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If $P,Q$ are $p-$subgroup of $G$, then so is $PQ$. [closed]
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Let $G$ be a finite group and $p$ divide $|G|$. Suppose $P$ and $Q$ are two subgroups s.t. $Psubset N_G(Q)$. Prove that $PQ$ is a $p-$subgroup of $G$.
I imagine that I have to use second isomorphism theorem, but I don't know how to do.
group-theory finite-groups
edited Jul 30 at 11:34
mesel
10.1k21644
asked Jul 30 at 11:28
Henri
304
closed as off-topic by Derek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, José Carlos Santos Aug 1 at 22:47
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." РDerek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, Jos̩ Carlos Santos
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up vote
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down vote
favorite
Let $G$ be a finite group and $p$ divide $|G|$. Suppose $P$ and $Q$ are two subgroups s.t. $Psubset N_G(Q)$. Prove that $PQ$ is a $p-$subgroup of $G$.
I imagine that I have to use second isomorphism theorem, but I don't know how to do.
group-theory finite-groups
edited Jul 30 at 11:34
mesel
10.1k21644
asked Jul 30 at 11:28
Henri
304
closed as off-topic by Derek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, José Carlos Santos Aug 1 at 22:47
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." РDerek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, Jos̩ Carlos Santos
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be a finite group and $p$ divide $|G|$. Suppose $P$ and $Q$ are two subgroups s.t. $Psubset N_G(Q)$. Prove that $PQ$ is a $p-$subgroup of $G$.
I imagine that I have to use second isomorphism theorem, but I don't know how to do.
group-theory finite-groups
edited Jul 30 at 11:34
mesel
10.1k21644
asked Jul 30 at 11:28
Henri
304
Let $G$ be a finite group and $p$ divide $|G|$. Suppose $P$ and $Q$ are two subgroups s.t. $Psubset N_G(Q)$. Prove that $PQ$ is a $p-$subgroup of $G$.
I imagine that I have to use second isomorphism theorem, but I don't know how to do.
group-theory finite-groups
edited Jul 30 at 11:34
mesel
10.1k21644
asked Jul 30 at 11:28
Henri
304
edited Jul 30 at 11:34
mesel
10.1k21644
edited Jul 30 at 11:34
mesel
10.1k21644
edited Jul 30 at 11:34
mesel
10.1k21644
10.1k21644
asked Jul 30 at 11:28
Henri
304
asked Jul 30 at 11:28
Henri
304
asked Jul 30 at 11:28
Henri
304
304
closed as off-topic by Derek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, José Carlos Santos Aug 1 at 22:47
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." РDerek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, Jos̩ Carlos Santos
closed as off-topic by Derek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, José Carlos Santos Aug 1 at 22:47
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." РDerek Holt, Simply Beautiful Art, John Ma, Mostafa Ayaz, Jos̩ Carlos Santos
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2 Answers
2
active
oldest
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up vote
3
down vote
accepted
You do not need second isomorphism theorem:
Recall that the equality always true $$|HK|=dfracK.$$
In general $HK$ need not to be a subgroup. Just notice that $HK$ is a group if $K$ normalizes $H$, that is, when $Kleq N_G(H)$.
answered Jul 30 at 11:32
mesel
10.1k21644
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
1
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
 |Â
show 3 more comments
up vote
0
down vote
Hint
Your idea by using 2nd isomorphism theorem is the right idea. You have that $P$ is a subgroup of $N_G(Q)$. Since $Qlhd N_G(Q)$ you have that $PQleq N_G(Q)$, and thus you can use 2nd isomorphism theorem in $N_G(Q)$.
answered Jul 30 at 11:30
Surb
36.2k84274
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You do not need second isomorphism theorem:
Recall that the equality always true $$|HK|=dfracK.$$
In general $HK$ need not to be a subgroup. Just notice that $HK$ is a group if $K$ normalizes $H$, that is, when $Kleq N_G(H)$.
answered Jul 30 at 11:32
mesel
10.1k21644
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
1
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
 |Â
show 3 more comments
up vote
3
down vote
accepted
You do not need second isomorphism theorem:
Recall that the equality always true $$|HK|=dfracK.$$
In general $HK$ need not to be a subgroup. Just notice that $HK$ is a group if $K$ normalizes $H$, that is, when $Kleq N_G(H)$.
answered Jul 30 at 11:32
mesel
10.1k21644
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
1
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
 |Â
show 3 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You do not need second isomorphism theorem:
Recall that the equality always true $$|HK|=dfracK.$$
In general $HK$ need not to be a subgroup. Just notice that $HK$ is a group if $K$ normalizes $H$, that is, when $Kleq N_G(H)$.
answered Jul 30 at 11:32
mesel
10.1k21644
You do not need second isomorphism theorem:
Recall that the equality always true $$|HK|=dfracK.$$
In general $HK$ need not to be a subgroup. Just notice that $HK$ is a group if $K$ normalizes $H$, that is, when $Kleq N_G(H)$.
answered Jul 30 at 11:32
mesel
10.1k21644
edited Jul 30 at 13:03
answered Jul 30 at 11:32
mesel
10.1k21644
answered Jul 30 at 11:32
mesel
10.1k21644
answered Jul 30 at 11:32
mesel
10.1k21644
10.1k21644
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
1
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
 |Â
show 3 more comments
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
1
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: It is an elementary fact and does not depend whether $HK$ is a subgroup or not. If you wish, I could write a proof for this fact.
– mesel
Jul 30 at 11:36
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
@Surb: you can also check from this: math.stackexchange.com/questions/168942/…
– mesel
Jul 30 at 11:38
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Just to be sure : In this case, $PQ$ is a subgroup of $G$ right ? Because $PQ$ is a subgroup of $N_G(Q)$ (or not necessarily) (thanks for the link)
– Surb
Jul 30 at 11:39
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
Yes, $PQ$ is a subgroup of $G$ in this case as $P$ normalize $Q$.
– mesel
Jul 30 at 11:41
1
1
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
It come from a peint of René Magritte "ceci n'est pas une pipe". I let you check on the internet :-)
– Surb
Jul 30 at 11:59
 |Â
show 3 more comments
up vote
0
down vote
Hint
Your idea by using 2nd isomorphism theorem is the right idea. You have that $P$ is a subgroup of $N_G(Q)$. Since $Qlhd N_G(Q)$ you have that $PQleq N_G(Q)$, and thus you can use 2nd isomorphism theorem in $N_G(Q)$.
answered Jul 30 at 11:30
Surb
36.2k84274
add a comment |Â
up vote
0
down vote
Hint
Your idea by using 2nd isomorphism theorem is the right idea. You have that $P$ is a subgroup of $N_G(Q)$. Since $Qlhd N_G(Q)$ you have that $PQleq N_G(Q)$, and thus you can use 2nd isomorphism theorem in $N_G(Q)$.
answered Jul 30 at 11:30
Surb
36.2k84274
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint
Your idea by using 2nd isomorphism theorem is the right idea. You have that $P$ is a subgroup of $N_G(Q)$. Since $Qlhd N_G(Q)$ you have that $PQleq N_G(Q)$, and thus you can use 2nd isomorphism theorem in $N_G(Q)$.
answered Jul 30 at 11:30
Surb
36.2k84274
Hint
Your idea by using 2nd isomorphism theorem is the right idea. You have that $P$ is a subgroup of $N_G(Q)$. Since $Qlhd N_G(Q)$ you have that $PQleq N_G(Q)$, and thus you can use 2nd isomorphism theorem in $N_G(Q)$.
answered Jul 30 at 11:30
Surb
36.2k84274
answered Jul 30 at 11:30
Surb
36.2k84274
answered Jul 30 at 11:30
Surb
36.2k84274
answered Jul 30 at 11:30
Surb
36.2k84274
36.2k84274
add a comment |Â
add a comment |Â