Question

Let M and N be two subgroups of a soluble group G. Then show that MN is also soluble.

Answer 1

3 Answer on Question #52576 – Math – Abstract Algebra

Question. Let $M$ and $N$ be two subgroups of a soluble group $G$. Then show that $MN$ is also soluble.

Solution. Recall that a group $G$ is called *soluble* if there exists an increasing finite sequence of subgroups of $G$

$\{1\}=G_{0}<G_{1}<\cdots<G_{n-1}<G_{n}=G,$

such that $G_{i}$ is normal in $G_{i+1}$ for $i=0,\ldots,n-1$, and the factor groups $G_{i+1}/G_{i}$ is abelian. This sequence is called *subnormal series*.

We claim that *every subgroup $H$ of a soluble group $G$ is soluble as well*. In particular, this will imply that so is $MN$.

Let

$\{1\}=G_{0}<G_{1}<\cdots<G_{n-1}<G_{n}=G,$

be a subnormal series for $G$. Denote

$H_{i}=H\cap G_{i}.$

Then

$H_{0}=H\cap G_{0}=H\cap\{1\}=\{1\},$

and

$H_{n}=H\cap G_{n}=H\cap G=H.$

We will show that then the sequence

$\{1\}=H_{0}<H_{1}<\cdots<H_{n-1}<H_{n}=H,$

is a the subnormal series for $H$. For this we should check the following two statements.

1) $H_{i}$ is normal subgroup of $H_{i+1}$, that is $hH_{i}h^{-1}=H_{i}$ for all $h\in H_{i+1}$.

Indeed, by assumption $G_{i}$ is a normal subgroup of $G_{i+1}$, so $hG_{i}h^{-1}=G_{i}$ for all $h\in G_{i+1}$. Hence if $h\in H_{i+1}=H\cap G_{i+1}$, then

$hH_{i}h^{-1}=h(H\cap G_{i})h^{-1}=hHh^{-1}\cap hG_{i}h^{-1}=H\cap G_{i}=H_{i}.$

2) $H_{i+1}/H_{i}$ is an abelan group.

Indeed, by assumption the quotient group $G_{i+1}/G_{i}$ is abelian. We will show that $H_{i+1}/H_{i}$ can be identified with a subgroup of $G_{i+1}/G_{i}$ which will imply that $H_{i+1}/H_{i}$ is abelian as well.

Let $j:H_{i+1}=H\cap G_{i+1}\subset G_{i+1}$ be the inclusion map and $p:G_{i+1}\rightarrow G_{i+1}/G_{i}$ be the quotient map. Consider the composition

$p\circ j:H_{i+1}\stackrel{{\scriptstyle j}}{{\longrightarrow}}G_{i+1}\stackrel{{\scriptstyle p}}{{\longrightarrow}}G_{i+1}/G_{i}.$

Then $\ker(p\circ j)=H\cap G_{i}=H_{i}$. Hence $p\circ j$ induces an isomorphism

$H_{i+1}/\ker(p\circ j)=H_{i+1}/H_{i}\ \cong\ p\circ j(H_{i+1}).$

In other words, $H_{i+1}/H_{i}$ is isomorphic with a subgroup of an abelian group $G_{i+1}/G_{i}$, whence $H_{i+1}/H_{i}$ is abelian as well.

Thus $\{1\}=H_{0}<H_{1}<\cdots<H_{n-1}<H_{n}=H$ is a subnormal series for $H$, and so $H$ is soluble.

In particular, if $M$ and $N$ are two subgroups of a soluble group $G$, then $M$, $N$ and $MN$ are soluble as well.

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