Question

Let H and K be subgroups of an infinite group G such that $K\subseteq H,[G:H]$ is finite and $[H:K]$is finite. Prove that $[G:K]$is finite and $[G:K]=[G:H][H:K]$.[Hint: Let $H{a}_1,H{a}_{2.........},H{a}_n$ be the distinct cosets of H in G and let be the distinct cosets of H in Gand let $K{b}_1,K{b}_2,......,K{b}_n,$ be the distinct cosets of Kin H. Show that $K{b}_i{a}_j$(with $1\le i\le m$and localid="1652344029730" $1\le j\le n$) are the distinct cosets of Kin G]

Short answer

We proved that $[G:K]$is finite and $[G:K]=[G:H][H:K]$

Step-by-step solution

To find distinct cosets of subgroups in a group

Let H and K be two subgroups of an infinite group G.

Suppose $K\subseteq H$and both $[G:H]$and $[H:K]$are finite.

Then we have to show that $[G:K]$is finite and $[G:K]=[G:H][H:K]$.

Let $[G:K]=n$and $[H:K]=m$.

Now, we write distinct cosets of H in G as $H{a}_1,H{a}_2,.......,H{a}_{n.}$ and distinct cosets of K in H as $K{b}_1,K{b}_2,......K{b}_n$

Then we have to prove that

$\{K{b}_i{a}_j:1\le i\le m,1\le j\le n\}$ are all distinct cosets of K in G.

In order to prove this, we need first to show that

(1) $K{b}_i{a}_j\cap K{b}_p{a}_q\ne \varphi ,$for $(i,j)\ne (p,q)$and $(i,j),(p,q)\in \{1,...,m\}\times \{1,....,n\}$

(2)$\cup k{b}_i{a}_j=G$

To prove part (1)

Let $(i,j),(p,q)\in \{1,....,m\}\times \{1,.....,n\}$ such that $(i,j)\ne (p,q)$.

Then we have either $i\ne p$ or $j\ne q$.

Now, if possible, suppose that, $K{b}_i{a}_j\cap k{b}_p{a}_q\ne \varphi ,$

Let $x\in K{b}_i{a}_j\cap b_p{a}_q$.

Then $\exists r,r\text{'}\in K$such that

$$\begin{gathered} x=r{b}_i{a}_j=r\text{'}{b}_p{a}_q \\ \Rightarrow r{b}_i{a}_j{a_q}^{-1}=r\text{'}{b}_p\in K{b}_p. \end{gathered}$$

Now if, $j=q$, then,

$K{b}_i\cap K{b}_p=\varphi r{b}_i\in K{b}_i\cap K{b}_p$

This is impossible since $i\ne p$.

Hence, $K{b}_i\cap K{b}_p=\varphi$

On the other hand, if $j\ne q$, then

$x=r{b}_i{a}_j=r\text{'}{b}_p{a}_q$

$\Rightarrow x=r{b}_i{a}_j\in H{a}_j$and $x=r{b}_p{a}_q\in H{a}_q$

$\Rightarrow x\in H{a}_j\cap H{a}_q$

This is impossible since $H{a}_j\cap H{a}_q\ne \varphi$

Thus we have shown that $K{b}_i{a}_j\cap K{b}_p{a}_q\ne \varphi ,$for$(i,j)\ne (p,q)$and$(i,j),(p,q)\in \{1,...,m\}\times \{1,....,n\}$