How big must the union of a group's Sylow p-subgroups be?
$begingroup$
For various orders $n$ it's a common exercise to prove that a finite group $G$ of order $n$ can't be simple by using the Sylow theorems to show that there is some prime $p mid n$ such that the number $n_p$ of Sylow $p$-subgroups equals $1$, so the unique Sylow $p$-subgroup is normal. One way these proofs can go is that you show that if $n_p$ isn't equal to $1$, then because $n_p equiv 1 bmod p$ it must be very large, so large that there isn't enough room in $G$ for all of its Sylow $p$-subgroups together plus the other Sylow subgroups.
I know how to run this argument if the exponent $a$ of $p$ in $n$ is $1$ and we can show that $n_p = frac{n}{p}$; in this case the Sylow $p$-subgroups are cyclic, so intersect only in the identity, which means that $G$ has at least $frac{n}{p}(p - 1)$ elements of order $p$, and hence only room for $frac{n}{p}$ elements of other orders.
However, I don't know how to run this argument if $a ge 2$; this came up when I was trying to answer this question and specifically trying to show that a group of order $|G| = 3 cdot 5 cdot 7^2 = 735$ can't be simple. The Sylow theorems give that if $n_7 neq 1$ then $n_7 = 15$, so $G$ has the maximum possible number of Sylow $7$-subgroups. The specific question this gave rise to is:
Specific question: What is the sharpest lower bound on the size of the union of these $15$ Sylow $7$-subgroups?
I wanted to use the Bonferroni inequalities to address this question, using the fact that any two Sylow $7$-subgroups intersect in at most $7$ elements, but something very funny happened: if I apply Bonferroni to all $15$ subgroups I get a lower bound of
$$15 cdot 49 - {15 choose 2} cdot 7 = 0.$$
The problem is that there are too many pairwise intersections between $15$ subgroups. If I instead apply Bonferroni with only $k$ of the $15$ subgroups I get a lower bound of
$$49k - 7 {k choose 2}$$
which turns out to be maximized when $k = 8$, giving a lower bound of $210$. Is it possible to do better than this? I'm ignoring $7$ of the Sylows!
So the general question is:
General question: What is the sharpest lower bound on the size of the union of the Sylow $p$-subgroups of a finite group $G$ which can be written as a function of the size $p^a$ of such a subgroup and the number $n_p$ of such subgroups? What if $G$ is assumed to be simple?
When $a = 1$ the union has size exactly $(p - 1) n_p + 1$. In general any two Sylows intersect in at most $p^{a-1}$ elements, so Bonferroni with $k$ of the Sylows gives a lower bound of
$$k p^a - {k choose 2} p^{a-1} = k p^{a-1} left( p - frac{k-1}{2} right)$$
which is maximized when $k approx p$ as above (or $k = n_p$, if $p$ is more than a little larger than $n_p$). But the smaller $p$ is compared to $n_p$ the less helpful of a bound this will be.
group-theory inequality finite-groups sylow-theory
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
$endgroup$
|
show 5 more comments
$begingroup$
For various orders $n$ it's a common exercise to prove that a finite group $G$ of order $n$ can't be simple by using the Sylow theorems to show that there is some prime $p mid n$ such that the number $n_p$ of Sylow $p$-subgroups equals $1$, so the unique Sylow $p$-subgroup is normal. One way these proofs can go is that you show that if $n_p$ isn't equal to $1$, then because $n_p equiv 1 bmod p$ it must be very large, so large that there isn't enough room in $G$ for all of its Sylow $p$-subgroups together plus the other Sylow subgroups.
I know how to run this argument if the exponent $a$ of $p$ in $n$ is $1$ and we can show that $n_p = frac{n}{p}$; in this case the Sylow $p$-subgroups are cyclic, so intersect only in the identity, which means that $G$ has at least $frac{n}{p}(p - 1)$ elements of order $p$, and hence only room for $frac{n}{p}$ elements of other orders.
However, I don't know how to run this argument if $a ge 2$; this came up when I was trying to answer this question and specifically trying to show that a group of order $|G| = 3 cdot 5 cdot 7^2 = 735$ can't be simple. The Sylow theorems give that if $n_7 neq 1$ then $n_7 = 15$, so $G$ has the maximum possible number of Sylow $7$-subgroups. The specific question this gave rise to is:
Specific question: What is the sharpest lower bound on the size of the union of these $15$ Sylow $7$-subgroups?
I wanted to use the Bonferroni inequalities to address this question, using the fact that any two Sylow $7$-subgroups intersect in at most $7$ elements, but something very funny happened: if I apply Bonferroni to all $15$ subgroups I get a lower bound of
$$15 cdot 49 - {15 choose 2} cdot 7 = 0.$$
The problem is that there are too many pairwise intersections between $15$ subgroups. If I instead apply Bonferroni with only $k$ of the $15$ subgroups I get a lower bound of
$$49k - 7 {k choose 2}$$
which turns out to be maximized when $k = 8$, giving a lower bound of $210$. Is it possible to do better than this? I'm ignoring $7$ of the Sylows!
So the general question is:
General question: What is the sharpest lower bound on the size of the union of the Sylow $p$-subgroups of a finite group $G$ which can be written as a function of the size $p^a$ of such a subgroup and the number $n_p$ of such subgroups? What if $G$ is assumed to be simple?
When $a = 1$ the union has size exactly $(p - 1) n_p + 1$. In general any two Sylows intersect in at most $p^{a-1}$ elements, so Bonferroni with $k$ of the Sylows gives a lower bound of
$$k p^a - {k choose 2} p^{a-1} = k p^{a-1} left( p - frac{k-1}{2} right)$$
which is maximized when $k approx p$ as above (or $k = n_p$, if $p$ is more than a little larger than $n_p$). But the smaller $p$ is compared to $n_p$ the less helpful of a bound this will be.
group-theory inequality finite-groups sylow-theory
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
$endgroup$
1
$begingroup$
I have no idea!
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 6:59
1
$begingroup$
The normalizer of the intersection has at least two $7$-Sylows (of order 49), so by Sylow it has to have 15, showing that the intersection is normal in the full group.
$endgroup$
– j.p.
Nov 14 '18 at 7:29
1
$begingroup$
@j.p. why 105 must be cyclic? There is a nonabelian one.
$endgroup$
– user10354138
Nov 14 '18 at 7:31
1
$begingroup$
@user10354138: Upps, $7bmod 3 = 1$. You're right! One only gets that a group of order 105 has a normal subgroup of order $7$, which shows that a group of order 735 has a normal subgroup of order 49.
$endgroup$
– j.p.
Nov 14 '18 at 7:50
4
$begingroup$
Burnside's Transfer Theorem implies that a group of order $735$ with $15$ Sylow $7$-subgroups would have a normal subgroup of order $15$. But the only group of order $15$ is cyclic, and that has no automorphism of order $7$, so this is impossible.
$endgroup$
– Derek Holt
Nov 14 '18 at 8:03
|
show 5 more comments
$begingroup$
For various orders $n$ it's a common exercise to prove that a finite group $G$ of order $n$ can't be simple by using the Sylow theorems to show that there is some prime $p mid n$ such that the number $n_p$ of Sylow $p$-subgroups equals $1$, so the unique Sylow $p$-subgroup is normal. One way these proofs can go is that you show that if $n_p$ isn't equal to $1$, then because $n_p equiv 1 bmod p$ it must be very large, so large that there isn't enough room in $G$ for all of its Sylow $p$-subgroups together plus the other Sylow subgroups.
I know how to run this argument if the exponent $a$ of $p$ in $n$ is $1$ and we can show that $n_p = frac{n}{p}$; in this case the Sylow $p$-subgroups are cyclic, so intersect only in the identity, which means that $G$ has at least $frac{n}{p}(p - 1)$ elements of order $p$, and hence only room for $frac{n}{p}$ elements of other orders.
However, I don't know how to run this argument if $a ge 2$; this came up when I was trying to answer this question and specifically trying to show that a group of order $|G| = 3 cdot 5 cdot 7^2 = 735$ can't be simple. The Sylow theorems give that if $n_7 neq 1$ then $n_7 = 15$, so $G$ has the maximum possible number of Sylow $7$-subgroups. The specific question this gave rise to is:
Specific question: What is the sharpest lower bound on the size of the union of these $15$ Sylow $7$-subgroups?
I wanted to use the Bonferroni inequalities to address this question, using the fact that any two Sylow $7$-subgroups intersect in at most $7$ elements, but something very funny happened: if I apply Bonferroni to all $15$ subgroups I get a lower bound of
$$15 cdot 49 - {15 choose 2} cdot 7 = 0.$$
The problem is that there are too many pairwise intersections between $15$ subgroups. If I instead apply Bonferroni with only $k$ of the $15$ subgroups I get a lower bound of
$$49k - 7 {k choose 2}$$
which turns out to be maximized when $k = 8$, giving a lower bound of $210$. Is it possible to do better than this? I'm ignoring $7$ of the Sylows!
So the general question is:
General question: What is the sharpest lower bound on the size of the union of the Sylow $p$-subgroups of a finite group $G$ which can be written as a function of the size $p^a$ of such a subgroup and the number $n_p$ of such subgroups? What if $G$ is assumed to be simple?
When $a = 1$ the union has size exactly $(p - 1) n_p + 1$. In general any two Sylows intersect in at most $p^{a-1}$ elements, so Bonferroni with $k$ of the Sylows gives a lower bound of
$$k p^a - {k choose 2} p^{a-1} = k p^{a-1} left( p - frac{k-1}{2} right)$$
which is maximized when $k approx p$ as above (or $k = n_p$, if $p$ is more than a little larger than $n_p$). But the smaller $p$ is compared to $n_p$ the less helpful of a bound this will be.
group-theory inequality finite-groups sylow-theory
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
$endgroup$
For various orders $n$ it's a common exercise to prove that a finite group $G$ of order $n$ can't be simple by using the Sylow theorems to show that there is some prime $p mid n$ such that the number $n_p$ of Sylow $p$-subgroups equals $1$, so the unique Sylow $p$-subgroup is normal. One way these proofs can go is that you show that if $n_p$ isn't equal to $1$, then because $n_p equiv 1 bmod p$ it must be very large, so large that there isn't enough room in $G$ for all of its Sylow $p$-subgroups together plus the other Sylow subgroups.
I know how to run this argument if the exponent $a$ of $p$ in $n$ is $1$ and we can show that $n_p = frac{n}{p}$; in this case the Sylow $p$-subgroups are cyclic, so intersect only in the identity, which means that $G$ has at least $frac{n}{p}(p - 1)$ elements of order $p$, and hence only room for $frac{n}{p}$ elements of other orders.
However, I don't know how to run this argument if $a ge 2$; this came up when I was trying to answer this question and specifically trying to show that a group of order $|G| = 3 cdot 5 cdot 7^2 = 735$ can't be simple. The Sylow theorems give that if $n_7 neq 1$ then $n_7 = 15$, so $G$ has the maximum possible number of Sylow $7$-subgroups. The specific question this gave rise to is:
Specific question: What is the sharpest lower bound on the size of the union of these $15$ Sylow $7$-subgroups?
I wanted to use the Bonferroni inequalities to address this question, using the fact that any two Sylow $7$-subgroups intersect in at most $7$ elements, but something very funny happened: if I apply Bonferroni to all $15$ subgroups I get a lower bound of
$$15 cdot 49 - {15 choose 2} cdot 7 = 0.$$
The problem is that there are too many pairwise intersections between $15$ subgroups. If I instead apply Bonferroni with only $k$ of the $15$ subgroups I get a lower bound of
$$49k - 7 {k choose 2}$$
which turns out to be maximized when $k = 8$, giving a lower bound of $210$. Is it possible to do better than this? I'm ignoring $7$ of the Sylows!
So the general question is:
General question: What is the sharpest lower bound on the size of the union of the Sylow $p$-subgroups of a finite group $G$ which can be written as a function of the size $p^a$ of such a subgroup and the number $n_p$ of such subgroups? What if $G$ is assumed to be simple?
When $a = 1$ the union has size exactly $(p - 1) n_p + 1$. In general any two Sylows intersect in at most $p^{a-1}$ elements, so Bonferroni with $k$ of the Sylows gives a lower bound of
$$k p^a - {k choose 2} p^{a-1} = k p^{a-1} left( p - frac{k-1}{2} right)$$
which is maximized when $k approx p$ as above (or $k = n_p$, if $p$ is more than a little larger than $n_p$). But the smaller $p$ is compared to $n_p$ the less helpful of a bound this will be.
group-theory inequality finite-groups sylow-theory
group-theory inequality finite-groups sylow-theory
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
asked Nov 14 '18 at 5:21
Qiaochu YuanQiaochu Yuan
278k32585923
278k32585923
1
$begingroup$
I have no idea!
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 6:59
1
$begingroup$
The normalizer of the intersection has at least two $7$-Sylows (of order 49), so by Sylow it has to have 15, showing that the intersection is normal in the full group.
$endgroup$
– j.p.
Nov 14 '18 at 7:29
1
$begingroup$
@j.p. why 105 must be cyclic? There is a nonabelian one.
$endgroup$
– user10354138
Nov 14 '18 at 7:31
1
$begingroup$
@user10354138: Upps, $7bmod 3 = 1$. You're right! One only gets that a group of order 105 has a normal subgroup of order $7$, which shows that a group of order 735 has a normal subgroup of order 49.
$endgroup$
– j.p.
Nov 14 '18 at 7:50
4
$begingroup$
Burnside's Transfer Theorem implies that a group of order $735$ with $15$ Sylow $7$-subgroups would have a normal subgroup of order $15$. But the only group of order $15$ is cyclic, and that has no automorphism of order $7$, so this is impossible.
$endgroup$
– Derek Holt
Nov 14 '18 at 8:03
|
show 5 more comments
1
$begingroup$
I have no idea!
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 6:59
1
$begingroup$
The normalizer of the intersection has at least two $7$-Sylows (of order 49), so by Sylow it has to have 15, showing that the intersection is normal in the full group.
$endgroup$
– j.p.
Nov 14 '18 at 7:29
1
$begingroup$
@j.p. why 105 must be cyclic? There is a nonabelian one.
$endgroup$
– user10354138
Nov 14 '18 at 7:31
1
$begingroup$
@user10354138: Upps, $7bmod 3 = 1$. You're right! One only gets that a group of order 105 has a normal subgroup of order $7$, which shows that a group of order 735 has a normal subgroup of order 49.
$endgroup$
– j.p.
Nov 14 '18 at 7:50
4
$begingroup$
Burnside's Transfer Theorem implies that a group of order $735$ with $15$ Sylow $7$-subgroups would have a normal subgroup of order $15$. But the only group of order $15$ is cyclic, and that has no automorphism of order $7$, so this is impossible.
$endgroup$
– Derek Holt
Nov 14 '18 at 8:03
1
1
$begingroup$
I have no idea!
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 6:59
$begingroup$
I have no idea!
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 6:59
1
1
$begingroup$
The normalizer of the intersection has at least two $7$-Sylows (of order 49), so by Sylow it has to have 15, showing that the intersection is normal in the full group.
$endgroup$
– j.p.
Nov 14 '18 at 7:29
$begingroup$
The normalizer of the intersection has at least two $7$-Sylows (of order 49), so by Sylow it has to have 15, showing that the intersection is normal in the full group.
$endgroup$
– j.p.
Nov 14 '18 at 7:29
1
1
$begingroup$
@j.p. why 105 must be cyclic? There is a nonabelian one.
$endgroup$
– user10354138
Nov 14 '18 at 7:31
$begingroup$
@j.p. why 105 must be cyclic? There is a nonabelian one.
$endgroup$
– user10354138
Nov 14 '18 at 7:31
1
1
$begingroup$
@user10354138: Upps, $7bmod 3 = 1$. You're right! One only gets that a group of order 105 has a normal subgroup of order $7$, which shows that a group of order 735 has a normal subgroup of order 49.
$endgroup$
– j.p.
Nov 14 '18 at 7:50
$begingroup$
@user10354138: Upps, $7bmod 3 = 1$. You're right! One only gets that a group of order 105 has a normal subgroup of order $7$, which shows that a group of order 735 has a normal subgroup of order 49.
$endgroup$
– j.p.
Nov 14 '18 at 7:50
4
4
$begingroup$
Burnside's Transfer Theorem implies that a group of order $735$ with $15$ Sylow $7$-subgroups would have a normal subgroup of order $15$. But the only group of order $15$ is cyclic, and that has no automorphism of order $7$, so this is impossible.
$endgroup$
– Derek Holt
Nov 14 '18 at 8:03
$begingroup$
Burnside's Transfer Theorem implies that a group of order $735$ with $15$ Sylow $7$-subgroups would have a normal subgroup of order $15$. But the only group of order $15$ is cyclic, and that has no automorphism of order $7$, so this is impossible.
$endgroup$
– Derek Holt
Nov 14 '18 at 8:03
|
show 5 more comments
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1
$begingroup$
I have no idea!
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 6:59
1
$begingroup$
The normalizer of the intersection has at least two $7$-Sylows (of order 49), so by Sylow it has to have 15, showing that the intersection is normal in the full group.
$endgroup$
– j.p.
Nov 14 '18 at 7:29
1
$begingroup$
@j.p. why 105 must be cyclic? There is a nonabelian one.
$endgroup$
– user10354138
Nov 14 '18 at 7:31
1
$begingroup$
@user10354138: Upps, $7bmod 3 = 1$. You're right! One only gets that a group of order 105 has a normal subgroup of order $7$, which shows that a group of order 735 has a normal subgroup of order 49.
$endgroup$
– j.p.
Nov 14 '18 at 7:50
4
$begingroup$
Burnside's Transfer Theorem implies that a group of order $735$ with $15$ Sylow $7$-subgroups would have a normal subgroup of order $15$. But the only group of order $15$ is cyclic, and that has no automorphism of order $7$, so this is impossible.
$endgroup$
– Derek Holt
Nov 14 '18 at 8:03