Equivalent conditions for a Group $G$ with order $p^2q$ ( with $p>q$ both prime) be abelian.
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I saw this homework many times, but always asks in the statament that $p^2 notequiv 1$ (mod $q$) and $q notequiv 1$ (mod $p$).
But today in a book text of Galois theory I Saw a similar example that the statament just asks that $q$ doesn't divides $p^2$.
Just using this fact i Tried to prove that exists just one subgroup with order $p^2$ and just one with order $q$
By the first Sylow theorem groups with such order exists, and by the third theorem the number of this groups are of the form $n_{p^2}=1+kp^2, kin mathbb{N}$ and $n_{q}=1+k'q, k'in mathbb{N}$ with boths dividing the order of $G$.
$n_{p^2}$ divides $p^2$ just for $k=0$ and by the hyphotesis $qnmid p^2$ how Could I conclude that there is no $kneq 0$ such that $n_{p^2}nmid q$?
Analogously how to use this to prove $n_{q}nmid p^2$?
group-theory prime-numbers abelian-groups sylow-theory
asked Nov 16 at 15:27
Eduardo Silva
67739
|
show 3 more comments
up vote
0
down vote
favorite
I saw this homework many times, but always asks in the statament that $p^2 notequiv 1$ (mod $q$) and $q notequiv 1$ (mod $p$).
But today in a book text of Galois theory I Saw a similar example that the statament just asks that $q$ doesn't divides $p^2$.
Just using this fact i Tried to prove that exists just one subgroup with order $p^2$ and just one with order $q$
By the first Sylow theorem groups with such order exists, and by the third theorem the number of this groups are of the form $n_{p^2}=1+kp^2, kin mathbb{N}$ and $n_{q}=1+k'q, k'in mathbb{N}$ with boths dividing the order of $G$.
$n_{p^2}$ divides $p^2$ just for $k=0$ and by the hyphotesis $qnmid p^2$ how Could I conclude that there is no $kneq 0$ such that $n_{p^2}nmid q$?
Analogously how to use this to prove $n_{q}nmid p^2$?
group-theory prime-numbers abelian-groups sylow-theory
asked Nov 16 at 15:27
Eduardo Silva
67739
1
Not sure the question is clear. The statement (or question?) in the header is false, clearly. Take the direct product of $S_3$ and the cyclic group of order $3$, say.
– lulu
Nov 16 at 15:30
1
@Eduardo Silva : (math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf)
– Yadati Kiran
Nov 16 at 15:41
1
@EduardoSilva : It s because of Sylow theorem: $n_pequiv1mod p$ and $n_p|: |G|$ in particular $n_p|qimplies 1+kp|q$. Here we see that it is possible for $k=1$. Hence the argument follows.
– Yadati Kiran
Nov 16 at 16:18
1
It should be $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$. Also $ :qnmid p^{2}-1$ therefore $n_{q} = 1+kq neq p,: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So - $G cong mathbb{Z}_{p^2} times mathbb{Z}_{q}$
– Yadati Kiran
Nov 16 at 16:41
1
Please edit your post to ask a clear, coherent question. The condition that $q$ shouldn't divide $p^2$ doesn't even make sense...the only prime that divides $p^2$ is $p$ and we know $qneq p$.
– lulu
Nov 16 at 16:41
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I saw this homework many times, but always asks in the statament that $p^2 notequiv 1$ (mod $q$) and $q notequiv 1$ (mod $p$).
But today in a book text of Galois theory I Saw a similar example that the statament just asks that $q$ doesn't divides $p^2$.
Just using this fact i Tried to prove that exists just one subgroup with order $p^2$ and just one with order $q$
By the first Sylow theorem groups with such order exists, and by the third theorem the number of this groups are of the form $n_{p^2}=1+kp^2, kin mathbb{N}$ and $n_{q}=1+k'q, k'in mathbb{N}$ with boths dividing the order of $G$.
$n_{p^2}$ divides $p^2$ just for $k=0$ and by the hyphotesis $qnmid p^2$ how Could I conclude that there is no $kneq 0$ such that $n_{p^2}nmid q$?
Analogously how to use this to prove $n_{q}nmid p^2$?
group-theory prime-numbers abelian-groups sylow-theory
asked Nov 16 at 15:27
Eduardo Silva
67739
I saw this homework many times, but always asks in the statament that $p^2 notequiv 1$ (mod $q$) and $q notequiv 1$ (mod $p$).
But today in a book text of Galois theory I Saw a similar example that the statament just asks that $q$ doesn't divides $p^2$.
Just using this fact i Tried to prove that exists just one subgroup with order $p^2$ and just one with order $q$
By the first Sylow theorem groups with such order exists, and by the third theorem the number of this groups are of the form $n_{p^2}=1+kp^2, kin mathbb{N}$ and $n_{q}=1+k'q, k'in mathbb{N}$ with boths dividing the order of $G$.
$n_{p^2}$ divides $p^2$ just for $k=0$ and by the hyphotesis $qnmid p^2$ how Could I conclude that there is no $kneq 0$ such that $n_{p^2}nmid q$?
Analogously how to use this to prove $n_{q}nmid p^2$?
group-theory prime-numbers abelian-groups sylow-theory
group-theory prime-numbers abelian-groups sylow-theory
asked Nov 16 at 15:27
Eduardo Silva
67739
asked Nov 16 at 15:27
Eduardo Silva
67739
edited Nov 16 at 20:30
asked Nov 16 at 15:27
Eduardo Silva
67739
asked Nov 16 at 15:27
Eduardo Silva
67739
asked Nov 16 at 15:27
Eduardo Silva
67739
67739
1
Not sure the question is clear. The statement (or question?) in the header is false, clearly. Take the direct product of $S_3$ and the cyclic group of order $3$, say.
– lulu
Nov 16 at 15:30
1
@Eduardo Silva : (math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf)
– Yadati Kiran
Nov 16 at 15:41
1
@EduardoSilva : It s because of Sylow theorem: $n_pequiv1mod p$ and $n_p|: |G|$ in particular $n_p|qimplies 1+kp|q$. Here we see that it is possible for $k=1$. Hence the argument follows.
– Yadati Kiran
Nov 16 at 16:18
1
It should be $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$. Also $ :qnmid p^{2}-1$ therefore $n_{q} = 1+kq neq p,: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So - $G cong mathbb{Z}_{p^2} times mathbb{Z}_{q}$
– Yadati Kiran
Nov 16 at 16:41
1
Please edit your post to ask a clear, coherent question. The condition that $q$ shouldn't divide $p^2$ doesn't even make sense...the only prime that divides $p^2$ is $p$ and we know $qneq p$.
– lulu
Nov 16 at 16:41
|
show 3 more comments
1
Not sure the question is clear. The statement (or question?) in the header is false, clearly. Take the direct product of $S_3$ and the cyclic group of order $3$, say.
– lulu
Nov 16 at 15:30
1
@Eduardo Silva : (math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf)
– Yadati Kiran
Nov 16 at 15:41
1
@EduardoSilva : It s because of Sylow theorem: $n_pequiv1mod p$ and $n_p|: |G|$ in particular $n_p|qimplies 1+kp|q$. Here we see that it is possible for $k=1$. Hence the argument follows.
– Yadati Kiran
Nov 16 at 16:18
1
It should be $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$. Also $ :qnmid p^{2}-1$ therefore $n_{q} = 1+kq neq p,: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So - $G cong mathbb{Z}_{p^2} times mathbb{Z}_{q}$
– Yadati Kiran
Nov 16 at 16:41
1
Please edit your post to ask a clear, coherent question. The condition that $q$ shouldn't divide $p^2$ doesn't even make sense...the only prime that divides $p^2$ is $p$ and we know $qneq p$.
– lulu
Nov 16 at 16:41
1
1
Not sure the question is clear. The statement (or question?) in the header is false, clearly. Take the direct product of $S_3$ and the cyclic group of order $3$, say.
– lulu
Nov 16 at 15:30
Not sure the question is clear. The statement (or question?) in the header is false, clearly. Take the direct product of $S_3$ and the cyclic group of order $3$, say.
– lulu
Nov 16 at 15:30
1
1
@Eduardo Silva : (math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf)
– Yadati Kiran
Nov 16 at 15:41
@Eduardo Silva : (math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf)
– Yadati Kiran
Nov 16 at 15:41
1
1
@EduardoSilva : It s because of Sylow theorem: $n_pequiv1mod p$ and $n_p|: |G|$ in particular $n_p|qimplies 1+kp|q$. Here we see that it is possible for $k=1$. Hence the argument follows.
– Yadati Kiran
Nov 16 at 16:18
@EduardoSilva : It s because of Sylow theorem: $n_pequiv1mod p$ and $n_p|: |G|$ in particular $n_p|qimplies 1+kp|q$. Here we see that it is possible for $k=1$. Hence the argument follows.
– Yadati Kiran
Nov 16 at 16:18
1
1
It should be $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$. Also $ :qnmid p^{2}-1$ therefore $n_{q} = 1+kq neq p,: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So - $G cong mathbb{Z}_{p^2} times mathbb{Z}_{q}$
– Yadati Kiran
Nov 16 at 16:41
It should be $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$. Also $ :qnmid p^{2}-1$ therefore $n_{q} = 1+kq neq p,: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So - $G cong mathbb{Z}_{p^2} times mathbb{Z}_{q}$
– Yadati Kiran
Nov 16 at 16:41
1
1
Please edit your post to ask a clear, coherent question. The condition that $q$ shouldn't divide $p^2$ doesn't even make sense...the only prime that divides $p^2$ is $p$ and we know $qneq p$.
– lulu
Nov 16 at 16:41
Please edit your post to ask a clear, coherent question. The condition that $q$ shouldn't divide $p^2$ doesn't even make sense...the only prime that divides $p^2$ is $p$ and we know $qneq p$.
– lulu
Nov 16 at 16:41
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
@Eduardo Silva: $q$ doesn't divide $p^2$ is not enough (in fact, it is equivalent to $p neq q$), and so Iulu example works as a counterexample. I guess the only thing you can say with your hypoteses is that there is only one Sylow p-Subgroup (since $p>q$, $n_p$ must be $1$, for the next available value is $p+1$, which cannot divide $q$).
Are you sure there is no other assumption?
answered Nov 16 at 16:35
Pietro Gheri
762
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
@Eduardo Silva: $q$ doesn't divide $p^2$ is not enough (in fact, it is equivalent to $p neq q$), and so Iulu example works as a counterexample. I guess the only thing you can say with your hypoteses is that there is only one Sylow p-Subgroup (since $p>q$, $n_p$ must be $1$, for the next available value is $p+1$, which cannot divide $q$).
Are you sure there is no other assumption?
answered Nov 16 at 16:35
Pietro Gheri
762
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
add a comment |
up vote
1
down vote
accepted
@Eduardo Silva: $q$ doesn't divide $p^2$ is not enough (in fact, it is equivalent to $p neq q$), and so Iulu example works as a counterexample. I guess the only thing you can say with your hypoteses is that there is only one Sylow p-Subgroup (since $p>q$, $n_p$ must be $1$, for the next available value is $p+1$, which cannot divide $q$).
Are you sure there is no other assumption?
answered Nov 16 at 16:35
Pietro Gheri
762
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
@Eduardo Silva: $q$ doesn't divide $p^2$ is not enough (in fact, it is equivalent to $p neq q$), and so Iulu example works as a counterexample. I guess the only thing you can say with your hypoteses is that there is only one Sylow p-Subgroup (since $p>q$, $n_p$ must be $1$, for the next available value is $p+1$, which cannot divide $q$).
Are you sure there is no other assumption?
answered Nov 16 at 16:35
Pietro Gheri
762
@Eduardo Silva: $q$ doesn't divide $p^2$ is not enough (in fact, it is equivalent to $p neq q$), and so Iulu example works as a counterexample. I guess the only thing you can say with your hypoteses is that there is only one Sylow p-Subgroup (since $p>q$, $n_p$ must be $1$, for the next available value is $p+1$, which cannot divide $q$).
Are you sure there is no other assumption?
answered Nov 16 at 16:35
Pietro Gheri
762
answered Nov 16 at 16:35
Pietro Gheri
762
answered Nov 16 at 16:35
Pietro Gheri
762
answered Nov 16 at 16:35
Pietro Gheri
762
762
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
add a comment |
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
you are right, the exercise must be wrong, i guess that the only condition nedded is that $qnmid p^2-1$, so we can assume that this was a type mistake
– Eduardo Silva
Nov 16 at 20:29
add a comment |
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1
Not sure the question is clear. The statement (or question?) in the header is false, clearly. Take the direct product of $S_3$ and the cyclic group of order $3$, say.
– lulu
Nov 16 at 15:30
1
@Eduardo Silva : (math.purdue.edu/~lipman/5532011/order-p%5E2q.pdf)
– Yadati Kiran
Nov 16 at 15:41
1
@EduardoSilva : It s because of Sylow theorem: $n_pequiv1mod p$ and $n_p|: |G|$ in particular $n_p|qimplies 1+kp|q$. Here we see that it is possible for $k=1$. Hence the argument follows.
– Yadati Kiran
Nov 16 at 16:18
1
It should be $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$. Also $ :qnmid p^{2}-1$ therefore $n_{q} = 1+kq neq p,: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So - $G cong mathbb{Z}_{p^2} times mathbb{Z}_{q}$
– Yadati Kiran
Nov 16 at 16:41
1
Please edit your post to ask a clear, coherent question. The condition that $q$ shouldn't divide $p^2$ doesn't even make sense...the only prime that divides $p^2$ is $p$ and we know $qneq p$.
– lulu
Nov 16 at 16:41