2
$\begingroup$

Let $G$ be a group and $H$ be a subgroup of $G$. I want to prove that if $[G:H]<\infty$, then $H$ contains a normal subgroup $N$ of $G$ such that $[G:N]<\infty$.

Professor gave me the following sketch of proof : Since $H\leq N_G(H)\leq G$, $[G:N_G(H)]< \infty$. Let $[G:N_G(H)]=t$. Then there exists $g_1, g_2, \cdots, g_t\in G$ such that $G=N_G(H)g_1\cup \cdots \cup N_G(H)g_t$. Then $N=\bigcap_{i=1}^{t}g_iHg_i^{-1}$ is a normal subgroup of G such that $[G:N]<\infty$.

But I cannot complete the proof because the last line does not clear for me. Please give me the detail. Thanks.

$\endgroup$
2
$\begingroup$

There are two claims in the last sentence: that $N$ is finite-index and that $N$ is normal. The first claim follows from Does the intersection of two finite index subgroups have finite index? and induction on $t$ (note that $gHg^{-1}$ is a subgroup of $G$ of index $[G:gHg^{-1}]=[G:H]$ for each $g\in G$).

For the second claim, choose $g\in G$ and $n\in N$. We wish to show that $gNg^{-1} = N$. We have: $$ gNg^{-1} = g \left( \bigcap_i g_i H g_i^{-1}\right) g^{-1} = \bigcap_i (g g_i) H (g g_i)^{-1} $$ It would be easier if instead of what you had written, the $g_i$s are coset representatives on the other side, and so that's what I will assume: $G = g_1 N_G(H) \cup \dots \cup g_t N_G(H)$. Then for each $i$ we may factor $gg_i = g_{j(i)}m_i$ for some $j$ depending on $i$ and $m_i \in N_G(H)$. Then $$ (g g_i) H (g g_i)^{-1} = g_{j(i)}m_i H m_i^{-1} g_{j(i)}^{-1} = g_{j(i)} H g_{j(i)}^{-1} $$ since $m_i$ normalizes $H$. Therefore: $$ \bigcap_i (g g_i) H (g g_i)^{-1} = \bigcap_i g_{j(i)} H g_{j(i)}^{-1} \supseteq \bigcap_j g_j H g_j^{-1} $$ A priori, the last containment might not be an equality: the intersection over $i$ is equivalent to the intersection over only those $j$ of the form $j(i)$ (the function $j(i)$ depends on the choice of $g$), whereas the right-hand side intersects over possibly more $j$s, and so can result in something smaller.

But this does prove that $gNg^{-1} \supseteq N$ for all $g\in G$. In particular, $g^{-1}Ng \supseteq N$, but conjugating by $g$ shows $N \supseteq gNg^{-1}$. This completes the proof.

$\endgroup$
  • $\begingroup$ Does [G:gHg−1]=[G:H] hold for any cases? $\endgroup$ – user112018 Mar 4 '14 at 1:09
  • $\begingroup$ Yes. The "real" reason is that $gGg^{-1} = G$, and $[G:H] = [gGg^{-1}:gHg^{-1}]$. But more hands on, suppose that $\{g_i\}_{i\in I}$ is a (possibly infinite) set of coset representatives for $G/H$. Then $\{gg_ig^{-1}\}_{i\in I}$ is a set of coset representatives for $G/gHg^{-1}$. $\endgroup$ – Theo Johnson-Freyd Mar 4 '14 at 2:39
4
$\begingroup$

There is an easy alternative for this proof (which upon inspection is probably just another way to say the same thing). By hypothesis there are finitely many cosets $gH$ in $G/H$. The group $G$ acts by left multiplication on this set of cosets, defining a group morphism $\rho:G\to S(G/H)$, the codomain being the finite group of permutations of the cosets. Then $\ker\rho$ is a normal subgroup (like any kernel) of finite index in$~G$ (the index is equal to the order of the image of$~\rho$, a subgroup of $S(G/H)$). Also $\ker\rho< H$ since its left multiplication by any element of$~\ker\rho$ must in particular stabilise the coset $eH$.

$\endgroup$
  • $\begingroup$ I do not accept this answer because this is not an answer that I asked for but it is very easy alternative proof as you said! $\endgroup$ – user112018 Mar 4 '14 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.