Mixing
Showing that $lvert H Krvert = frac K $ [duplicate]
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Order of a product of subgroups. Prove that $o(HK) = fraco(H)o(K)o(H cap K)$.
5 answers
I have seen a proof that for two subgroups $H$ and $K$ of a finite group $G$,
$$
|H K| = frac
$$
The proof counts the number of elements $hk$ that are the same as $h'k'$ for $h,h'in H$ and $k,k'in K$. I got confused with this proof.
My question is if there is another way to make things like this more clear. For example, I know that if one has a surjective homomorphism $f:G to H$ then $lvert Grvert / lvert ker frvert = lvert Hrvert$. That is, one can count the number of elements in something by essentially finding the order of the kernel of a map. This looks like what is going on with the product $HK$. But I guess that $HK$ in general isn't a group, so one can't talk about a group homomorphism.
Is there a way to find the "kernel" of a general function and get something like the first isomorphism theorem? All I would be interested in is how this can be used to count things.
abstract-algebra combinatorics group-theory finite-groups
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
asked Oct 12 '17 at 17:06
John Doe
25311242
marked as duplicate by Guy Fsone, Rolf Hoyer, Claude Leibovici, JonMark Perry, mechanodroid Nov 7 '17 at 8:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
7
down vote
favorite
This question already has an answer here:
Order of a product of subgroups. Prove that $o(HK) = fraco(H)o(K)o(H cap K)$.
5 answers
I have seen a proof that for two subgroups $H$ and $K$ of a finite group $G$,
$$
|H K| = frac
$$
The proof counts the number of elements $hk$ that are the same as $h'k'$ for $h,h'in H$ and $k,k'in K$. I got confused with this proof.
My question is if there is another way to make things like this more clear. For example, I know that if one has a surjective homomorphism $f:G to H$ then $lvert Grvert / lvert ker frvert = lvert Hrvert$. That is, one can count the number of elements in something by essentially finding the order of the kernel of a map. This looks like what is going on with the product $HK$. But I guess that $HK$ in general isn't a group, so one can't talk about a group homomorphism.
Is there a way to find the "kernel" of a general function and get something like the first isomorphism theorem? All I would be interested in is how this can be used to count things.
abstract-algebra combinatorics group-theory finite-groups
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
asked Oct 12 '17 at 17:06
John Doe
25311242
marked as duplicate by Guy Fsone, Rolf Hoyer, Claude Leibovici, JonMark Perry, mechanodroid Nov 7 '17 at 8:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
This question already has an answer here:
Order of a product of subgroups. Prove that $o(HK) = fraco(H)o(K)o(H cap K)$.
5 answers
I have seen a proof that for two subgroups $H$ and $K$ of a finite group $G$,
$$
|H K| = frac
$$
The proof counts the number of elements $hk$ that are the same as $h'k'$ for $h,h'in H$ and $k,k'in K$. I got confused with this proof.
My question is if there is another way to make things like this more clear. For example, I know that if one has a surjective homomorphism $f:G to H$ then $lvert Grvert / lvert ker frvert = lvert Hrvert$. That is, one can count the number of elements in something by essentially finding the order of the kernel of a map. This looks like what is going on with the product $HK$. But I guess that $HK$ in general isn't a group, so one can't talk about a group homomorphism.
Is there a way to find the "kernel" of a general function and get something like the first isomorphism theorem? All I would be interested in is how this can be used to count things.
abstract-algebra combinatorics group-theory finite-groups
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
asked Oct 12 '17 at 17:06
John Doe
25311242
This question already has an answer here:
Order of a product of subgroups. Prove that $o(HK) = fraco(H)o(K)o(H cap K)$.
5 answers
I have seen a proof that for two subgroups $H$ and $K$ of a finite group $G$,
$$
|H K| = frac
$$
The proof counts the number of elements $hk$ that are the same as $h'k'$ for $h,h'in H$ and $k,k'in K$. I got confused with this proof.
My question is if there is another way to make things like this more clear. For example, I know that if one has a surjective homomorphism $f:G to H$ then $lvert Grvert / lvert ker frvert = lvert Hrvert$. That is, one can count the number of elements in something by essentially finding the order of the kernel of a map. This looks like what is going on with the product $HK$. But I guess that $HK$ in general isn't a group, so one can't talk about a group homomorphism.
Is there a way to find the "kernel" of a general function and get something like the first isomorphism theorem? All I would be interested in is how this can be used to count things.
This question already has an answer here:
Order of a product of subgroups. Prove that $o(HK) = fraco(H)o(K)o(H cap K)$.
5 answers
abstract-algebra combinatorics group-theory finite-groups
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
asked Oct 12 '17 at 17:06
John Doe
25311242
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
edited Oct 23 '17 at 16:01
Guy Fsone
16.8k42671
16.8k42671
asked Oct 12 '17 at 17:06
John Doe
25311242
asked Oct 12 '17 at 17:06
John Doe
25311242
asked Oct 12 '17 at 17:06
John Doe
25311242
25311242
marked as duplicate by Guy Fsone, Rolf Hoyer, Claude Leibovici, JonMark Perry, mechanodroid Nov 7 '17 at 8:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Guy Fsone, Rolf Hoyer, Claude Leibovici, JonMark Perry, mechanodroid Nov 7 '17 at 8:23
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
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accepted
$Htimes K$ acts on $G$ (on the right) via $gcdot(h,k)=h^-1gk$. Then $HK$ is the orbit of the identity element $e$, and so its size is $|Htimes K|/|S_e|$ where $S_e$ is the stabiliser of $e$. Then $S_e=(h,h):kin Hcap K$.
This method can also count the general double coset $HxK$.
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
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up vote
10
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Lemma: $colorblue Htimes K/mathcal R$ has $n$ elements that is,
$colorblue $ and we have,
$$ colorblue =fracHcap K $$
This is a consequence of $E_1$ and $E_2$ see below for all the details.
Consider the map
beginsplit
phi :&& Htimes Kto HK\
&& (h,k)mapsto hk
endsplit
Clearly, $phi $ is onto (surjective ). Now we consider the relation,
$$colorred(h,k)mathcal R(h',k')Longleftrightarrow hk=h'k'Longleftrightarrow phi(h,k)=phi(h',k')$$
It is easy to check that $mathcal R$ is an equivalent relation on $Htimes K$.
Fact.I. Let denote by $[h,k]_mathcal R$ the class of an element $(h,k) in Htimes K.$
Let $nin mathbb N$ be the number of classes of $ Htimes K$ with respect to $mathcal R.$ Also we denote by $colorblue Htimes K/mathcal R$ the set of class. Precisely we have,
$$colorblue Htimes K/mathcal R= [h_1,k_1]_mathcal R, [h_2,k_2]_mathcal Rcdots, [h_n,k_n]_mathcal R $$
Where, $(h_j,k_j)_j$ is a set of representative class of $Htimes K/mathcal R $.
First Equality: We consider the $overlinephi$ the quotient map of $phi$ w.r.t $mathcal R.$ defines as follows:
beginsplit
overlinephi :&& Htimes K/mathcal R to HK\
&& [h,k]_mathcal R mapsto phi(h,k) = hk
endsplit
$overlinephi $ is well define since from the red line above we have,
$$colorred(h',k') in [h,k]_mathcal RLongleftrightarrow (h,k)mathcal R(h',k')\Longleftrightarrow hk=h'k'Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)tagEq.$$
- It is also easy to check that $overlinephi$ is one-to-one(injective) and Hence bijective since $phi$ is onto.
$$colorred[h',k']_mathcal R = [h,k]_mathcal RLongleftrightarrow (h',k') in [h,k]_mathcal R Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)$$
conclusion $overlinephi$ is a bijection and therfore,
$$colorbluetag$E_1$ $$
we are jumping to the second equality, starting from the following observation.
Fact. II Since $mathcal R$ is an equivalent relation, we know that $colorred([h_j,k_j]_mathcal R)_1le jle n$ is a partition of $Htimes K$ that is,
$$colorred $$
Claim:(see the proof Below)
$$colorred = $$
Second Equality: Since for any finte sets A and B we have $|Atimes B| =|A|times|B|,$ using the claim and the foregoing relations, we get that
$$colorblue$$
Then $$ colorbluen= tag$E_2$ $$
Proof of the claim:
Now we would like to investigate $|[h,k]_mathcal R|$.
$$colorblue(h',k')in [h,k]_mathcal R Longleftrightarrow hk=h'k'Longleftrightarrow h'^-1h=k'k^-1in Hcap K .$$
Consider the map
beginsplit
f :&& [h,k]_mathcal R to Hcap K\
&& (h',k')mapsto h'^-1h=k'k^-1
endsplit
The above relation shows that $f$ is well defined as a map. We will show that $f$ is a bijective map to conclude.
$f$ is onto(surjective): Let $sin Hcap K $. If we let
$ k' = h s^-1~~~textand~~~ k'=sk$
then $colorredh'k' = hs^-1 sk =hkLongleftrightarrow (h',k')mathcal R (h,k) implies (h',k') in [h,k]_mathcal R$
and hence $$colorredf(h',k')= f(hs^-1, sk) = s.$$
this prove that $f$ is onto.
$f$ is one-to-one(injective): let, $(a,b), (x,y)in [h,k]_mathcal R $ such that $f(a,b)=f(x,y)$.
We have
$f(a,b) =a^-1h =bk^-1 ~~~~textand~~~f(x,y) =x^-1h =yk^-1$
then,
beginsplit
f(a,b)=f(x,y)&implies& colorbluea^-1h =bk^-1 =colorredx^-1h =yk^-1\
&implies& colorbluea^-1h =x^-1h ~~~~textand~~~~colorred bk^-1=yk^-1\
&implies& colorreda =x ~~~~textand~~~~colorred b=y\
endsplit
This proves that $f$ is bijective,s then the claim follows
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
$Htimes K$ acts on $G$ (on the right) via $gcdot(h,k)=h^-1gk$. Then $HK$ is the orbit of the identity element $e$, and so its size is $|Htimes K|/|S_e|$ where $S_e$ is the stabiliser of $e$. Then $S_e=(h,h):kin Hcap K$.
This method can also count the general double coset $HxK$.
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
add a comment |Â
up vote
13
down vote
accepted
$Htimes K$ acts on $G$ (on the right) via $gcdot(h,k)=h^-1gk$. Then $HK$ is the orbit of the identity element $e$, and so its size is $|Htimes K|/|S_e|$ where $S_e$ is the stabiliser of $e$. Then $S_e=(h,h):kin Hcap K$.
This method can also count the general double coset $HxK$.
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
add a comment |Â
up vote
13
down vote
accepted
up vote
13
down vote
accepted
$Htimes K$ acts on $G$ (on the right) via $gcdot(h,k)=h^-1gk$. Then $HK$ is the orbit of the identity element $e$, and so its size is $|Htimes K|/|S_e|$ where $S_e$ is the stabiliser of $e$. Then $S_e=(h,h):kin Hcap K$.
This method can also count the general double coset $HxK$.
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
$Htimes K$ acts on $G$ (on the right) via $gcdot(h,k)=h^-1gk$. Then $HK$ is the orbit of the identity element $e$, and so its size is $|Htimes K|/|S_e|$ where $S_e$ is the stabiliser of $e$. Then $S_e=(h,h):kin Hcap K$.
This method can also count the general double coset $HxK$.
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
answered Oct 12 '17 at 17:13
Lord Shark the Unknown
85.5k951112
85.5k951112
add a comment |Â
add a comment |Â
up vote
10
down vote
Lemma: $colorblue Htimes K/mathcal R$ has $n$ elements that is,
$colorblue $ and we have,
$$ colorblue =fracHcap K $$
This is a consequence of $E_1$ and $E_2$ see below for all the details.
Consider the map
beginsplit
phi :&& Htimes Kto HK\
&& (h,k)mapsto hk
endsplit
Clearly, $phi $ is onto (surjective ). Now we consider the relation,
$$colorred(h,k)mathcal R(h',k')Longleftrightarrow hk=h'k'Longleftrightarrow phi(h,k)=phi(h',k')$$
It is easy to check that $mathcal R$ is an equivalent relation on $Htimes K$.
Fact.I. Let denote by $[h,k]_mathcal R$ the class of an element $(h,k) in Htimes K.$
Let $nin mathbb N$ be the number of classes of $ Htimes K$ with respect to $mathcal R.$ Also we denote by $colorblue Htimes K/mathcal R$ the set of class. Precisely we have,
$$colorblue Htimes K/mathcal R= [h_1,k_1]_mathcal R, [h_2,k_2]_mathcal Rcdots, [h_n,k_n]_mathcal R $$
Where, $(h_j,k_j)_j$ is a set of representative class of $Htimes K/mathcal R $.
First Equality: We consider the $overlinephi$ the quotient map of $phi$ w.r.t $mathcal R.$ defines as follows:
beginsplit
overlinephi :&& Htimes K/mathcal R to HK\
&& [h,k]_mathcal R mapsto phi(h,k) = hk
endsplit
$overlinephi $ is well define since from the red line above we have,
$$colorred(h',k') in [h,k]_mathcal RLongleftrightarrow (h,k)mathcal R(h',k')\Longleftrightarrow hk=h'k'Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)tagEq.$$
- It is also easy to check that $overlinephi$ is one-to-one(injective) and Hence bijective since $phi$ is onto.
$$colorred[h',k']_mathcal R = [h,k]_mathcal RLongleftrightarrow (h',k') in [h,k]_mathcal R Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)$$
conclusion $overlinephi$ is a bijection and therfore,
$$colorbluetag$E_1$ $$
we are jumping to the second equality, starting from the following observation.
Fact. II Since $mathcal R$ is an equivalent relation, we know that $colorred([h_j,k_j]_mathcal R)_1le jle n$ is a partition of $Htimes K$ that is,
$$colorred $$
Claim:(see the proof Below)
$$colorred = $$
Second Equality: Since for any finte sets A and B we have $|Atimes B| =|A|times|B|,$ using the claim and the foregoing relations, we get that
$$colorblue$$
Then $$ colorbluen= tag$E_2$ $$
Proof of the claim:
Now we would like to investigate $|[h,k]_mathcal R|$.
$$colorblue(h',k')in [h,k]_mathcal R Longleftrightarrow hk=h'k'Longleftrightarrow h'^-1h=k'k^-1in Hcap K .$$
Consider the map
beginsplit
f :&& [h,k]_mathcal R to Hcap K\
&& (h',k')mapsto h'^-1h=k'k^-1
endsplit
The above relation shows that $f$ is well defined as a map. We will show that $f$ is a bijective map to conclude.
$f$ is onto(surjective): Let $sin Hcap K $. If we let
$ k' = h s^-1~~~textand~~~ k'=sk$
then $colorredh'k' = hs^-1 sk =hkLongleftrightarrow (h',k')mathcal R (h,k) implies (h',k') in [h,k]_mathcal R$
and hence $$colorredf(h',k')= f(hs^-1, sk) = s.$$
this prove that $f$ is onto.
$f$ is one-to-one(injective): let, $(a,b), (x,y)in [h,k]_mathcal R $ such that $f(a,b)=f(x,y)$.
We have
$f(a,b) =a^-1h =bk^-1 ~~~~textand~~~f(x,y) =x^-1h =yk^-1$
then,
beginsplit
f(a,b)=f(x,y)&implies& colorbluea^-1h =bk^-1 =colorredx^-1h =yk^-1\
&implies& colorbluea^-1h =x^-1h ~~~~textand~~~~colorred bk^-1=yk^-1\
&implies& colorreda =x ~~~~textand~~~~colorred b=y\
endsplit
This proves that $f$ is bijective,s then the claim follows
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
add a comment |Â
up vote
10
down vote
Lemma: $colorblue Htimes K/mathcal R$ has $n$ elements that is,
$colorblue $ and we have,
$$ colorblue =fracHcap K $$
This is a consequence of $E_1$ and $E_2$ see below for all the details.
Consider the map
beginsplit
phi :&& Htimes Kto HK\
&& (h,k)mapsto hk
endsplit
Clearly, $phi $ is onto (surjective ). Now we consider the relation,
$$colorred(h,k)mathcal R(h',k')Longleftrightarrow hk=h'k'Longleftrightarrow phi(h,k)=phi(h',k')$$
It is easy to check that $mathcal R$ is an equivalent relation on $Htimes K$.
Fact.I. Let denote by $[h,k]_mathcal R$ the class of an element $(h,k) in Htimes K.$
Let $nin mathbb N$ be the number of classes of $ Htimes K$ with respect to $mathcal R.$ Also we denote by $colorblue Htimes K/mathcal R$ the set of class. Precisely we have,
$$colorblue Htimes K/mathcal R= [h_1,k_1]_mathcal R, [h_2,k_2]_mathcal Rcdots, [h_n,k_n]_mathcal R $$
Where, $(h_j,k_j)_j$ is a set of representative class of $Htimes K/mathcal R $.
First Equality: We consider the $overlinephi$ the quotient map of $phi$ w.r.t $mathcal R.$ defines as follows:
beginsplit
overlinephi :&& Htimes K/mathcal R to HK\
&& [h,k]_mathcal R mapsto phi(h,k) = hk
endsplit
$overlinephi $ is well define since from the red line above we have,
$$colorred(h',k') in [h,k]_mathcal RLongleftrightarrow (h,k)mathcal R(h',k')\Longleftrightarrow hk=h'k'Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)tagEq.$$
- It is also easy to check that $overlinephi$ is one-to-one(injective) and Hence bijective since $phi$ is onto.
$$colorred[h',k']_mathcal R = [h,k]_mathcal RLongleftrightarrow (h',k') in [h,k]_mathcal R Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)$$
conclusion $overlinephi$ is a bijection and therfore,
$$colorbluetag$E_1$ $$
we are jumping to the second equality, starting from the following observation.
Fact. II Since $mathcal R$ is an equivalent relation, we know that $colorred([h_j,k_j]_mathcal R)_1le jle n$ is a partition of $Htimes K$ that is,
$$colorred $$
Claim:(see the proof Below)
$$colorred = $$
Second Equality: Since for any finte sets A and B we have $|Atimes B| =|A|times|B|,$ using the claim and the foregoing relations, we get that
$$colorblue$$
Then $$ colorbluen= tag$E_2$ $$
Proof of the claim:
Now we would like to investigate $|[h,k]_mathcal R|$.
$$colorblue(h',k')in [h,k]_mathcal R Longleftrightarrow hk=h'k'Longleftrightarrow h'^-1h=k'k^-1in Hcap K .$$
Consider the map
beginsplit
f :&& [h,k]_mathcal R to Hcap K\
&& (h',k')mapsto h'^-1h=k'k^-1
endsplit
The above relation shows that $f$ is well defined as a map. We will show that $f$ is a bijective map to conclude.
$f$ is onto(surjective): Let $sin Hcap K $. If we let
$ k' = h s^-1~~~textand~~~ k'=sk$
then $colorredh'k' = hs^-1 sk =hkLongleftrightarrow (h',k')mathcal R (h,k) implies (h',k') in [h,k]_mathcal R$
and hence $$colorredf(h',k')= f(hs^-1, sk) = s.$$
this prove that $f$ is onto.
$f$ is one-to-one(injective): let, $(a,b), (x,y)in [h,k]_mathcal R $ such that $f(a,b)=f(x,y)$.
We have
$f(a,b) =a^-1h =bk^-1 ~~~~textand~~~f(x,y) =x^-1h =yk^-1$
then,
beginsplit
f(a,b)=f(x,y)&implies& colorbluea^-1h =bk^-1 =colorredx^-1h =yk^-1\
&implies& colorbluea^-1h =x^-1h ~~~~textand~~~~colorred bk^-1=yk^-1\
&implies& colorreda =x ~~~~textand~~~~colorred b=y\
endsplit
This proves that $f$ is bijective,s then the claim follows
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
add a comment |Â
up vote
10
down vote
up vote
10
down vote
Lemma: $colorblue Htimes K/mathcal R$ has $n$ elements that is,
$colorblue $ and we have,
$$ colorblue =fracHcap K $$
This is a consequence of $E_1$ and $E_2$ see below for all the details.
Consider the map
beginsplit
phi :&& Htimes Kto HK\
&& (h,k)mapsto hk
endsplit
Clearly, $phi $ is onto (surjective ). Now we consider the relation,
$$colorred(h,k)mathcal R(h',k')Longleftrightarrow hk=h'k'Longleftrightarrow phi(h,k)=phi(h',k')$$
It is easy to check that $mathcal R$ is an equivalent relation on $Htimes K$.
Fact.I. Let denote by $[h,k]_mathcal R$ the class of an element $(h,k) in Htimes K.$
Let $nin mathbb N$ be the number of classes of $ Htimes K$ with respect to $mathcal R.$ Also we denote by $colorblue Htimes K/mathcal R$ the set of class. Precisely we have,
$$colorblue Htimes K/mathcal R= [h_1,k_1]_mathcal R, [h_2,k_2]_mathcal Rcdots, [h_n,k_n]_mathcal R $$
Where, $(h_j,k_j)_j$ is a set of representative class of $Htimes K/mathcal R $.
First Equality: We consider the $overlinephi$ the quotient map of $phi$ w.r.t $mathcal R.$ defines as follows:
beginsplit
overlinephi :&& Htimes K/mathcal R to HK\
&& [h,k]_mathcal R mapsto phi(h,k) = hk
endsplit
$overlinephi $ is well define since from the red line above we have,
$$colorred(h',k') in [h,k]_mathcal RLongleftrightarrow (h,k)mathcal R(h',k')\Longleftrightarrow hk=h'k'Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)tagEq.$$
- It is also easy to check that $overlinephi$ is one-to-one(injective) and Hence bijective since $phi$ is onto.
$$colorred[h',k']_mathcal R = [h,k]_mathcal RLongleftrightarrow (h',k') in [h,k]_mathcal R Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)$$
conclusion $overlinephi$ is a bijection and therfore,
$$colorbluetag$E_1$ $$
we are jumping to the second equality, starting from the following observation.
Fact. II Since $mathcal R$ is an equivalent relation, we know that $colorred([h_j,k_j]_mathcal R)_1le jle n$ is a partition of $Htimes K$ that is,
$$colorred $$
Claim:(see the proof Below)
$$colorred = $$
Second Equality: Since for any finte sets A and B we have $|Atimes B| =|A|times|B|,$ using the claim and the foregoing relations, we get that
$$colorblue$$
Then $$ colorbluen= tag$E_2$ $$
Proof of the claim:
Now we would like to investigate $|[h,k]_mathcal R|$.
$$colorblue(h',k')in [h,k]_mathcal R Longleftrightarrow hk=h'k'Longleftrightarrow h'^-1h=k'k^-1in Hcap K .$$
Consider the map
beginsplit
f :&& [h,k]_mathcal R to Hcap K\
&& (h',k')mapsto h'^-1h=k'k^-1
endsplit
The above relation shows that $f$ is well defined as a map. We will show that $f$ is a bijective map to conclude.
$f$ is onto(surjective): Let $sin Hcap K $. If we let
$ k' = h s^-1~~~textand~~~ k'=sk$
then $colorredh'k' = hs^-1 sk =hkLongleftrightarrow (h',k')mathcal R (h,k) implies (h',k') in [h,k]_mathcal R$
and hence $$colorredf(h',k')= f(hs^-1, sk) = s.$$
this prove that $f$ is onto.
$f$ is one-to-one(injective): let, $(a,b), (x,y)in [h,k]_mathcal R $ such that $f(a,b)=f(x,y)$.
We have
$f(a,b) =a^-1h =bk^-1 ~~~~textand~~~f(x,y) =x^-1h =yk^-1$
then,
beginsplit
f(a,b)=f(x,y)&implies& colorbluea^-1h =bk^-1 =colorredx^-1h =yk^-1\
&implies& colorbluea^-1h =x^-1h ~~~~textand~~~~colorred bk^-1=yk^-1\
&implies& colorreda =x ~~~~textand~~~~colorred b=y\
endsplit
This proves that $f$ is bijective,s then the claim follows
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
Lemma: $colorblue Htimes K/mathcal R$ has $n$ elements that is,
$colorblue $ and we have,
$$ colorblue =fracHcap K $$
This is a consequence of $E_1$ and $E_2$ see below for all the details.
Consider the map
beginsplit
phi :&& Htimes Kto HK\
&& (h,k)mapsto hk
endsplit
Clearly, $phi $ is onto (surjective ). Now we consider the relation,
$$colorred(h,k)mathcal R(h',k')Longleftrightarrow hk=h'k'Longleftrightarrow phi(h,k)=phi(h',k')$$
It is easy to check that $mathcal R$ is an equivalent relation on $Htimes K$.
Fact.I. Let denote by $[h,k]_mathcal R$ the class of an element $(h,k) in Htimes K.$
Let $nin mathbb N$ be the number of classes of $ Htimes K$ with respect to $mathcal R.$ Also we denote by $colorblue Htimes K/mathcal R$ the set of class. Precisely we have,
$$colorblue Htimes K/mathcal R= [h_1,k_1]_mathcal R, [h_2,k_2]_mathcal Rcdots, [h_n,k_n]_mathcal R $$
Where, $(h_j,k_j)_j$ is a set of representative class of $Htimes K/mathcal R $.
First Equality: We consider the $overlinephi$ the quotient map of $phi$ w.r.t $mathcal R.$ defines as follows:
beginsplit
overlinephi :&& Htimes K/mathcal R to HK\
&& [h,k]_mathcal R mapsto phi(h,k) = hk
endsplit
$overlinephi $ is well define since from the red line above we have,
$$colorred(h',k') in [h,k]_mathcal RLongleftrightarrow (h,k)mathcal R(h',k')\Longleftrightarrow hk=h'k'Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)tagEq.$$
- It is also easy to check that $overlinephi$ is one-to-one(injective) and Hence bijective since $phi$ is onto.
$$colorred[h',k']_mathcal R = [h,k]_mathcal RLongleftrightarrow (h',k') in [h,k]_mathcal R Longleftrightarrow overlinephi([h,k]_mathcal R)=overlinephi([h',k']_mathcal R)$$
conclusion $overlinephi$ is a bijection and therfore,
$$colorbluetag$E_1$ $$
we are jumping to the second equality, starting from the following observation.
Fact. II Since $mathcal R$ is an equivalent relation, we know that $colorred([h_j,k_j]_mathcal R)_1le jle n$ is a partition of $Htimes K$ that is,
$$colorred $$
Claim:(see the proof Below)
$$colorred = $$
Second Equality: Since for any finte sets A and B we have $|Atimes B| =|A|times|B|,$ using the claim and the foregoing relations, we get that
$$colorblue$$
Then $$ colorbluen= tag$E_2$ $$
Proof of the claim:
Now we would like to investigate $|[h,k]_mathcal R|$.
$$colorblue(h',k')in [h,k]_mathcal R Longleftrightarrow hk=h'k'Longleftrightarrow h'^-1h=k'k^-1in Hcap K .$$
Consider the map
beginsplit
f :&& [h,k]_mathcal R to Hcap K\
&& (h',k')mapsto h'^-1h=k'k^-1
endsplit
The above relation shows that $f$ is well defined as a map. We will show that $f$ is a bijective map to conclude.
$f$ is onto(surjective): Let $sin Hcap K $. If we let
$ k' = h s^-1~~~textand~~~ k'=sk$
then $colorredh'k' = hs^-1 sk =hkLongleftrightarrow (h',k')mathcal R (h,k) implies (h',k') in [h,k]_mathcal R$
and hence $$colorredf(h',k')= f(hs^-1, sk) = s.$$
this prove that $f$ is onto.
$f$ is one-to-one(injective): let, $(a,b), (x,y)in [h,k]_mathcal R $ such that $f(a,b)=f(x,y)$.
We have
$f(a,b) =a^-1h =bk^-1 ~~~~textand~~~f(x,y) =x^-1h =yk^-1$
then,
beginsplit
f(a,b)=f(x,y)&implies& colorbluea^-1h =bk^-1 =colorredx^-1h =yk^-1\
&implies& colorbluea^-1h =x^-1h ~~~~textand~~~~colorred bk^-1=yk^-1\
&implies& colorreda =x ~~~~textand~~~~colorred b=y\
endsplit
This proves that $f$ is bijective,s then the claim follows
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
edited Nov 21 '17 at 20:04
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
answered Oct 12 '17 at 17:37
Guy Fsone
16.8k42671
16.8k42671
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