Exercise 14.2.4

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

If $G,H$ are solvable groups, then $G 	imes H$ is solvable.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\medskip

{\it Proof of Lemma.} We have subgroups 
\begin{align*}
&\{e\} \subset G_n \subset \cdots \subset G_1 \subset G_0 = G\
&\{e'\} \subset H_m \subset \cdots \subset H_1 \subset H_0 = H
\end{align*}
such that $G_i$ is normal in $G_{i-1}$ and $G_{i-1}/G_i$ is Abelian for $i=1,\ldots,n$, and $H_i$ is normal in $H_{i-1}$ and $H_{i-1}/H_i$ is Abelian for $i=1,\ldots,m$.

If $n >m$, we can define $H_{m+1} = H_{m+2} =\cdots=H_n=\{e'\}$,  and proceed similarly if $n<m$, so we can assume that $n=m$: 
\begin{align*}
&\{e\} \subset G_n \subset \cdots \subset G_1 \subset G_0 = G\
&\{e'\} \subset H_n \subset \cdots \subset H_1 \subset H_0 = H
\end{align*}
Then
$$
\{(e,e')\} =G_n	imes H_n \subset \cdots \subset G_1	imes H_1 \subset G_0	imes H_0 = G 	imes H.
$$
We prove 
$$(G_{i-1}	imes H_{i-1}) / (G_i 	imes H_i) \simeq G_{i-1}/G_i 	imes H_{i-1}/H_i.$$
Indeed, 
$$
\psi
\left\{
\begin{array}{ccc}
G_{i-1} 	imes H_{i-1} & 	o  & G_{i-1}/G_i 	imes H_{i-1}/H_i\
(g,h) & \mapsto & (gG_i, hH_i)
\end{array}
ight.
$$
is surjective, and its kernel is $G_i 	imes H_i$. This proves our assertion.

Therefore $(G_{i-1}	imes H_{i-1}) / (G_i 	imes H_i)$ is Abelian. Then Exercise 8.1.8 shows that $G	imes H $ is solvable.
\qed
\begin{proof} {\it (of Ex.14.2.4.)}
Let 
$$
\varphi
\left\{
\begin{array}{ccc}
A \wr B & 	o & A\
(	au; \mu_1,\ldots,\mu_k) & \mapsto &	au.
\end{array}
ight.
$$
By Lemma 14.2.8, $\varphi$ is onto, and its kernel $H = \ker(\varphi)$ is isomorphic to $B^k$. Then $B^k$ is solvable by induction with the above Lemma, so that $H$ is solvable, and $(A \wr B) / H  = (A \wr B)/\ker(\varphi) \simeq A$ is solvable. By Theorem 8.1.4, $A\wr B$ is solvable.
\end{proof}

Exercise 14.2.4

Answers

Comments

Exercise 14.2.4

Answers

Comments

Add answer