Let $H$ and $K$ be a subgroup of inifinte group $G$ s.t $K \subset H$, $[G:H]$ is finite and $[H:K]$ is finite

prove that

  1. $[G:K]$ is finite
  2. $[G:K]=[G:H][H:K]$


Let $$H a_1 , \dots H a_n$$ be distinct cosets of $H$ in G

and let $$K b_1 , \dots K b_m $$ be disticn cosets of K in $H$

show $K b_i a_J$ (with $1 \leq i \leq m$ and $1 \leq J \leq m$) be distinct cosets k in $G$

Letting $[G:H]$ is finite, let it be equal to $n$ so there are distinct $a_1 , \dots, a_n \in G$ where

$$ Ha_1 \cup \dots \cup Ha_n =G$$

Also, $[H:K]=m$ so $\exists b_1, \dots , b_m \in H$ s.t $$ Kb_1 \cup \dots \cup Kb_m =H$$

not sure where to go from there guessing that $[G:K]=m*n$ has to do with partition

  • $\begingroup$ Well off the top of my head, by the 3rd Isomorphism Theorem $(G/H)/(K/H) \cong G/K$ so that would definitely take care of part 1 (just have to show why it will). Let me think about the second part. $\endgroup$
    – FofX
    Feb 22 '17 at 21:21
  • $\begingroup$ For some reason I can't edit my previous comment so I'll just add that I would look that the double quotient, if you think about what its elements are that should help out with part 2. $\endgroup$
    – FofX
    Feb 22 '17 at 21:34
  • $\begingroup$ @FofX I think you mean $(G/K)/(H/K) \approx G/H$, because we have $K \subset H \subset G$ rather than $H \subset K \subset G$. $\endgroup$
    – joeb
    Feb 22 '17 at 21:46
  • $\begingroup$ Ahh yes you are right, thanks for catching that. $\endgroup$
    – FofX
    Feb 22 '17 at 23:01

Let $G$ be the disjoint union of the cosets $Ha_1,...,Ha_n$, and furthermore, let $H$ be the disjoint union of the cosets $Kb_1,...,Kb_m$. For brevity, let $K_{i,j} := Kb_ia_j$ and

$\Phi = \{K_{i,j}: 1 \leqslant i \leqslant m,\,\, 1 \leqslant j \leqslant n\}$

First show that the collection $\Phi$ covers $G$. Supposing $x \in G$, there is some $j$ such that $x \in Ha_j$. In particular, $x = ha_j$ for some $h \in H$. Moreover, because $h \in H$ we also know there is some $i$ such that $h \in Kb_i$, which is to say $h = kb_i$ for some $k \in K$. But then $x = ha_j = kb_ia_j \in K(b_ia_j) = K_{i,j}$. This proves that $\Phi$ covers $G$.

Next, we will show $| \Phi | = mn$. To that end suppose $K_{p,q}=K_{r,s}$. We will show that $p=r$ and $q=s$. Now, because $Kb_p,Kb_r \subset H$, we know that $K_{p,q} \subset Ha_q$ and $K_{r,s} \subset Ha_s$. But by construction $Ha_q,Ha_s$ are disjoint when $q \neq s$. So it must be that $q = s$ and $a_q = a_s := a$. Consequently, we have

$Kb_p = (Kb_p)(a_qa^{-1}) = (K_{p,q})a^{-1} = (K_{r,s})a^{-1} = (Kb_r)(a_sa^{-1}) = Kb_r$.

So again by construction, it must be that $p = r$. This concludes the proof that $| \Phi | = mn$

Finish up with

$[G:K] = |\Phi| = mn = [H:K][G:H] = [G:H][H:K]$.

Also, you have not mentioned anything regarding normality of $H,K$. If you knew $H,K$ were normal subgroups of $G$, then as a @FofX noted, the third isomorphism theorem is applicable:

$(G/K) / (H/K) \cong G/H$.


$[G:K] $

$=|G/K| $

$= [G/K:H/K] \cdot |H/K| $

$= |(G/K)/(H/K)| \cdot |H/K| $

$= | G/H| \cdot |H/K| $

$= [G:H] \cdot [H:K]$

Where the fourth inequality follows from $|(G/K)/(H/K)| = |G/H|$ (which in turn is a direct result of the third isomorphism theorem.)

  • $\begingroup$ The third isomorphism theorem is only applicable if $H$ and $K$ are normal subgroups. $\endgroup$
    – ryanblack
    Feb 22 '17 at 22:07
  • $\begingroup$ True. I'll edit to reflect this fact $\endgroup$
    – joeb
    Feb 22 '17 at 22:09

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