Mixing
If $H,K$ are subgroups of finite group $G,$ then $|G:(H cap K)|leq |G:H||G:K|$
If $H,K$ are subgroups of finite group $G,$ then $|G:(H cap K)|leq |G:H||G:K|$
May I know if my proof is correct? Thank you v. much.
Proof:
$$|G:(H cap K)| leq |G:H||G:K| Longleftrightarrow frac{|G|}{|H cap K|}frac{|H|}{|G|} = frac{|H|}{|H cap K|}leq frac{|G|}{|K|}$$
So it suffices to prove: $|H/(Hcap K)| leq |G/K|,$ where $G$ is not necessarily finite. Given $h in H,$ let $$ h(Hcap K) mapsto hK$$
The mapping is well-defined:
$$h'(H cap K) = h(H cap K) implies exists m in H cap K: h'=hm implies h'K = hmK= hK$$
The mapping is injective:
$$hK = h'K implies exists k in K: h=h'k implies k in H cap K implies h(Hcap K) = h'(Hcap K)$$
group-theory finite-groups proof-verification
edited Nov 26 '13 at 7:40
mrs
1
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
add a comment |
If $H,K$ are subgroups of finite group $G,$ then $|G:(H cap K)|leq |G:H||G:K|$
May I know if my proof is correct? Thank you v. much.
Proof:
$$|G:(H cap K)| leq |G:H||G:K| Longleftrightarrow frac{|G|}{|H cap K|}frac{|H|}{|G|} = frac{|H|}{|H cap K|}leq frac{|G|}{|K|}$$
So it suffices to prove: $|H/(Hcap K)| leq |G/K|,$ where $G$ is not necessarily finite. Given $h in H,$ let $$ h(Hcap K) mapsto hK$$
The mapping is well-defined:
$$h'(H cap K) = h(H cap K) implies exists m in H cap K: h'=hm implies h'K = hmK= hK$$
The mapping is injective:
$$hK = h'K implies exists k in K: h=h'k implies k in H cap K implies h(Hcap K) = h'(Hcap K)$$
group-theory finite-groups proof-verification
edited Nov 26 '13 at 7:40
mrs
1
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
5
I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding.
– Saal Hardali
Nov 26 '13 at 8:17
Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $;forall,,,exists;,;implies;,;iff;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of.
– DonAntonio
Nov 26 '13 at 13:56
Although many mathematicians do it anyway, it is technically agrammatical to use $Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $text{P}$ be true, then $text{Q}$ is true so $text{R}$ is true so $text{S}$ is true so $text{T}$ is true..."
– Alexander Gruber♦
Nov 26 '13 at 21:40
add a comment |
If $H,K$ are subgroups of finite group $G,$ then $|G:(H cap K)|leq |G:H||G:K|$
May I know if my proof is correct? Thank you v. much.
Proof:
$$|G:(H cap K)| leq |G:H||G:K| Longleftrightarrow frac{|G|}{|H cap K|}frac{|H|}{|G|} = frac{|H|}{|H cap K|}leq frac{|G|}{|K|}$$
So it suffices to prove: $|H/(Hcap K)| leq |G/K|,$ where $G$ is not necessarily finite. Given $h in H,$ let $$ h(Hcap K) mapsto hK$$
The mapping is well-defined:
$$h'(H cap K) = h(H cap K) implies exists m in H cap K: h'=hm implies h'K = hmK= hK$$
The mapping is injective:
$$hK = h'K implies exists k in K: h=h'k implies k in H cap K implies h(Hcap K) = h'(Hcap K)$$
group-theory finite-groups proof-verification
edited Nov 26 '13 at 7:40
mrs
1
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
If $H,K$ are subgroups of finite group $G,$ then $|G:(H cap K)|leq |G:H||G:K|$
May I know if my proof is correct? Thank you v. much.
Proof:
$$|G:(H cap K)| leq |G:H||G:K| Longleftrightarrow frac{|G|}{|H cap K|}frac{|H|}{|G|} = frac{|H|}{|H cap K|}leq frac{|G|}{|K|}$$
So it suffices to prove: $|H/(Hcap K)| leq |G/K|,$ where $G$ is not necessarily finite. Given $h in H,$ let $$ h(Hcap K) mapsto hK$$
The mapping is well-defined:
$$h'(H cap K) = h(H cap K) implies exists m in H cap K: h'=hm implies h'K = hmK= hK$$
The mapping is injective:
$$hK = h'K implies exists k in K: h=h'k implies k in H cap K implies h(Hcap K) = h'(Hcap K)$$
group-theory finite-groups proof-verification
group-theory finite-groups proof-verification
edited Nov 26 '13 at 7:40
mrs
1
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
edited Nov 26 '13 at 7:40
mrs
1
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
edited Nov 26 '13 at 7:40
mrs
1
edited Nov 26 '13 at 7:40
mrs
1
edited Nov 26 '13 at 7:40
mrs
1
1
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
asked Nov 26 '13 at 7:33
Alexy VincenzoAlexy Vincenzo
2,1753925
2,1753925
5
I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding.
– Saal Hardali
Nov 26 '13 at 8:17
Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $;forall,,,exists;,;implies;,;iff;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of.
– DonAntonio
Nov 26 '13 at 13:56
Although many mathematicians do it anyway, it is technically agrammatical to use $Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $text{P}$ be true, then $text{Q}$ is true so $text{R}$ is true so $text{S}$ is true so $text{T}$ is true..."
– Alexander Gruber♦
Nov 26 '13 at 21:40
add a comment |
5
I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding.
– Saal Hardali
Nov 26 '13 at 8:17
Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $;forall,,,exists;,;implies;,;iff;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of.
– DonAntonio
Nov 26 '13 at 13:56
Although many mathematicians do it anyway, it is technically agrammatical to use $Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $text{P}$ be true, then $text{Q}$ is true so $text{R}$ is true so $text{S}$ is true so $text{T}$ is true..."
– Alexander Gruber♦
Nov 26 '13 at 21:40
5
5
I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding.
– Saal Hardali
Nov 26 '13 at 8:17
I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding.
– Saal Hardali
Nov 26 '13 at 8:17
Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $;forall,,,exists;,;implies;,;iff;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of.
– DonAntonio
Nov 26 '13 at 13:56
Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $;forall,,,exists;,;implies;,;iff;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of.
– DonAntonio
Nov 26 '13 at 13:56
Although many mathematicians do it anyway, it is technically agrammatical to use $Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $text{P}$ be true, then $text{Q}$ is true so $text{R}$ is true so $text{S}$ is true so $text{T}$ is true..."
– Alexander Gruber♦
Nov 26 '13 at 21:40
Although many mathematicians do it anyway, it is technically agrammatical to use $Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $text{P}$ be true, then $text{Q}$ is true so $text{R}$ is true so $text{S}$ is true so $text{T}$ is true..."
– Alexander Gruber♦
Nov 26 '13 at 21:40
add a comment |
2 Answers
2
active
oldest
votes
$renewcommand{phi}{varphi}$I think it is preferable to prove first a result of general interest, that is
$$
lvert H K : K rvert = lvert H : H cap K rvert.
$$
Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h in H$.
This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H to H K$ given by $h mapsto h cdot 1$ with the map $HK to HK/K$ given by $g mapsto gK$ to get the map $phi: H mapsto HK/K$ given by $h mapsto h K$. This is clearly onto, and for $a, b in H$ one has
begin{align}
phi(a) = phi(b)
text{ iff }&
a K = b K
\text{iff } &
a^{-1} b in H cap K
\text{iff }&
a (H cap K) = b (H cap K).
end{align}
Thus $phi$ induces a bijection $H/H cap K to HK/K$.
Now argue as you did
$$
lvert G : H cap K rvert
=
lvert G : H rvert cdot lvert H : H cap K rvert
=
lvert G : H rvert cdot lvert HK : K rvert
le
lvert G : H rvert cdot lvert G : K rvert.
$$
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
add a comment |
A simpler solution is to prove that the map
$
qquad G/(Hcap K) to G/H times G/K$
given by
$
g(Hcap K) mapsto (gH, gK)
$
is well defined and injective.
(The quotients here are sets of cosets.)
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
add a comment |
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2 Answers
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2 Answers
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active
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oldest
votes
$renewcommand{phi}{varphi}$I think it is preferable to prove first a result of general interest, that is
$$
lvert H K : K rvert = lvert H : H cap K rvert.
$$
Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h in H$.
This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H to H K$ given by $h mapsto h cdot 1$ with the map $HK to HK/K$ given by $g mapsto gK$ to get the map $phi: H mapsto HK/K$ given by $h mapsto h K$. This is clearly onto, and for $a, b in H$ one has
begin{align}
phi(a) = phi(b)
text{ iff }&
a K = b K
\text{iff } &
a^{-1} b in H cap K
\text{iff }&
a (H cap K) = b (H cap K).
end{align}
Thus $phi$ induces a bijection $H/H cap K to HK/K$.
Now argue as you did
$$
lvert G : H cap K rvert
=
lvert G : H rvert cdot lvert H : H cap K rvert
=
lvert G : H rvert cdot lvert HK : K rvert
le
lvert G : H rvert cdot lvert G : K rvert.
$$
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
add a comment |
$renewcommand{phi}{varphi}$I think it is preferable to prove first a result of general interest, that is
$$
lvert H K : K rvert = lvert H : H cap K rvert.
$$
Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h in H$.
This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H to H K$ given by $h mapsto h cdot 1$ with the map $HK to HK/K$ given by $g mapsto gK$ to get the map $phi: H mapsto HK/K$ given by $h mapsto h K$. This is clearly onto, and for $a, b in H$ one has
begin{align}
phi(a) = phi(b)
text{ iff }&
a K = b K
\text{iff } &
a^{-1} b in H cap K
\text{iff }&
a (H cap K) = b (H cap K).
end{align}
Thus $phi$ induces a bijection $H/H cap K to HK/K$.
Now argue as you did
$$
lvert G : H cap K rvert
=
lvert G : H rvert cdot lvert H : H cap K rvert
=
lvert G : H rvert cdot lvert HK : K rvert
le
lvert G : H rvert cdot lvert G : K rvert.
$$
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
add a comment |
$renewcommand{phi}{varphi}$I think it is preferable to prove first a result of general interest, that is
$$
lvert H K : K rvert = lvert H : H cap K rvert.
$$
Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h in H$.
This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H to H K$ given by $h mapsto h cdot 1$ with the map $HK to HK/K$ given by $g mapsto gK$ to get the map $phi: H mapsto HK/K$ given by $h mapsto h K$. This is clearly onto, and for $a, b in H$ one has
begin{align}
phi(a) = phi(b)
text{ iff }&
a K = b K
\text{iff } &
a^{-1} b in H cap K
\text{iff }&
a (H cap K) = b (H cap K).
end{align}
Thus $phi$ induces a bijection $H/H cap K to HK/K$.
Now argue as you did
$$
lvert G : H cap K rvert
=
lvert G : H rvert cdot lvert H : H cap K rvert
=
lvert G : H rvert cdot lvert HK : K rvert
le
lvert G : H rvert cdot lvert G : K rvert.
$$
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
$renewcommand{phi}{varphi}$I think it is preferable to prove first a result of general interest, that is
$$
lvert H K : K rvert = lvert H : H cap K rvert.
$$
Here $HK$ is not necessarily a subgroup, but it is definitely a union of right cosets $h K$ of $K$, for $h in H$.
This I believe you have implicitly proved, anyway it goes as follows. Compose the map $H to H K$ given by $h mapsto h cdot 1$ with the map $HK to HK/K$ given by $g mapsto gK$ to get the map $phi: H mapsto HK/K$ given by $h mapsto h K$. This is clearly onto, and for $a, b in H$ one has
begin{align}
phi(a) = phi(b)
text{ iff }&
a K = b K
\text{iff } &
a^{-1} b in H cap K
\text{iff }&
a (H cap K) = b (H cap K).
end{align}
Thus $phi$ induces a bijection $H/H cap K to HK/K$.
Now argue as you did
$$
lvert G : H cap K rvert
=
lvert G : H rvert cdot lvert H : H cap K rvert
=
lvert G : H rvert cdot lvert HK : K rvert
le
lvert G : H rvert cdot lvert G : K rvert.
$$
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
edited Nov 26 '13 at 21:43
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
answered Nov 26 '13 at 21:36
Andreas CarantiAndreas Caranti
56.2k34295
56.2k34295
add a comment |
add a comment |
A simpler solution is to prove that the map
$
qquad G/(Hcap K) to G/H times G/K$
given by
$
g(Hcap K) mapsto (gH, gK)
$
is well defined and injective.
(The quotients here are sets of cosets.)
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
add a comment |
A simpler solution is to prove that the map
$
qquad G/(Hcap K) to G/H times G/K$
given by
$
g(Hcap K) mapsto (gH, gK)
$
is well defined and injective.
(The quotients here are sets of cosets.)
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
add a comment |
A simpler solution is to prove that the map
$
qquad G/(Hcap K) to G/H times G/K$
given by
$
g(Hcap K) mapsto (gH, gK)
$
is well defined and injective.
(The quotients here are sets of cosets.)
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
A simpler solution is to prove that the map
$
qquad G/(Hcap K) to G/H times G/K$
given by
$
g(Hcap K) mapsto (gH, gK)
$
is well defined and injective.
(The quotients here are sets of cosets.)
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
answered Nov 21 '18 at 23:59
lhflhf
163k10167388
163k10167388
add a comment |
add a comment |
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5
I would recommend getting used to writing proof with much less logic notation. I had it too and it's a bad habit. Additional i found that writing something in clear english demands a higher level of understanding than writing it logically so it's another good way of testing your understanding.
– Saal Hardali
Nov 26 '13 at 8:17
Less logic notation than the used in the OP?? I don't agree: usual, standard logical symbols (e.g. $;forall,,,exists;,;implies;,;iff;$, etc.) make things much clearer and shorter, something most of us mathematicians cannot have too much of.
– DonAntonio
Nov 26 '13 at 13:56
Although many mathematicians do it anyway, it is technically agrammatical to use $Rightarrow$ as a synonym for "thus." What it really means is "implies." If you really want to write "thus" as a symbol, you should use ∴ instead. Personally I'm a big fan of "so" for long calculations, e.g. "Let $text{P}$ be true, then $text{Q}$ is true so $text{R}$ is true so $text{S}$ is true so $text{T}$ is true..."
– Alexander Gruber♦
Nov 26 '13 at 21:40