_{i_1}	imes G_{i_2} 	imes \cdots 	imes G_{i_k} & 	o & G_I\
(g_{i_1}, g_{i_2}, \ldots,g_{i_k}) & \mapsto &(h_1,h_2,\ldots,h_n)
\end{array}
ight.
$$
where
\begin{align*}
\begin{cases}
h_i = 1 & 	ext{if } i \in J,\
h_i = g_{i_r} & 	ext{if } i = i_r \in I.
\end{cases}
\end{align*}
Then $\varphi \circ \psi = \mathrm{id}$ and $\psi \circ \varphi = \mathrm{id}$, so $\varphi$ is bijective. Moreover  $\varphi$ is a homomorphism, thus $\varphi$ is an isomorphism, and so
$$G_I \simeq G_{i_1} 	imes G_{i_2}	imes \cdots  	imes G_{i_k}.$$

\item[(b)] Consider the map
$$
\chi
\left\{
\begin{array}{ccl}
G & 	o & G_J\
(g_1,g_2,\ldots,g_n) & \mapsto & (g_{j_1}, \ldots, g_{j_l}),
\end{array}
ight.
$$
Then $\chi$ is a homomorphism, whose kernel is $\ker(\chi) = G_I$, so $G_I \unlhd G$. Moreover $\chi$ is surjective: every $(g_{j_1}, \ldots, g_{j_l})\in G_J$ is the image of $(g_1,g_2,\ldots,g_n)$, where $g_j= 1$ if $j \in I$ and $g_j = g_{j_s}$ if $j = j_s \in J$. The First Isomorphism Theorem gives
$$G/G_I \simeq G_J.$$

\item[(c)] Consider the map defined by
$$
\lambda
\left\{
\begin{array}{ccl}
G & 	o & G_I 	imes G_J\
(g_1, g_2, \ldots, g_n) & \mapsto & ((u_1,\ldots,u_n), (v_1, v_2,\ldots, v_n))
\end{array}
ight.
$$
where
\begin{align*}
u_i = g_i, \ v_i = 1 \qquad 	ext{if } i \in I,\
u_1 = 1,\ v_i = g_i \qquad 	ext{if } i \in J.
\end{align*}
Then $\lambda$ is an isomorphism, and so
$$G \simeq G_I 	imes G_J.$$
\end{proof}

Exercise 5.1.2 (Decomposition of direct products)

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Exercise 5.1.2 (Decomposition of direct products)

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