4

If $H$ is of finite index in $G$ prove that there is a subgroup $N$ of $G$, contained in $H$, and of finite index in $G$ such that $aNa^{-1}=N$ for all $a \in G$.

This problem is from Herstein's introductory exercise to subgroups. There seems to be a solution involving quotienting (pg.10, Q.20), but Herstein has yet not introduced quotienting in the text. Is there an alternate solution?

user26857
  • 53,190
  • 1
    I don't see where it involves quotients. Problem 12 shows the intersection of two finite-index subgroups is a finite-index subgroup, simply by counting with cosets, and by induction this extends to the intersection of any finite number of subgroups. Problem 19 shows there are only finitely many conjugates of a finite-index subgroup, again simply by counting. Finally, we intersect all the conjugates to get the normal subgroup in Problem 20. – anon Jul 03 '14 at 10:23
  • @blue I was looking at a random solution manual on the net and it made use of quotienting, I am yet to learn quotient groups. Yes, but using the result of 12th and 19th we can only conclude that the intersection of all $aNa^{-1}$ is a finite index subgroup, how does one show it is also equal to $N$? –  Jul 03 '14 at 10:37
  • You mean if $N=\bigcap aHa^{-1}$ then $N=gNg^{-1}$ for all $g\in G$. This is actually fairly straightforward to show. Have you tried? Do you have any thoughts or ideas? – anon Jul 03 '14 at 10:39
  • "Intersect" is a wrong word, I am sorry for that. I meant that we need to show $aNa^{-1} = N$ for all $a \in G$, how do you show that? For a start I thought if we could equate the left and right cosets, then probably we could show it is abelian and then $aNa^{-1}$ becomes $Naa^{-1}$ which is N but I couldn't do it. –  Jul 03 '14 at 10:52
  • $g(\bigcap_{a\in G} aNa^{-1})g^{-1}=\bigcap_{a\in G}gaN(ga)^{-1}$. Can you continue? – anon Jul 03 '14 at 14:16
  • See also https://math.stackexchange.com/questions/88719/a-group-g-with-a-subgroup-h-of-index-n-has-a-normal-subgroup-k-subset-h – lhf Jan 24 '22 at 11:52

2 Answers2

3

One approach might be to let $G$ act on $G/H$ by left-multiplication. We can think of the action as a group homomorphism from $G$ to the symmetric group on the finite set $G/H$. Let $N$ be the kernel of this homomorphism. It is a normal subgroup of $G$ since it is a kernel. Also, since every element of $N$ fixes the identity coset, we see that $N$ is a subgroup of $H$. Also, the group $G/N$ is isomorphic to a subgroup of the (finite) symmetric group of $G/H$ by the First Isomorphism Theorem, so $G/N$ is finite. In other words, the index of $N$ is finite.

Peter Crooks
  • 2,111
0

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 20 on p.48 in Herstein's book.
I solved this problem as follows:

I referred to "An Introduction to Algebraic Systems" (in Japanese) by Kazuo Matsuzaka.
The following problem is in this Matsuzaka's book:

Problem 3 (on p.60)
Suppose $K,L$ be subgroups of $G$ and $i_G(L)<+\infty$.
Then, $i_K(K\cap L)<+\infty$ and $i_K(K\cap L)\leq i_G(L)$ holds.

My solution to Problem 3 in Matsuzaka's book:
$K/K\cap L\ni k(K\cap L)\mapsto kL\in G/L$ is well-defined since $k^{-1}k^{'}\in K\cap L$ implies $k^{-1}k^{'}\in L$.
Suppose that $kL=k^{'}L$.
Then $k^{-1}k^{'}\in L$.
Since $k,k^{'}\in K$, $k^{-1}k^{'}\in K\cap L$.
So, $k(K\cap L)=k^{'}(K\cap L)$.
So, this well-defined mapping is injective.
So, $i_K(K\cap L)<+\infty$ and $i_K(K\cap L)\leq i_G(L)$ must hold.

My solution to Problem 20 in Herstein's book:
We use the result of Problem 18 on p.48 and Problem 19 on p.48 in Herstein's book.
By Problem 19 on p.48, there are only a finite number of distinct subgroups in $G$ of the form $aHa^{-1}$.
Let $\{H,a_1Ha_{1}^{-1},\dots,a_{n-1}Ha_{n-1}^{-1}\}$ be the set of all distinct subgroups in $G$ of the form $aHa^{-1}$.
$n=\#G/N(H)$.
Let $H_0:=H$ and $H_i:=a_iHa_{i}^{-1}$ for $i\in\{1,\dots,n-1\}$.
Obviously, $i_G(H_i)=i_G(H)$ for all $i\in\{0,\dots,n-1\}$.
Then, $i_G(\cap_{i=0}^k H_i) < +\infty$ holds for $k\in\{0,1,\dots,n-1\}$.
We prove this.
$i_G(\cap_{i=0}^0 H_i) < +\infty$ holds by assumption of Problem 20.
Assume that $i_G(\cap_{i=0}^k H_i) < +\infty$ holds.
Then, by Problem 3 in Matsuzaka's book, $i_{\cap_{i=0}^k H_i}(\cap_{i=0}^{k+1} H_i)<+\infty$ holds since $i_G(H_{k+1})<+\infty$.
(Let $K:=\cap_{i=0}^{k} H_i$ and $L:=H_{k+1}$ in Problem 3.)
$i_G(\cap_{i=0}^{k} H_i)<+\infty$ holds by assumption.
So, $i_G(\cap_{i=0}^{k+1} H_i)=i_G(\cap_{i=0}^{k} H_i)i_{\cap_{i=0}^k H_i}(\cap_{i=0}^{k+1} H_i)<+\infty$ holds.

$N:=\cap_{i=0}^{n-1} H_i$ is a normal subgroup of $G$ by Problem 18 on p.48 in Herstein's book.

$N\subset H$ is obvious.

So, $N$ is a normal subgroup of $G$ and $i_G(N)<+\infty$ and $N\subset H$.

And $i_G(H)^{i_G(N(H))}$ is an upper bound for $i_G(N)$.

tchappy ha
  • 9,894