Question about second Isomorphism Theorem
$begingroup$
I came across this theorem that says:
If H and K are subgroups of a group G, then $[H:Hcap K]leq[G:K]$. If $[G:K]$ is finite then $[H:Hcap K]=[G:K]$ if and only if $G=KH.$
And if K is normal, then KH is a subgroup, so we will have $[H:Hcap K] = [G:K]$ and hence $|H|/|H cap K| = |HK|/|K|$
However, I know that second isomorphism theorem says that if we have an extra condition saying K is a normal subgroup of G. Then we have $H/(H cap K)$ isomorphic to $HK/K$. And this will mean that $|H|/|Hcap K| = |HK|/|K|$. Why then do we need G = KH?
My question is what are the difference between these two? I find the first theorem redundant?
abstract-algebra group-theory
edited Jan 13 at 19:21
Matt Samuel
38k63666
asked Nov 24 '14 at 14:18
user10024395user10024395
1
$endgroup$
add a comment |
$begingroup$
I came across this theorem that says:
If H and K are subgroups of a group G, then $[H:Hcap K]leq[G:K]$. If $[G:K]$ is finite then $[H:Hcap K]=[G:K]$ if and only if $G=KH.$
And if K is normal, then KH is a subgroup, so we will have $[H:Hcap K] = [G:K]$ and hence $|H|/|H cap K| = |HK|/|K|$
However, I know that second isomorphism theorem says that if we have an extra condition saying K is a normal subgroup of G. Then we have $H/(H cap K)$ isomorphic to $HK/K$. And this will mean that $|H|/|Hcap K| = |HK|/|K|$. Why then do we need G = KH?
My question is what are the difference between these two? I find the first theorem redundant?
abstract-algebra group-theory
edited Jan 13 at 19:21
Matt Samuel
38k63666
asked Nov 24 '14 at 14:18
user10024395user10024395
1
$endgroup$
add a comment |
$begingroup$
I came across this theorem that says:
If H and K are subgroups of a group G, then $[H:Hcap K]leq[G:K]$. If $[G:K]$ is finite then $[H:Hcap K]=[G:K]$ if and only if $G=KH.$
And if K is normal, then KH is a subgroup, so we will have $[H:Hcap K] = [G:K]$ and hence $|H|/|H cap K| = |HK|/|K|$
However, I know that second isomorphism theorem says that if we have an extra condition saying K is a normal subgroup of G. Then we have $H/(H cap K)$ isomorphic to $HK/K$. And this will mean that $|H|/|Hcap K| = |HK|/|K|$. Why then do we need G = KH?
My question is what are the difference between these two? I find the first theorem redundant?
abstract-algebra group-theory
edited Jan 13 at 19:21
Matt Samuel
38k63666
asked Nov 24 '14 at 14:18
user10024395user10024395
1
$endgroup$
I came across this theorem that says:
If H and K are subgroups of a group G, then $[H:Hcap K]leq[G:K]$. If $[G:K]$ is finite then $[H:Hcap K]=[G:K]$ if and only if $G=KH.$
And if K is normal, then KH is a subgroup, so we will have $[H:Hcap K] = [G:K]$ and hence $|H|/|H cap K| = |HK|/|K|$
However, I know that second isomorphism theorem says that if we have an extra condition saying K is a normal subgroup of G. Then we have $H/(H cap K)$ isomorphic to $HK/K$. And this will mean that $|H|/|Hcap K| = |HK|/|K|$. Why then do we need G = KH?
My question is what are the difference between these two? I find the first theorem redundant?
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 13 at 19:21
Matt Samuel
38k63666
asked Nov 24 '14 at 14:18
user10024395user10024395
1
edited Jan 13 at 19:21
Matt Samuel
38k63666
asked Nov 24 '14 at 14:18
user10024395user10024395
1
edited Jan 13 at 19:21
Matt Samuel
38k63666
edited Jan 13 at 19:21
Matt Samuel
38k63666
edited Jan 13 at 19:21
Matt Samuel
38k63666
38k63666
asked Nov 24 '14 at 14:18
user10024395user10024395
1
asked Nov 24 '14 at 14:18
user10024395user10024395
1
asked Nov 24 '14 at 14:18
user10024395user10024395
1
1
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
The fact that $HK=KH$ is a subgroup is not sufficient to conclude that $[H:Hcap K]=[G:K]$ as you have written. In that case we can only conclude that $[H:Hcap K]=[HK:K]$, and if $HKneq G$ this will not be equal to $ [G:K]$.
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
$endgroup$
add a comment |
$begingroup$
First of all, if $G$ is finite then $|HK| = frac{|H||K|}{|H cap K|}$ for any two subgroups $H$ and $K$ of $G$. But here $HK$ need not be a subgroup of $G$. If $K$ is normal, then $HK$ is a subgroup of $G$. Now if $G = HK$ then $[H : H cap K] = [G : K].$ On the other hand $[H : H cap K] = [G : K] Rightarrow |HK| = |G| Rightarrow G = HK.$
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
The fact that $HK=KH$ is a subgroup is not sufficient to conclude that $[H:Hcap K]=[G:K]$ as you have written. In that case we can only conclude that $[H:Hcap K]=[HK:K]$, and if $HKneq G$ this will not be equal to $ [G:K]$.
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
$endgroup$
add a comment |
$begingroup$
The fact that $HK=KH$ is a subgroup is not sufficient to conclude that $[H:Hcap K]=[G:K]$ as you have written. In that case we can only conclude that $[H:Hcap K]=[HK:K]$, and if $HKneq G$ this will not be equal to $ [G:K]$.
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
$endgroup$
add a comment |
$begingroup$
The fact that $HK=KH$ is a subgroup is not sufficient to conclude that $[H:Hcap K]=[G:K]$ as you have written. In that case we can only conclude that $[H:Hcap K]=[HK:K]$, and if $HKneq G$ this will not be equal to $ [G:K]$.
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
$endgroup$
The fact that $HK=KH$ is a subgroup is not sufficient to conclude that $[H:Hcap K]=[G:K]$ as you have written. In that case we can only conclude that $[H:Hcap K]=[HK:K]$, and if $HKneq G$ this will not be equal to $ [G:K]$.
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
answered Nov 24 '14 at 14:34
Matt SamuelMatt Samuel
38k63666
38k63666
add a comment |
add a comment |
$begingroup$
First of all, if $G$ is finite then $|HK| = frac{|H||K|}{|H cap K|}$ for any two subgroups $H$ and $K$ of $G$. But here $HK$ need not be a subgroup of $G$. If $K$ is normal, then $HK$ is a subgroup of $G$. Now if $G = HK$ then $[H : H cap K] = [G : K].$ On the other hand $[H : H cap K] = [G : K] Rightarrow |HK| = |G| Rightarrow G = HK.$
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
$endgroup$
add a comment |
$begingroup$
First of all, if $G$ is finite then $|HK| = frac{|H||K|}{|H cap K|}$ for any two subgroups $H$ and $K$ of $G$. But here $HK$ need not be a subgroup of $G$. If $K$ is normal, then $HK$ is a subgroup of $G$. Now if $G = HK$ then $[H : H cap K] = [G : K].$ On the other hand $[H : H cap K] = [G : K] Rightarrow |HK| = |G| Rightarrow G = HK.$
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
$endgroup$
add a comment |
$begingroup$
First of all, if $G$ is finite then $|HK| = frac{|H||K|}{|H cap K|}$ for any two subgroups $H$ and $K$ of $G$. But here $HK$ need not be a subgroup of $G$. If $K$ is normal, then $HK$ is a subgroup of $G$. Now if $G = HK$ then $[H : H cap K] = [G : K].$ On the other hand $[H : H cap K] = [G : K] Rightarrow |HK| = |G| Rightarrow G = HK.$
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
$endgroup$
First of all, if $G$ is finite then $|HK| = frac{|H||K|}{|H cap K|}$ for any two subgroups $H$ and $K$ of $G$. But here $HK$ need not be a subgroup of $G$. If $K$ is normal, then $HK$ is a subgroup of $G$. Now if $G = HK$ then $[H : H cap K] = [G : K].$ On the other hand $[H : H cap K] = [G : K] Rightarrow |HK| = |G| Rightarrow G = HK.$
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
answered Nov 24 '14 at 14:42
KrishKrish
6,31411020
6,31411020
add a comment |
add a comment |
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