Application of the Sylow Theorems to groups of order $p^2q$
$begingroup$
I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.
In the case $p>q$ I have no problem.. By Sylow $n_p|q$, so $n_p$ is either $1$ or $q$. But we also have $n_pequiv1mod p$ which rules out $q$. So in the case $p>q$, $n_p=1$ and the unique p-Sylow subgroup is normal.
In the case $p<q$ however, I run into a problem.. Again we have $n_q|p^2$ (so $n_qin{1,p,p^2}$). Again, the condition $n_qequiv1mod q$ rules out $p$.
Now I am attempting to rule out the case $n_q=p^2$: Assume $n_q=p^2$. Then $p^2equiv1mod qimplies (p+1)(p-1)equiv0mod qimplies q$ must divide $(p+1)$ since $p<q$ and $q$ prime. But since $p<q$ and $q|(p+1)$ we see $q=p+1$. For primes this only happens when $p=2, q=3$.
Does this mean I need to check groups of order $2^2cdot3=12$ or did I miss something along the way that lets me conclude $n_qneq p^2$ and thus that $n_q=1$.
abstract-algebra group-theory finite-groups
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
$endgroup$
add a comment |
$begingroup$
I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.
In the case $p>q$ I have no problem.. By Sylow $n_p|q$, so $n_p$ is either $1$ or $q$. But we also have $n_pequiv1mod p$ which rules out $q$. So in the case $p>q$, $n_p=1$ and the unique p-Sylow subgroup is normal.
In the case $p<q$ however, I run into a problem.. Again we have $n_q|p^2$ (so $n_qin{1,p,p^2}$). Again, the condition $n_qequiv1mod q$ rules out $p$.
Now I am attempting to rule out the case $n_q=p^2$: Assume $n_q=p^2$. Then $p^2equiv1mod qimplies (p+1)(p-1)equiv0mod qimplies q$ must divide $(p+1)$ since $p<q$ and $q$ prime. But since $p<q$ and $q|(p+1)$ we see $q=p+1$. For primes this only happens when $p=2, q=3$.
Does this mean I need to check groups of order $2^2cdot3=12$ or did I miss something along the way that lets me conclude $n_qneq p^2$ and thus that $n_q=1$.
abstract-algebra group-theory finite-groups
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
$endgroup$
add a comment |
$begingroup$
I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.
In the case $p>q$ I have no problem.. By Sylow $n_p|q$, so $n_p$ is either $1$ or $q$. But we also have $n_pequiv1mod p$ which rules out $q$. So in the case $p>q$, $n_p=1$ and the unique p-Sylow subgroup is normal.
In the case $p<q$ however, I run into a problem.. Again we have $n_q|p^2$ (so $n_qin{1,p,p^2}$). Again, the condition $n_qequiv1mod q$ rules out $p$.
Now I am attempting to rule out the case $n_q=p^2$: Assume $n_q=p^2$. Then $p^2equiv1mod qimplies (p+1)(p-1)equiv0mod qimplies q$ must divide $(p+1)$ since $p<q$ and $q$ prime. But since $p<q$ and $q|(p+1)$ we see $q=p+1$. For primes this only happens when $p=2, q=3$.
Does this mean I need to check groups of order $2^2cdot3=12$ or did I miss something along the way that lets me conclude $n_qneq p^2$ and thus that $n_q=1$.
abstract-algebra group-theory finite-groups
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
$endgroup$
I am trying to show that any group of order $p^2q$ has a normal Sylow subgroup where $p$ and $q$ are distinct primes.
In the case $p>q$ I have no problem.. By Sylow $n_p|q$, so $n_p$ is either $1$ or $q$. But we also have $n_pequiv1mod p$ which rules out $q$. So in the case $p>q$, $n_p=1$ and the unique p-Sylow subgroup is normal.
In the case $p<q$ however, I run into a problem.. Again we have $n_q|p^2$ (so $n_qin{1,p,p^2}$). Again, the condition $n_qequiv1mod q$ rules out $p$.
Now I am attempting to rule out the case $n_q=p^2$: Assume $n_q=p^2$. Then $p^2equiv1mod qimplies (p+1)(p-1)equiv0mod qimplies q$ must divide $(p+1)$ since $p<q$ and $q$ prime. But since $p<q$ and $q|(p+1)$ we see $q=p+1$. For primes this only happens when $p=2, q=3$.
Does this mean I need to check groups of order $2^2cdot3=12$ or did I miss something along the way that lets me conclude $n_qneq p^2$ and thus that $n_q=1$.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
asked Aug 16 '11 at 23:35
RHPRHP
1,40811525
1,40811525
add a comment |
add a comment |
3 Answers
3
active
oldest
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$begingroup$
Here's another approach: suppose $n_q=p^2$. The $p^2$ $q$-Sylows each have $q-1$ nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has $p^2(q-1)$ elements of order $q$, and so only $p^2$ elements of any other order. There's thus only room for one $p$-Sylow, which is thus normal.
This possibiity does in fact occur: $A_4$ has order 12, has the Klein group as a normal $2$-Sylow, and all $8$ other elements are $3$-cycles.
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
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add a comment |
$begingroup$
Everything you've written seems to be correct. But you don't necessarily have to know the groups of order $12$.
To finish questions like this one off, it's often helpful to count elements. Here, if a group of order $12$ has more than one Sylow $3$-subgroup then there are four, and these intersect trivially. If you remove the elements of period $3$ from the group, then you aren't left with much!
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
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add a comment |
$begingroup$
Consider the case of $q > p$. Then,
$n_q = p^2$ or $1$ and
$n_p = q$ or $1$.
Let G have no non-trivial normal subgroup. Then,
$n_q = p^2$ and $n_p=q$. Then there are $p^2 (q-1) $ elements of order $q$ and $q(p-1) $ elements of order $p$. So, we have the following ,
begin{align}
p^2q &geq p^2(q-1) + q(p-1) + 1 \
& geq p^2q+pq-p^2-q+1 \
& geq p^2q + (p-1) (q-1)
end{align}
$$implies (p-1)(q-1) = 0 implies p=1 text{ or } q=1$$
But $p$ and $q$ are primes.
Hence $G$ has a non-trivial normal subgroup.
answered Jan 8 at 7:08
pavanpavan
314
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's another approach: suppose $n_q=p^2$. The $p^2$ $q$-Sylows each have $q-1$ nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has $p^2(q-1)$ elements of order $q$, and so only $p^2$ elements of any other order. There's thus only room for one $p$-Sylow, which is thus normal.
This possibiity does in fact occur: $A_4$ has order 12, has the Klein group as a normal $2$-Sylow, and all $8$ other elements are $3$-cycles.
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
$endgroup$
add a comment |
$begingroup$
Here's another approach: suppose $n_q=p^2$. The $p^2$ $q$-Sylows each have $q-1$ nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has $p^2(q-1)$ elements of order $q$, and so only $p^2$ elements of any other order. There's thus only room for one $p$-Sylow, which is thus normal.
This possibiity does in fact occur: $A_4$ has order 12, has the Klein group as a normal $2$-Sylow, and all $8$ other elements are $3$-cycles.
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
$endgroup$
add a comment |
$begingroup$
Here's another approach: suppose $n_q=p^2$. The $p^2$ $q$-Sylows each have $q-1$ nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has $p^2(q-1)$ elements of order $q$, and so only $p^2$ elements of any other order. There's thus only room for one $p$-Sylow, which is thus normal.
This possibiity does in fact occur: $A_4$ has order 12, has the Klein group as a normal $2$-Sylow, and all $8$ other elements are $3$-cycles.
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
$endgroup$
Here's another approach: suppose $n_q=p^2$. The $p^2$ $q$-Sylows each have $q-1$ nonidentity elements, and since a group of prime order is generated by any of its nonidentity elements, these sets are disjoint. So our group has $p^2(q-1)$ elements of order $q$, and so only $p^2$ elements of any other order. There's thus only room for one $p$-Sylow, which is thus normal.
This possibiity does in fact occur: $A_4$ has order 12, has the Klein group as a normal $2$-Sylow, and all $8$ other elements are $3$-cycles.
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
answered Aug 16 '11 at 23:56
Chris EagleChris Eagle
29k26997
29k26997
add a comment |
add a comment |
$begingroup$
Everything you've written seems to be correct. But you don't necessarily have to know the groups of order $12$.
To finish questions like this one off, it's often helpful to count elements. Here, if a group of order $12$ has more than one Sylow $3$-subgroup then there are four, and these intersect trivially. If you remove the elements of period $3$ from the group, then you aren't left with much!
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
$endgroup$
add a comment |
$begingroup$
Everything you've written seems to be correct. But you don't necessarily have to know the groups of order $12$.
To finish questions like this one off, it's often helpful to count elements. Here, if a group of order $12$ has more than one Sylow $3$-subgroup then there are four, and these intersect trivially. If you remove the elements of period $3$ from the group, then you aren't left with much!
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
$endgroup$
add a comment |
$begingroup$
Everything you've written seems to be correct. But you don't necessarily have to know the groups of order $12$.
To finish questions like this one off, it's often helpful to count elements. Here, if a group of order $12$ has more than one Sylow $3$-subgroup then there are four, and these intersect trivially. If you remove the elements of period $3$ from the group, then you aren't left with much!
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
$endgroup$
Everything you've written seems to be correct. But you don't necessarily have to know the groups of order $12$.
To finish questions like this one off, it's often helpful to count elements. Here, if a group of order $12$ has more than one Sylow $3$-subgroup then there are four, and these intersect trivially. If you remove the elements of period $3$ from the group, then you aren't left with much!
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
edited Aug 17 '11 at 0:06
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
answered Aug 16 '11 at 23:55
Dylan MorelandDylan Moreland
16.8k23563
16.8k23563
add a comment |
add a comment |
$begingroup$
Consider the case of $q > p$. Then,
$n_q = p^2$ or $1$ and
$n_p = q$ or $1$.
Let G have no non-trivial normal subgroup. Then,
$n_q = p^2$ and $n_p=q$. Then there are $p^2 (q-1) $ elements of order $q$ and $q(p-1) $ elements of order $p$. So, we have the following ,
begin{align}
p^2q &geq p^2(q-1) + q(p-1) + 1 \
& geq p^2q+pq-p^2-q+1 \
& geq p^2q + (p-1) (q-1)
end{align}
$$implies (p-1)(q-1) = 0 implies p=1 text{ or } q=1$$
But $p$ and $q$ are primes.
Hence $G$ has a non-trivial normal subgroup.
answered Jan 8 at 7:08
pavanpavan
314
$endgroup$
add a comment |
$begingroup$
Consider the case of $q > p$. Then,
$n_q = p^2$ or $1$ and
$n_p = q$ or $1$.
Let G have no non-trivial normal subgroup. Then,
$n_q = p^2$ and $n_p=q$. Then there are $p^2 (q-1) $ elements of order $q$ and $q(p-1) $ elements of order $p$. So, we have the following ,
begin{align}
p^2q &geq p^2(q-1) + q(p-1) + 1 \
& geq p^2q+pq-p^2-q+1 \
& geq p^2q + (p-1) (q-1)
end{align}
$$implies (p-1)(q-1) = 0 implies p=1 text{ or } q=1$$
But $p$ and $q$ are primes.
Hence $G$ has a non-trivial normal subgroup.
answered Jan 8 at 7:08
pavanpavan
314
$endgroup$
add a comment |
$begingroup$
Consider the case of $q > p$. Then,
$n_q = p^2$ or $1$ and
$n_p = q$ or $1$.
Let G have no non-trivial normal subgroup. Then,
$n_q = p^2$ and $n_p=q$. Then there are $p^2 (q-1) $ elements of order $q$ and $q(p-1) $ elements of order $p$. So, we have the following ,
begin{align}
p^2q &geq p^2(q-1) + q(p-1) + 1 \
& geq p^2q+pq-p^2-q+1 \
& geq p^2q + (p-1) (q-1)
end{align}
$$implies (p-1)(q-1) = 0 implies p=1 text{ or } q=1$$
But $p$ and $q$ are primes.
Hence $G$ has a non-trivial normal subgroup.
answered Jan 8 at 7:08
pavanpavan
314
$endgroup$
Consider the case of $q > p$. Then,
$n_q = p^2$ or $1$ and
$n_p = q$ or $1$.
Let G have no non-trivial normal subgroup. Then,
$n_q = p^2$ and $n_p=q$. Then there are $p^2 (q-1) $ elements of order $q$ and $q(p-1) $ elements of order $p$. So, we have the following ,
begin{align}
p^2q &geq p^2(q-1) + q(p-1) + 1 \
& geq p^2q+pq-p^2-q+1 \
& geq p^2q + (p-1) (q-1)
end{align}
$$implies (p-1)(q-1) = 0 implies p=1 text{ or } q=1$$
But $p$ and $q$ are primes.
Hence $G$ has a non-trivial normal subgroup.
answered Jan 8 at 7:08
pavanpavan
314
answered Jan 8 at 7:08
pavanpavan
314
answered Jan 8 at 7:08
pavanpavan
314
answered Jan 8 at 7:08
pavanpavan
314
314
add a comment |
add a comment |
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