Regarding normal subgroup [closed]
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In a group $G$, $H$ is such a finite subgroup that there is no other subgroup containing $o(H)$ elements. Prove that $H$ is normal in $G$.
normal-subgroups
edited Aug 26 at 4:22
Guido A.
4,858728
asked Aug 26 at 4:20
SULAGNA BARAT
303
closed as off-topic by Shaun, Dietrich Burde, Brahadeesh, José Carlos Santos, Holo Sep 12 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РShaun, Dietrich Burde, Brahadeesh, Jos̩ Carlos Santos, Holo
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up vote
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In a group $G$, $H$ is such a finite subgroup that there is no other subgroup containing $o(H)$ elements. Prove that $H$ is normal in $G$.
normal-subgroups
edited Aug 26 at 4:22
Guido A.
4,858728
asked Aug 26 at 4:20
SULAGNA BARAT
303
closed as off-topic by Shaun, Dietrich Burde, Brahadeesh, José Carlos Santos, Holo Sep 12 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РShaun, Dietrich Burde, Brahadeesh, Jos̩ Carlos Santos, Holo
What is $o(H)$? What have you tried?
– Guido A.
Aug 26 at 4:22
It has only been mentioned as o(H). I was trying to approach with index proportion. But it has also not clearly been given whether G is of finite order or not. That's why I'm stuck & can't find any other alternative way.
– SULAGNA BARAT
Aug 26 at 4:25
Does $o(H)$ mean orbit of $H$ ??
– Anik Bhowmick
Aug 26 at 4:34
I'm assuming $o(H)$ is the order of $H$.
– matt stokes
Aug 26 at 4:37
1
Yeah o(H) is nothing but the number of distinct elements in H.
– SULAGNA BARAT
Aug 26 at 4:39
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
In a group $G$, $H$ is such a finite subgroup that there is no other subgroup containing $o(H)$ elements. Prove that $H$ is normal in $G$.
normal-subgroups
edited Aug 26 at 4:22
Guido A.
4,858728
asked Aug 26 at 4:20
SULAGNA BARAT
303
In a group $G$, $H$ is such a finite subgroup that there is no other subgroup containing $o(H)$ elements. Prove that $H$ is normal in $G$.
normal-subgroups
normal-subgroups
edited Aug 26 at 4:22
Guido A.
4,858728
asked Aug 26 at 4:20
SULAGNA BARAT
303
edited Aug 26 at 4:22
Guido A.
4,858728
asked Aug 26 at 4:20
SULAGNA BARAT
303
edited Aug 26 at 4:22
Guido A.
4,858728
edited Aug 26 at 4:22
Guido A.
4,858728
edited Aug 26 at 4:22
Guido A.
4,858728
4,858728
asked Aug 26 at 4:20
SULAGNA BARAT
303
asked Aug 26 at 4:20
SULAGNA BARAT
303
asked Aug 26 at 4:20
SULAGNA BARAT
303
303
closed as off-topic by Shaun, Dietrich Burde, Brahadeesh, José Carlos Santos, Holo Sep 12 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РShaun, Dietrich Burde, Brahadeesh, Jos̩ Carlos Santos, Holo
closed as off-topic by Shaun, Dietrich Burde, Brahadeesh, José Carlos Santos, Holo Sep 12 at 20:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РShaun, Dietrich Burde, Brahadeesh, Jos̩ Carlos Santos, Holo
What is $o(H)$? What have you tried?
– Guido A.
Aug 26 at 4:22
It has only been mentioned as o(H). I was trying to approach with index proportion. But it has also not clearly been given whether G is of finite order or not. That's why I'm stuck & can't find any other alternative way.
– SULAGNA BARAT
Aug 26 at 4:25
Does $o(H)$ mean orbit of $H$ ??
– Anik Bhowmick
Aug 26 at 4:34
I'm assuming $o(H)$ is the order of $H$.
– matt stokes
Aug 26 at 4:37
1
Yeah o(H) is nothing but the number of distinct elements in H.
– SULAGNA BARAT
Aug 26 at 4:39
add a comment |Â
What is $o(H)$? What have you tried?
– Guido A.
Aug 26 at 4:22
It has only been mentioned as o(H). I was trying to approach with index proportion. But it has also not clearly been given whether G is of finite order or not. That's why I'm stuck & can't find any other alternative way.
– SULAGNA BARAT
Aug 26 at 4:25
Does $o(H)$ mean orbit of $H$ ??
– Anik Bhowmick
Aug 26 at 4:34
I'm assuming $o(H)$ is the order of $H$.
– matt stokes
Aug 26 at 4:37
1
Yeah o(H) is nothing but the number of distinct elements in H.
– SULAGNA BARAT
Aug 26 at 4:39
What is $o(H)$? What have you tried?
– Guido A.
Aug 26 at 4:22
What is $o(H)$? What have you tried?
– Guido A.
Aug 26 at 4:22
It has only been mentioned as o(H). I was trying to approach with index proportion. But it has also not clearly been given whether G is of finite order or not. That's why I'm stuck & can't find any other alternative way.
– SULAGNA BARAT
Aug 26 at 4:25
It has only been mentioned as o(H). I was trying to approach with index proportion. But it has also not clearly been given whether G is of finite order or not. That's why I'm stuck & can't find any other alternative way.
– SULAGNA BARAT
Aug 26 at 4:25
Does $o(H)$ mean orbit of $H$ ??
– Anik Bhowmick
Aug 26 at 4:34
Does $o(H)$ mean orbit of $H$ ??
– Anik Bhowmick
Aug 26 at 4:34
I'm assuming $o(H)$ is the order of $H$.
– matt stokes
Aug 26 at 4:37
I'm assuming $o(H)$ is the order of $H$.
– matt stokes
Aug 26 at 4:37
1
1
Yeah o(H) is nothing but the number of distinct elements in H.
– SULAGNA BARAT
Aug 26 at 4:39
Yeah o(H) is nothing but the number of distinct elements in H.
– SULAGNA BARAT
Aug 26 at 4:39
add a comment |Â
2 Answers
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up vote
5
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Hint: Let $g in G$ and show that $gHg^-1 = ghg^-1 ,:,h in H$ is a subgroup of $G$. What is its order?
answered Aug 26 at 4:30
matt stokes
62529
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
add a comment |Â
up vote
0
down vote
A little bit detailed version of the previous answer:
Let any $g in G$. We know $ g in G:ghg^-1=K_g$ forms a subgroup of G.
[
Proof: $geg^-1=e in K_g$ ,i.e $K_g$ is non empty. Now, $[gh_1g^-1][gh_2g^-1]^-1=gh_1h_2^-1g^-1=gh_3g^-1 in K_g$ which implies that it is indeed a subgroup of G
]
To prove that $o(H)=o(K_g)$, we take arbitrary elements in $K_g$. Now $gh_ig^-1=gh_jg^-1 implies h_i=h_j$. Hence, we can see, that for each element $g in G$, we have $K_g= H$ [ As $o(H)$ is unique] hence, $gHg^-1=H, forall g in G$, which implies the normality of the subgroup $H$.
answered Aug 26 at 5:33
Subhasis Biswas
313111
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Hint: Let $g in G$ and show that $gHg^-1 = ghg^-1 ,:,h in H$ is a subgroup of $G$. What is its order?
answered Aug 26 at 4:30
matt stokes
62529
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
add a comment |Â
up vote
5
down vote
Hint: Let $g in G$ and show that $gHg^-1 = ghg^-1 ,:,h in H$ is a subgroup of $G$. What is its order?
answered Aug 26 at 4:30
matt stokes
62529
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Hint: Let $g in G$ and show that $gHg^-1 = ghg^-1 ,:,h in H$ is a subgroup of $G$. What is its order?
answered Aug 26 at 4:30
matt stokes
62529
Hint: Let $g in G$ and show that $gHg^-1 = ghg^-1 ,:,h in H$ is a subgroup of $G$. What is its order?
answered Aug 26 at 4:30
matt stokes
62529
answered Aug 26 at 4:30
matt stokes
62529
answered Aug 26 at 4:30
matt stokes
62529
answered Aug 26 at 4:30
matt stokes
62529
62529
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
add a comment |Â
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
It's order is equal to o(H)
– SULAGNA BARAT
Aug 26 at 4:33
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Yes, and what does that tell you about how $H$ is related to $gHg^-1$?
– matt stokes
Aug 26 at 4:34
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
Since there is no other subgroup of G containing o(H) elements, it is a must that gHg^(-1)=H. Hence proved. Ohh! got it..thank you:)
– SULAGNA BARAT
Aug 26 at 4:38
add a comment |Â
up vote
0
down vote
A little bit detailed version of the previous answer:
Let any $g in G$. We know $ g in G:ghg^-1=K_g$ forms a subgroup of G.
[
Proof: $geg^-1=e in K_g$ ,i.e $K_g$ is non empty. Now, $[gh_1g^-1][gh_2g^-1]^-1=gh_1h_2^-1g^-1=gh_3g^-1 in K_g$ which implies that it is indeed a subgroup of G
]
To prove that $o(H)=o(K_g)$, we take arbitrary elements in $K_g$. Now $gh_ig^-1=gh_jg^-1 implies h_i=h_j$. Hence, we can see, that for each element $g in G$, we have $K_g= H$ [ As $o(H)$ is unique] hence, $gHg^-1=H, forall g in G$, which implies the normality of the subgroup $H$.
answered Aug 26 at 5:33
Subhasis Biswas
313111
add a comment |Â
up vote
0
down vote
A little bit detailed version of the previous answer:
Let any $g in G$. We know $ g in G:ghg^-1=K_g$ forms a subgroup of G.
[
Proof: $geg^-1=e in K_g$ ,i.e $K_g$ is non empty. Now, $[gh_1g^-1][gh_2g^-1]^-1=gh_1h_2^-1g^-1=gh_3g^-1 in K_g$ which implies that it is indeed a subgroup of G
]
To prove that $o(H)=o(K_g)$, we take arbitrary elements in $K_g$. Now $gh_ig^-1=gh_jg^-1 implies h_i=h_j$. Hence, we can see, that for each element $g in G$, we have $K_g= H$ [ As $o(H)$ is unique] hence, $gHg^-1=H, forall g in G$, which implies the normality of the subgroup $H$.
answered Aug 26 at 5:33
Subhasis Biswas
313111
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A little bit detailed version of the previous answer:
Let any $g in G$. We know $ g in G:ghg^-1=K_g$ forms a subgroup of G.
[
Proof: $geg^-1=e in K_g$ ,i.e $K_g$ is non empty. Now, $[gh_1g^-1][gh_2g^-1]^-1=gh_1h_2^-1g^-1=gh_3g^-1 in K_g$ which implies that it is indeed a subgroup of G
]
To prove that $o(H)=o(K_g)$, we take arbitrary elements in $K_g$. Now $gh_ig^-1=gh_jg^-1 implies h_i=h_j$. Hence, we can see, that for each element $g in G$, we have $K_g= H$ [ As $o(H)$ is unique] hence, $gHg^-1=H, forall g in G$, which implies the normality of the subgroup $H$.
answered Aug 26 at 5:33
Subhasis Biswas
313111
A little bit detailed version of the previous answer:
Let any $g in G$. We know $ g in G:ghg^-1=K_g$ forms a subgroup of G.
[
Proof: $geg^-1=e in K_g$ ,i.e $K_g$ is non empty. Now, $[gh_1g^-1][gh_2g^-1]^-1=gh_1h_2^-1g^-1=gh_3g^-1 in K_g$ which implies that it is indeed a subgroup of G
]
To prove that $o(H)=o(K_g)$, we take arbitrary elements in $K_g$. Now $gh_ig^-1=gh_jg^-1 implies h_i=h_j$. Hence, we can see, that for each element $g in G$, we have $K_g= H$ [ As $o(H)$ is unique] hence, $gHg^-1=H, forall g in G$, which implies the normality of the subgroup $H$.
answered Aug 26 at 5:33
Subhasis Biswas
313111
edited Aug 26 at 5:39
answered Aug 26 at 5:33
Subhasis Biswas
313111
answered Aug 26 at 5:33
Subhasis Biswas
313111
answered Aug 26 at 5:33
Subhasis Biswas
313111
313111
add a comment |Â
add a comment |Â
What is $o(H)$? What have you tried?
– Guido A.
Aug 26 at 4:22
It has only been mentioned as o(H). I was trying to approach with index proportion. But it has also not clearly been given whether G is of finite order or not. That's why I'm stuck & can't find any other alternative way.
– SULAGNA BARAT
Aug 26 at 4:25
Does $o(H)$ mean orbit of $H$ ??
– Anik Bhowmick
Aug 26 at 4:34
I'm assuming $o(H)$ is the order of $H$.
– matt stokes
Aug 26 at 4:37
1
Yeah o(H) is nothing but the number of distinct elements in H.
– SULAGNA BARAT
Aug 26 at 4:39