Mixing
Does the intersection of two finite index subgroups have finite index?
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Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $Hcap K$ also a subgroup of finite index? I feel like need that $[Gcolon(Hcap K)]$ is a divisor of $[Gcolon H]cdot[Gcolon K]$, but I dont't know when this holds.
Can somebody help me out?
abstract-algebra group-theory
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
$endgroup$
|
show 1 more comment
$begingroup$
Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $Hcap K$ also a subgroup of finite index? I feel like need that $[Gcolon(Hcap K)]$ is a divisor of $[Gcolon H]cdot[Gcolon K]$, but I dont't know when this holds.
Can somebody help me out?
abstract-algebra group-theory
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
$endgroup$
$begingroup$
Step 1: Show that $H cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H cap K][Hcap K : H]$.
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– Clive Newstead
Apr 5 '12 at 22:41
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@MartQ. You don't have to show that it's a divisor. It suffices to show that $[Gcolon(Hcap K)]leq [Gcolon H]cdot[Gcolon K].$ You can try to find an injection from a certain set to another.
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– user23211
Apr 5 '12 at 23:02
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A related question.
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– user23211
Apr 5 '12 at 23:06
4
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@CliveNewstead The formula $[G:H] = [G:H cap K][Hcap K : H]$ seems incorrect to me. What is $[Hcap K : H]?$ Did you mean $[G:Hcap K]=[G:H][H:Hcap K]?$
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– user23211
Apr 5 '12 at 23:23
1
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@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.
$endgroup$
– Clive Newstead
Apr 7 '12 at 11:45
|
show 1 more comment
$begingroup$
Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $Hcap K$ also a subgroup of finite index? I feel like need that $[Gcolon(Hcap K)]$ is a divisor of $[Gcolon H]cdot[Gcolon K]$, but I dont't know when this holds.
Can somebody help me out?
abstract-algebra group-theory
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
$endgroup$
Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $Hcap K$ also a subgroup of finite index? I feel like need that $[Gcolon(Hcap K)]$ is a divisor of $[Gcolon H]cdot[Gcolon K]$, but I dont't know when this holds.
Can somebody help me out?
abstract-algebra group-theory
abstract-algebra group-theory
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
edited Jul 14 '12 at 22:56
Dylan Moreland
16.9k23564
16.9k23564
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
asked Apr 5 '12 at 22:36
Mart Q.Mart Q.
7113
7113
$begingroup$
Step 1: Show that $H cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H cap K][Hcap K : H]$.
$endgroup$
– Clive Newstead
Apr 5 '12 at 22:41
$begingroup$
@MartQ. You don't have to show that it's a divisor. It suffices to show that $[Gcolon(Hcap K)]leq [Gcolon H]cdot[Gcolon K].$ You can try to find an injection from a certain set to another.
$endgroup$
– user23211
Apr 5 '12 at 23:02
$begingroup$
A related question.
$endgroup$
– user23211
Apr 5 '12 at 23:06
4
$begingroup$
@CliveNewstead The formula $[G:H] = [G:H cap K][Hcap K : H]$ seems incorrect to me. What is $[Hcap K : H]?$ Did you mean $[G:Hcap K]=[G:H][H:Hcap K]?$
$endgroup$
– user23211
Apr 5 '12 at 23:23
1
$begingroup$
@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.
$endgroup$
– Clive Newstead
Apr 7 '12 at 11:45
|
show 1 more comment
$begingroup$
Step 1: Show that $H cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H cap K][Hcap K : H]$.
$endgroup$
– Clive Newstead
Apr 5 '12 at 22:41
$begingroup$
@MartQ. You don't have to show that it's a divisor. It suffices to show that $[Gcolon(Hcap K)]leq [Gcolon H]cdot[Gcolon K].$ You can try to find an injection from a certain set to another.
$endgroup$
– user23211
Apr 5 '12 at 23:02
$begingroup$
A related question.
$endgroup$
– user23211
Apr 5 '12 at 23:06
4
$begingroup$
@CliveNewstead The formula $[G:H] = [G:H cap K][Hcap K : H]$ seems incorrect to me. What is $[Hcap K : H]?$ Did you mean $[G:Hcap K]=[G:H][H:Hcap K]?$
$endgroup$
– user23211
Apr 5 '12 at 23:23
1
$begingroup$
@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.
$endgroup$
– Clive Newstead
Apr 7 '12 at 11:45
$begingroup$
Step 1: Show that $H cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H cap K][Hcap K : H]$.
$endgroup$
– Clive Newstead
Apr 5 '12 at 22:41
$begingroup$
Step 1: Show that $H cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H cap K][Hcap K : H]$.
$endgroup$
– Clive Newstead
Apr 5 '12 at 22:41
$begingroup$
@MartQ. You don't have to show that it's a divisor. It suffices to show that $[Gcolon(Hcap K)]leq [Gcolon H]cdot[Gcolon K].$ You can try to find an injection from a certain set to another.
$endgroup$
– user23211
Apr 5 '12 at 23:02
$begingroup$
@MartQ. You don't have to show that it's a divisor. It suffices to show that $[Gcolon(Hcap K)]leq [Gcolon H]cdot[Gcolon K].$ You can try to find an injection from a certain set to another.
$endgroup$
– user23211
Apr 5 '12 at 23:02
$begingroup$
A related question.
$endgroup$
– user23211
Apr 5 '12 at 23:06
$begingroup$
A related question.
$endgroup$
– user23211
Apr 5 '12 at 23:06
4
4
$begingroup$
@CliveNewstead The formula $[G:H] = [G:H cap K][Hcap K : H]$ seems incorrect to me. What is $[Hcap K : H]?$ Did you mean $[G:Hcap K]=[G:H][H:Hcap K]?$
$endgroup$
– user23211
Apr 5 '12 at 23:23
$begingroup$
@CliveNewstead The formula $[G:H] = [G:H cap K][Hcap K : H]$ seems incorrect to me. What is $[Hcap K : H]?$ Did you mean $[G:Hcap K]=[G:H][H:Hcap K]?$
$endgroup$
– user23211
Apr 5 '12 at 23:23
1
1
$begingroup$
@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.
$endgroup$
– Clive Newstead
Apr 7 '12 at 11:45
$begingroup$
@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.
$endgroup$
– Clive Newstead
Apr 7 '12 at 11:45
|
show 1 more comment
7 Answers
7
active
oldest
votes
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Proof $1$: $quad[G:Hcap K]=[G:H][H:Hcap K]=[G:H][HK:K]le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:Hcap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $Hcap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/Htimes G/K$. The latter is finite so the orbit of $Htimes K$ is finite, and the stabilizer of it is simply $Hcap K$. Invoke orbit-stabilizer.
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
$endgroup$
$begingroup$
There's probably a thread in which your second equality is proved, too. I'll go digging.
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– Dylan Moreland
Jul 14 '12 at 23:10
2
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Found one: math.stackexchange.com/questions/168942/…
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:15
$begingroup$
The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
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– Wolfgang
Sep 23 '18 at 21:08
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@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
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– anon
Sep 24 '18 at 0:29
add a comment |
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Let ${H_1,dots,H_m}$ be the left cosets of $H$, and let ${K_1,dots,K_n}$ be the left cosets of $K$. For each $xin G$ there are unique $h(x)in{1,dots,m}$ and $k(x)in{1,dots,n}$ such that $xin H_{h(x)}$ and $xin K_{k(x)}$. Let $p(x)=langle h(x),k(x)rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:
Proposition: If $x$ and $y$ are in different left cosets of $Hcap K$, then $p(x)ne p(y)$.
It follows immediately that $Hcap K$ can have at most $mn$ left cosets.
It may be easier to consider the contrapositive of the proposition:
If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $Hcap K$.
You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}yin H$.
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
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Please, professor, could you clarify what does the $p(x)$ mean?
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– Danilo Gregorin
Sep 4 '17 at 1:20
1
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@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
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– Shahab
Sep 5 '17 at 15:32
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@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
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– Danilo Gregorin
Sep 5 '17 at 19:52
add a comment |
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Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 subset H$ and $N_2 subset K$. Then $G/N_1 times G/N_2$ is a finite group; do you see why this implies that $N_1 cap N_2$ has finite index in $G$?
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
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1
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I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
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– anon
Jul 14 '12 at 22:59
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@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
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– Dylan Moreland
Jul 14 '12 at 23:03
add a comment |
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$H, K$ be subgroups of $G$. Any $A in frac{G}{H cap K}$ can be written as $A = B cap C$, where $B in G/H$ and $C in G/K$ as follows. For any $g in G$
$$g(Hcap K) = gH cap gK$$.
Hence,
$$[G:H cap K] leq [G:H] [G:K]$$.
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
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add a comment |
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Let $C$ be the set of left cosets of $S = H cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C to C_1 times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| leq |C_1| cdot |C_2|$, which proves that $[G: S]$ is finite.
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
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add a comment |
$begingroup$
Let $l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l in G$ and $G$ be partitioned as $$G=h_1H cup ... cup h_lH = k_1K cup ... cup k_mK$$
I will find $a_1,...a_n in G$ s.t. $G$ is partitioned as $$G=a_1(H cap K) cup ... cup a_n(H cap K)$$ which means $n=[G:H cap K] < infty$.
Let $b_1 in G$. Then $exists i_1 in {1,...,m}, j_i in {1,...,l}$ s.t. $b_1 in h_{i_1}H cap k_{j_1}K=b_1H cap b_1K = b_1(H cap K)$. Next, let $b_2 in G setminus b_1(H cap K)$. Then $b_2 in b_2(H cap K)$ where $b_2(H cap K) cap b_1(H cap K) = emptyset$ because $$b_1H cap b_2H = h_{i_1}H cap h_{i_2}H = emptyset = k_{j_1}K cap k_{j_2}K = b_1K cap b_2K$$
where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 in {1,...,m} setminus {i_1}, j_2 in {1,...,l} setminus {j_1}$$ This process continues at most $lm$ times for $a_p=b_p, p in {1,2,...,n}$ Thus, $n le lm < infty.$
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I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $H$ and $K$ are finite index subgroups of $G$, we can consider the right cosets $G/H$ and this is finite by hypothesis. In particular $G$ acts (transitively) on $G/H$ by right multiplication, i.e.
$$
G/H times G to G, qquad
(Hx, g) mapsto Hxg.
$$
If we now restrict the action to $K$, and consider the $K$-orbit of $H$, call it $mathcal{O}_K(H)$, then it must be finite since it's contained in $G/H$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality
$$
|mathcal{O}_K(H)| = [K : text{Stab}_K(H)],
$$
where $text{Stab}_K(H)$ is the stabiliser of $H$ under the $K$-action. Notice that $text{Stab}_K(H) = K cap text{Stab}_G(H)$, and what's the stabiliser of $H$ for the $G$-action? But of course $H$ itself, so $text{Stab}_K(H) = K cap H$, and finally we conclude that $[K : K cap H]$ is finite. Hence $[G : H cap K] = [G : K][K : K cap H]$ must be finite as well.
answered Jan 27 at 2:54
fmnqfmnq
11
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add a comment |
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7 Answers
7
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7 Answers
7
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$begingroup$
Proof $1$: $quad[G:Hcap K]=[G:H][H:Hcap K]=[G:H][HK:K]le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:Hcap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $Hcap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/Htimes G/K$. The latter is finite so the orbit of $Htimes K$ is finite, and the stabilizer of it is simply $Hcap K$. Invoke orbit-stabilizer.
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
$endgroup$
$begingroup$
There's probably a thread in which your second equality is proved, too. I'll go digging.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:10
2
$begingroup$
Found one: math.stackexchange.com/questions/168942/…
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:15
$begingroup$
The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
$endgroup$
– Wolfgang
Sep 23 '18 at 21:08
$begingroup$
@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
$endgroup$
– anon
Sep 24 '18 at 0:29
add a comment |
$begingroup$
Proof $1$: $quad[G:Hcap K]=[G:H][H:Hcap K]=[G:H][HK:K]le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:Hcap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $Hcap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/Htimes G/K$. The latter is finite so the orbit of $Htimes K$ is finite, and the stabilizer of it is simply $Hcap K$. Invoke orbit-stabilizer.
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
$endgroup$
$begingroup$
There's probably a thread in which your second equality is proved, too. I'll go digging.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:10
2
$begingroup$
Found one: math.stackexchange.com/questions/168942/…
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:15
$begingroup$
The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
$endgroup$
– Wolfgang
Sep 23 '18 at 21:08
$begingroup$
@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
$endgroup$
– anon
Sep 24 '18 at 0:29
add a comment |
$begingroup$
Proof $1$: $quad[G:Hcap K]=[G:H][H:Hcap K]=[G:H][HK:K]le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:Hcap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $Hcap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/Htimes G/K$. The latter is finite so the orbit of $Htimes K$ is finite, and the stabilizer of it is simply $Hcap K$. Invoke orbit-stabilizer.
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
$endgroup$
Proof $1$: $quad[G:Hcap K]=[G:H][H:Hcap K]=[G:H][HK:K]le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:Hcap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $Hcap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/Htimes G/K$. The latter is finite so the orbit of $Htimes K$ is finite, and the stabilizer of it is simply $Hcap K$. Invoke orbit-stabilizer.
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
edited Sep 22 '15 at 5:48
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
answered Jul 14 '12 at 23:08
anonanon
72.5k5111214
72.5k5111214
$begingroup$
There's probably a thread in which your second equality is proved, too. I'll go digging.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:10
2
$begingroup$
Found one: math.stackexchange.com/questions/168942/…
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:15
$begingroup$
The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
$endgroup$
– Wolfgang
Sep 23 '18 at 21:08
$begingroup$
@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
$endgroup$
– anon
Sep 24 '18 at 0:29
add a comment |
$begingroup$
There's probably a thread in which your second equality is proved, too. I'll go digging.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:10
2
$begingroup$
Found one: math.stackexchange.com/questions/168942/…
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– Dylan Moreland
Jul 14 '12 at 23:15
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The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
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– Wolfgang
Sep 23 '18 at 21:08
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@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
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– anon
Sep 24 '18 at 0:29
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There's probably a thread in which your second equality is proved, too. I'll go digging.
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– Dylan Moreland
Jul 14 '12 at 23:10
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There's probably a thread in which your second equality is proved, too. I'll go digging.
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– Dylan Moreland
Jul 14 '12 at 23:10
2
2
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Found one: math.stackexchange.com/questions/168942/…
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– Dylan Moreland
Jul 14 '12 at 23:15
$begingroup$
Found one: math.stackexchange.com/questions/168942/…
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:15
$begingroup$
The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
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– Wolfgang
Sep 23 '18 at 21:08
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The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of ${gK mid g in HK}$. But then $g=hk$ and so any element of ${gK mid g in HK}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense?
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– Wolfgang
Sep 23 '18 at 21:08
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@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
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– anon
Sep 24 '18 at 0:29
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@AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $hin H$.
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– anon
Sep 24 '18 at 0:29
add a comment |
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Let ${H_1,dots,H_m}$ be the left cosets of $H$, and let ${K_1,dots,K_n}$ be the left cosets of $K$. For each $xin G$ there are unique $h(x)in{1,dots,m}$ and $k(x)in{1,dots,n}$ such that $xin H_{h(x)}$ and $xin K_{k(x)}$. Let $p(x)=langle h(x),k(x)rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:
Proposition: If $x$ and $y$ are in different left cosets of $Hcap K$, then $p(x)ne p(y)$.
It follows immediately that $Hcap K$ can have at most $mn$ left cosets.
It may be easier to consider the contrapositive of the proposition:
If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $Hcap K$.
You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}yin H$.
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
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Please, professor, could you clarify what does the $p(x)$ mean?
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– Danilo Gregorin
Sep 4 '17 at 1:20
1
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@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
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– Shahab
Sep 5 '17 at 15:32
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@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
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– Danilo Gregorin
Sep 5 '17 at 19:52
add a comment |
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Let ${H_1,dots,H_m}$ be the left cosets of $H$, and let ${K_1,dots,K_n}$ be the left cosets of $K$. For each $xin G$ there are unique $h(x)in{1,dots,m}$ and $k(x)in{1,dots,n}$ such that $xin H_{h(x)}$ and $xin K_{k(x)}$. Let $p(x)=langle h(x),k(x)rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:
Proposition: If $x$ and $y$ are in different left cosets of $Hcap K$, then $p(x)ne p(y)$.
It follows immediately that $Hcap K$ can have at most $mn$ left cosets.
It may be easier to consider the contrapositive of the proposition:
If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $Hcap K$.
You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}yin H$.
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
$endgroup$
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Please, professor, could you clarify what does the $p(x)$ mean?
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– Danilo Gregorin
Sep 4 '17 at 1:20
1
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@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
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– Shahab
Sep 5 '17 at 15:32
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@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
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– Danilo Gregorin
Sep 5 '17 at 19:52
add a comment |
$begingroup$
Let ${H_1,dots,H_m}$ be the left cosets of $H$, and let ${K_1,dots,K_n}$ be the left cosets of $K$. For each $xin G$ there are unique $h(x)in{1,dots,m}$ and $k(x)in{1,dots,n}$ such that $xin H_{h(x)}$ and $xin K_{k(x)}$. Let $p(x)=langle h(x),k(x)rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:
Proposition: If $x$ and $y$ are in different left cosets of $Hcap K$, then $p(x)ne p(y)$.
It follows immediately that $Hcap K$ can have at most $mn$ left cosets.
It may be easier to consider the contrapositive of the proposition:
If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $Hcap K$.
You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}yin H$.
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
$endgroup$
Let ${H_1,dots,H_m}$ be the left cosets of $H$, and let ${K_1,dots,K_n}$ be the left cosets of $K$. For each $xin G$ there are unique $h(x)in{1,dots,m}$ and $k(x)in{1,dots,n}$ such that $xin H_{h(x)}$ and $xin K_{k(x)}$. Let $p(x)=langle h(x),k(x)rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:
Proposition: If $x$ and $y$ are in different left cosets of $Hcap K$, then $p(x)ne p(y)$.
It follows immediately that $Hcap K$ can have at most $mn$ left cosets.
It may be easier to consider the contrapositive of the proposition:
If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $Hcap K$.
You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}yin H$.
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
answered Apr 5 '12 at 23:18
Brian M. ScottBrian M. Scott
459k39515917
459k39515917
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Please, professor, could you clarify what does the $p(x)$ mean?
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– Danilo Gregorin
Sep 4 '17 at 1:20
1
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@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
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– Shahab
Sep 5 '17 at 15:32
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@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
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– Danilo Gregorin
Sep 5 '17 at 19:52
add a comment |
$begingroup$
Please, professor, could you clarify what does the $p(x)$ mean?
$endgroup$
– Danilo Gregorin
Sep 4 '17 at 1:20
1
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@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
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– Shahab
Sep 5 '17 at 15:32
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@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
$endgroup$
– Danilo Gregorin
Sep 5 '17 at 19:52
$begingroup$
Please, professor, could you clarify what does the $p(x)$ mean?
$endgroup$
– Danilo Gregorin
Sep 4 '17 at 1:20
$begingroup$
Please, professor, could you clarify what does the $p(x)$ mean?
$endgroup$
– Danilo Gregorin
Sep 4 '17 at 1:20
1
1
$begingroup$
@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
$endgroup$
– Shahab
Sep 5 '17 at 15:32
$begingroup$
@Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values.
$endgroup$
– Shahab
Sep 5 '17 at 15:32
$begingroup$
@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
$endgroup$
– Danilo Gregorin
Sep 5 '17 at 19:52
$begingroup$
@Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else.
$endgroup$
– Danilo Gregorin
Sep 5 '17 at 19:52
add a comment |
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Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 subset H$ and $N_2 subset K$. Then $G/N_1 times G/N_2$ is a finite group; do you see why this implies that $N_1 cap N_2$ has finite index in $G$?
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
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1
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I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
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– anon
Jul 14 '12 at 22:59
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@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
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– Dylan Moreland
Jul 14 '12 at 23:03
add a comment |
$begingroup$
Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 subset H$ and $N_2 subset K$. Then $G/N_1 times G/N_2$ is a finite group; do you see why this implies that $N_1 cap N_2$ has finite index in $G$?
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
$endgroup$
1
$begingroup$
I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
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– anon
Jul 14 '12 at 22:59
$begingroup$
@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:03
add a comment |
$begingroup$
Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 subset H$ and $N_2 subset K$. Then $G/N_1 times G/N_2$ is a finite group; do you see why this implies that $N_1 cap N_2$ has finite index in $G$?
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
$endgroup$
Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 subset H$ and $N_2 subset K$. Then $G/N_1 times G/N_2$ is a finite group; do you see why this implies that $N_1 cap N_2$ has finite index in $G$?
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
answered Jul 14 '12 at 22:56
Dylan MorelandDylan Moreland
16.9k23564
16.9k23564
1
$begingroup$
I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
$endgroup$
– anon
Jul 14 '12 at 22:59
$begingroup$
@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:03
add a comment |
1
$begingroup$
I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
$endgroup$
– anon
Jul 14 '12 at 22:59
$begingroup$
@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:03
1
1
$begingroup$
I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
$endgroup$
– anon
Jul 14 '12 at 22:59
$begingroup$
I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal.
$endgroup$
– anon
Jul 14 '12 at 22:59
$begingroup$
@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:03
$begingroup$
@anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind.
$endgroup$
– Dylan Moreland
Jul 14 '12 at 23:03
add a comment |
$begingroup$
$H, K$ be subgroups of $G$. Any $A in frac{G}{H cap K}$ can be written as $A = B cap C$, where $B in G/H$ and $C in G/K$ as follows. For any $g in G$
$$g(Hcap K) = gH cap gK$$.
Hence,
$$[G:H cap K] leq [G:H] [G:K]$$.
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
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add a comment |
$begingroup$
$H, K$ be subgroups of $G$. Any $A in frac{G}{H cap K}$ can be written as $A = B cap C$, where $B in G/H$ and $C in G/K$ as follows. For any $g in G$
$$g(Hcap K) = gH cap gK$$.
Hence,
$$[G:H cap K] leq [G:H] [G:K]$$.
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
$endgroup$
add a comment |
$begingroup$
$H, K$ be subgroups of $G$. Any $A in frac{G}{H cap K}$ can be written as $A = B cap C$, where $B in G/H$ and $C in G/K$ as follows. For any $g in G$
$$g(Hcap K) = gH cap gK$$.
Hence,
$$[G:H cap K] leq [G:H] [G:K]$$.
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
$endgroup$
$H, K$ be subgroups of $G$. Any $A in frac{G}{H cap K}$ can be written as $A = B cap C$, where $B in G/H$ and $C in G/K$ as follows. For any $g in G$
$$g(Hcap K) = gH cap gK$$.
Hence,
$$[G:H cap K] leq [G:H] [G:K]$$.
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
edited Feb 12 '18 at 18:30
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
answered Feb 9 '18 at 22:39
Awra DipAwra Dip
112
112
add a comment |
add a comment |
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Let $C$ be the set of left cosets of $S = H cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C to C_1 times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| leq |C_1| cdot |C_2|$, which proves that $[G: S]$ is finite.
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
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add a comment |
$begingroup$
Let $C$ be the set of left cosets of $S = H cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C to C_1 times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| leq |C_1| cdot |C_2|$, which proves that $[G: S]$ is finite.
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
$endgroup$
add a comment |
$begingroup$
Let $C$ be the set of left cosets of $S = H cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C to C_1 times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| leq |C_1| cdot |C_2|$, which proves that $[G: S]$ is finite.
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
$endgroup$
Let $C$ be the set of left cosets of $S = H cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C to C_1 times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| leq |C_1| cdot |C_2|$, which proves that $[G: S]$ is finite.
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
answered Jun 3 '18 at 23:12
Alan YanAlan Yan
655411
655411
add a comment |
add a comment |
$begingroup$
Let $l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l in G$ and $G$ be partitioned as $$G=h_1H cup ... cup h_lH = k_1K cup ... cup k_mK$$
I will find $a_1,...a_n in G$ s.t. $G$ is partitioned as $$G=a_1(H cap K) cup ... cup a_n(H cap K)$$ which means $n=[G:H cap K] < infty$.
Let $b_1 in G$. Then $exists i_1 in {1,...,m}, j_i in {1,...,l}$ s.t. $b_1 in h_{i_1}H cap k_{j_1}K=b_1H cap b_1K = b_1(H cap K)$. Next, let $b_2 in G setminus b_1(H cap K)$. Then $b_2 in b_2(H cap K)$ where $b_2(H cap K) cap b_1(H cap K) = emptyset$ because $$b_1H cap b_2H = h_{i_1}H cap h_{i_2}H = emptyset = k_{j_1}K cap k_{j_2}K = b_1K cap b_2K$$
where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 in {1,...,m} setminus {i_1}, j_2 in {1,...,l} setminus {j_1}$$ This process continues at most $lm$ times for $a_p=b_p, p in {1,2,...,n}$ Thus, $n le lm < infty.$
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add a comment |
$begingroup$
Let $l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l in G$ and $G$ be partitioned as $$G=h_1H cup ... cup h_lH = k_1K cup ... cup k_mK$$
I will find $a_1,...a_n in G$ s.t. $G$ is partitioned as $$G=a_1(H cap K) cup ... cup a_n(H cap K)$$ which means $n=[G:H cap K] < infty$.
Let $b_1 in G$. Then $exists i_1 in {1,...,m}, j_i in {1,...,l}$ s.t. $b_1 in h_{i_1}H cap k_{j_1}K=b_1H cap b_1K = b_1(H cap K)$. Next, let $b_2 in G setminus b_1(H cap K)$. Then $b_2 in b_2(H cap K)$ where $b_2(H cap K) cap b_1(H cap K) = emptyset$ because $$b_1H cap b_2H = h_{i_1}H cap h_{i_2}H = emptyset = k_{j_1}K cap k_{j_2}K = b_1K cap b_2K$$
where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 in {1,...,m} setminus {i_1}, j_2 in {1,...,l} setminus {j_1}$$ This process continues at most $lm$ times for $a_p=b_p, p in {1,2,...,n}$ Thus, $n le lm < infty.$
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add a comment |
$begingroup$
Let $l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l in G$ and $G$ be partitioned as $$G=h_1H cup ... cup h_lH = k_1K cup ... cup k_mK$$
I will find $a_1,...a_n in G$ s.t. $G$ is partitioned as $$G=a_1(H cap K) cup ... cup a_n(H cap K)$$ which means $n=[G:H cap K] < infty$.
Let $b_1 in G$. Then $exists i_1 in {1,...,m}, j_i in {1,...,l}$ s.t. $b_1 in h_{i_1}H cap k_{j_1}K=b_1H cap b_1K = b_1(H cap K)$. Next, let $b_2 in G setminus b_1(H cap K)$. Then $b_2 in b_2(H cap K)$ where $b_2(H cap K) cap b_1(H cap K) = emptyset$ because $$b_1H cap b_2H = h_{i_1}H cap h_{i_2}H = emptyset = k_{j_1}K cap k_{j_2}K = b_1K cap b_2K$$
where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 in {1,...,m} setminus {i_1}, j_2 in {1,...,l} setminus {j_1}$$ This process continues at most $lm$ times for $a_p=b_p, p in {1,2,...,n}$ Thus, $n le lm < infty.$
$endgroup$
Let $l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l in G$ and $G$ be partitioned as $$G=h_1H cup ... cup h_lH = k_1K cup ... cup k_mK$$
I will find $a_1,...a_n in G$ s.t. $G$ is partitioned as $$G=a_1(H cap K) cup ... cup a_n(H cap K)$$ which means $n=[G:H cap K] < infty$.
Let $b_1 in G$. Then $exists i_1 in {1,...,m}, j_i in {1,...,l}$ s.t. $b_1 in h_{i_1}H cap k_{j_1}K=b_1H cap b_1K = b_1(H cap K)$. Next, let $b_2 in G setminus b_1(H cap K)$. Then $b_2 in b_2(H cap K)$ where $b_2(H cap K) cap b_1(H cap K) = emptyset$ because $$b_1H cap b_2H = h_{i_1}H cap h_{i_2}H = emptyset = k_{j_1}K cap k_{j_2}K = b_1K cap b_2K$$
where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 in {1,...,m} setminus {i_1}, j_2 in {1,...,l} setminus {j_1}$$ This process continues at most $lm$ times for $a_p=b_p, p in {1,2,...,n}$ Thus, $n le lm < infty.$
answered Oct 14 '18 at 3:04
user198044
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$begingroup$
I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $H$ and $K$ are finite index subgroups of $G$, we can consider the right cosets $G/H$ and this is finite by hypothesis. In particular $G$ acts (transitively) on $G/H$ by right multiplication, i.e.
$$
G/H times G to G, qquad
(Hx, g) mapsto Hxg.
$$
If we now restrict the action to $K$, and consider the $K$-orbit of $H$, call it $mathcal{O}_K(H)$, then it must be finite since it's contained in $G/H$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality
$$
|mathcal{O}_K(H)| = [K : text{Stab}_K(H)],
$$
where $text{Stab}_K(H)$ is the stabiliser of $H$ under the $K$-action. Notice that $text{Stab}_K(H) = K cap text{Stab}_G(H)$, and what's the stabiliser of $H$ for the $G$-action? But of course $H$ itself, so $text{Stab}_K(H) = K cap H$, and finally we conclude that $[K : K cap H]$ is finite. Hence $[G : H cap K] = [G : K][K : K cap H]$ must be finite as well.
answered Jan 27 at 2:54
fmnqfmnq
11
$endgroup$
add a comment |
$begingroup$
I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $H$ and $K$ are finite index subgroups of $G$, we can consider the right cosets $G/H$ and this is finite by hypothesis. In particular $G$ acts (transitively) on $G/H$ by right multiplication, i.e.
$$
G/H times G to G, qquad
(Hx, g) mapsto Hxg.
$$
If we now restrict the action to $K$, and consider the $K$-orbit of $H$, call it $mathcal{O}_K(H)$, then it must be finite since it's contained in $G/H$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality
$$
|mathcal{O}_K(H)| = [K : text{Stab}_K(H)],
$$
where $text{Stab}_K(H)$ is the stabiliser of $H$ under the $K$-action. Notice that $text{Stab}_K(H) = K cap text{Stab}_G(H)$, and what's the stabiliser of $H$ for the $G$-action? But of course $H$ itself, so $text{Stab}_K(H) = K cap H$, and finally we conclude that $[K : K cap H]$ is finite. Hence $[G : H cap K] = [G : K][K : K cap H]$ must be finite as well.
answered Jan 27 at 2:54
fmnqfmnq
11
$endgroup$
add a comment |
$begingroup$
I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $H$ and $K$ are finite index subgroups of $G$, we can consider the right cosets $G/H$ and this is finite by hypothesis. In particular $G$ acts (transitively) on $G/H$ by right multiplication, i.e.
$$
G/H times G to G, qquad
(Hx, g) mapsto Hxg.
$$
If we now restrict the action to $K$, and consider the $K$-orbit of $H$, call it $mathcal{O}_K(H)$, then it must be finite since it's contained in $G/H$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality
$$
|mathcal{O}_K(H)| = [K : text{Stab}_K(H)],
$$
where $text{Stab}_K(H)$ is the stabiliser of $H$ under the $K$-action. Notice that $text{Stab}_K(H) = K cap text{Stab}_G(H)$, and what's the stabiliser of $H$ for the $G$-action? But of course $H$ itself, so $text{Stab}_K(H) = K cap H$, and finally we conclude that $[K : K cap H]$ is finite. Hence $[G : H cap K] = [G : K][K : K cap H]$ must be finite as well.
answered Jan 27 at 2:54
fmnqfmnq
11
$endgroup$
I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.
Since both $H$ and $K$ are finite index subgroups of $G$, we can consider the right cosets $G/H$ and this is finite by hypothesis. In particular $G$ acts (transitively) on $G/H$ by right multiplication, i.e.
$$
G/H times G to G, qquad
(Hx, g) mapsto Hxg.
$$
If we now restrict the action to $K$, and consider the $K$-orbit of $H$, call it $mathcal{O}_K(H)$, then it must be finite since it's contained in $G/H$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality
$$
|mathcal{O}_K(H)| = [K : text{Stab}_K(H)],
$$
where $text{Stab}_K(H)$ is the stabiliser of $H$ under the $K$-action. Notice that $text{Stab}_K(H) = K cap text{Stab}_G(H)$, and what's the stabiliser of $H$ for the $G$-action? But of course $H$ itself, so $text{Stab}_K(H) = K cap H$, and finally we conclude that $[K : K cap H]$ is finite. Hence $[G : H cap K] = [G : K][K : K cap H]$ must be finite as well.
answered Jan 27 at 2:54
fmnqfmnq
11
edited Mar 7 at 13:37
answered Jan 27 at 2:54
fmnqfmnq
11
answered Jan 27 at 2:54
fmnqfmnq
11
answered Jan 27 at 2:54
fmnqfmnq
11
11
add a comment |
add a comment |
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$begingroup$
Step 1: Show that $H cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H cap K][Hcap K : H]$.
$endgroup$
– Clive Newstead
Apr 5 '12 at 22:41
$begingroup$
@MartQ. You don't have to show that it's a divisor. It suffices to show that $[Gcolon(Hcap K)]leq [Gcolon H]cdot[Gcolon K].$ You can try to find an injection from a certain set to another.
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– user23211
Apr 5 '12 at 23:02
$begingroup$
A related question.
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– user23211
Apr 5 '12 at 23:06
4
$begingroup$
@CliveNewstead The formula $[G:H] = [G:H cap K][Hcap K : H]$ seems incorrect to me. What is $[Hcap K : H]?$ Did you mean $[G:Hcap K]=[G:H][H:Hcap K]?$
$endgroup$
– user23211
Apr 5 '12 at 23:23
1
$begingroup$
@ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.
$endgroup$
– Clive Newstead
Apr 7 '12 at 11:45