Mixing
Number of $p$-subgroups of finite group
Assume we have a finite group $G$, $|G|=p^cm$ where $pnotmid m$ is prime. Fix a subgroup $H$ of order $p^a$ and a number $ale ble c$. Prove that the number of subgroups of order $p^b$ which contain the subgroup $H$ is $equiv1pmod p$.
I tried to use Wielandt's proof of Sylow's theorem as in Wielandt's proof of Sylow's theorem., but if we set $S={Asubset G, |A|=p^b,Hsubset A}$, we can't have $G$ act on $S$ by right multiplication. How can I prove this statement?
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 22 '18 at 11:11
the_fox
2,47211431
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
add a comment |
Assume we have a finite group $G$, $|G|=p^cm$ where $pnotmid m$ is prime. Fix a subgroup $H$ of order $p^a$ and a number $ale ble c$. Prove that the number of subgroups of order $p^b$ which contain the subgroup $H$ is $equiv1pmod p$.
I tried to use Wielandt's proof of Sylow's theorem as in Wielandt's proof of Sylow's theorem., but if we set $S={Asubset G, |A|=p^b,Hsubset A}$, we can't have $G$ act on $S$ by right multiplication. How can I prove this statement?
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 22 '18 at 11:11
the_fox
2,47211431
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$.
– Derek Holt
Nov 8 '18 at 8:06
1
See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf.
– KCd
Nov 22 '18 at 11:28
add a comment |
Assume we have a finite group $G$, $|G|=p^cm$ where $pnotmid m$ is prime. Fix a subgroup $H$ of order $p^a$ and a number $ale ble c$. Prove that the number of subgroups of order $p^b$ which contain the subgroup $H$ is $equiv1pmod p$.
I tried to use Wielandt's proof of Sylow's theorem as in Wielandt's proof of Sylow's theorem., but if we set $S={Asubset G, |A|=p^b,Hsubset A}$, we can't have $G$ act on $S$ by right multiplication. How can I prove this statement?
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 22 '18 at 11:11
the_fox
2,47211431
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
Assume we have a finite group $G$, $|G|=p^cm$ where $pnotmid m$ is prime. Fix a subgroup $H$ of order $p^a$ and a number $ale ble c$. Prove that the number of subgroups of order $p^b$ which contain the subgroup $H$ is $equiv1pmod p$.
I tried to use Wielandt's proof of Sylow's theorem as in Wielandt's proof of Sylow's theorem., but if we set $S={Asubset G, |A|=p^b,Hsubset A}$, we can't have $G$ act on $S$ by right multiplication. How can I prove this statement?
abstract-algebra group-theory finite-groups sylow-theory
abstract-algebra group-theory finite-groups sylow-theory
edited Nov 22 '18 at 11:11
the_fox
2,47211431
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
edited Nov 22 '18 at 11:11
the_fox
2,47211431
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
edited Nov 22 '18 at 11:11
the_fox
2,47211431
edited Nov 22 '18 at 11:11
the_fox
2,47211431
edited Nov 22 '18 at 11:11
the_fox
2,47211431
2,47211431
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
asked Nov 7 '18 at 16:41
hctbhctb
1,010410
1,010410
I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$.
– Derek Holt
Nov 8 '18 at 8:06
1
See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf.
– KCd
Nov 22 '18 at 11:28
add a comment |
I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$.
– Derek Holt
Nov 8 '18 at 8:06
1
See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf.
– KCd
Nov 22 '18 at 11:28
I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$.
– Derek Holt
Nov 8 '18 at 8:06
I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$.
– Derek Holt
Nov 8 '18 at 8:06
1
1
See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf.
– KCd
Nov 22 '18 at 11:28
See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf.
– KCd
Nov 22 '18 at 11:28
add a comment |
1 Answer
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I suspect that this is something well-known, but I am not aware
of any specific references. Anyway, here is a proof.
Theorem. Let $H$ be a $p$-subgroup of a group $G$, and suppose $p^t geq |H|$. Then the
number of subgroups $X leq G$ such that $H leq X$ and $|X|=p^t$ is congruent to $1 bmod p$.
Lemma 1. The theorem is true in the case where $p^t$ is the order of a Sylow $p$-subgroup of $G$.
Proof. Let $H leq P in operatorname{Syl}_p(G)$ and induct on $|P:H|$.
The result is trivial if $H=P$, so we can assume that $H < P$ and we let $N = N_P(H)$.
Conjugation by $N$ permutes the Sylow $p$-subgroups that contain $H$.
If $S$ is one of these and $N nleq S$, then $N$ does not normalise $S$,
and so the $N$-orbit of $S$ has size divisible by $p$.
Also, all members of this orbit fail to contain $N$,
so we need only count the Sylow $p$-subgroups of $G$ that contain $N$,
and since $N > H$, this number is congruent to $1 bmod p$ by the inductive hypothesis.
Lemma 2. The theorem is true in the case where $G$ is a $p$-group.
Proof. Induct on $|G:H|$. The result is trivial if $H=G$ so we can assume that $H<G$.
We can also assume that $p^t < |G|$.
Let $mathcal{X}$ and $mathcal{M}$, respectively, be the set of subgroups of order $p^t$ that contain $H$ and
the set of maximal subgroups that contain $H$.
We count in two ways the number $N$ of ordered pairs $(X,M)$ with
$X in mathcal{X}$, $M in mathcal{M}$ and $X leq M$.
First, $N$ is the sum over $X in mathcal{X}$ of the number of maximal subgroups that contain $X$.
Also, $N$ is the sum over $M in mathcal{M}$ of the number of members of $mathcal{X}$ that are contained in $M$.
It is easy to see that the number of maximal subgroups of $G$ that contain any given proper subgroup of $G$ is congruent to $1 bmod p$.
The first formula for $N$ gives $N equiv |mathcal{X}| bmod p$ because the number of $M in mathcal{M}$
containing any given member of $mathcal{X}$ is congruent to $1 bmod p$.
By the inductive hypothesis, the number of members of $mathcal{X}$ contained in any given member of $mathcal{M}$ is congruent to $1$, so the second formula for $N$ gives $$N equiv |mathcal{M}| equiv 1 bmod p.$$
Thus $$|mathcal{X}| equiv N equiv 1 bmod p,$$ as wanted.
Proof of Theorem. Let $mathcal{X}$ be the set of subgroups of order $p^t$ that contain $H$ and let $mathcal{P}$ be the set of $P in operatorname{Syl}_p(G)$ that contain $H$. We count the number $S$ of pairs $(X,P)$, where $X in mathcal{X}$, $P in mathcal{P}$ and $X leq P$.
First, $S$ is the sum over $X in mathcal{X}$ of the number of $P in mathcal{P}$ such that $X leq P$. Also, $S$ is the sum over $P in mathcal{P}$ of the number of $X in mathcal{X}$ such that $X leq P$. Since the number of $P$ containing a given $X$ is congruent to $1 bmod p$ by Lemma $mathbf{1}$, the first formula for $S$ yields $S equiv |mathcal{X}| bmod p$. The number of $X$ contained in a given $P$ is congruent to $1 bmod p$ by Lemma $mathbf{2}$, so the second formula for $S$ yields $S equiv |mathcal{P}| bmod p$.
Also, $$|mathcal{P}| equiv 1 bmod p$$ by Lemma $mathbf{1}$, so we get
$$|mathcal{X}| equiv S equiv 1 bmod p,$$
which concludes our proof.
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
add a comment |
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I suspect that this is something well-known, but I am not aware
of any specific references. Anyway, here is a proof.
Theorem. Let $H$ be a $p$-subgroup of a group $G$, and suppose $p^t geq |H|$. Then the
number of subgroups $X leq G$ such that $H leq X$ and $|X|=p^t$ is congruent to $1 bmod p$.
Lemma 1. The theorem is true in the case where $p^t$ is the order of a Sylow $p$-subgroup of $G$.
Proof. Let $H leq P in operatorname{Syl}_p(G)$ and induct on $|P:H|$.
The result is trivial if $H=P$, so we can assume that $H < P$ and we let $N = N_P(H)$.
Conjugation by $N$ permutes the Sylow $p$-subgroups that contain $H$.
If $S$ is one of these and $N nleq S$, then $N$ does not normalise $S$,
and so the $N$-orbit of $S$ has size divisible by $p$.
Also, all members of this orbit fail to contain $N$,
so we need only count the Sylow $p$-subgroups of $G$ that contain $N$,
and since $N > H$, this number is congruent to $1 bmod p$ by the inductive hypothesis.
Lemma 2. The theorem is true in the case where $G$ is a $p$-group.
Proof. Induct on $|G:H|$. The result is trivial if $H=G$ so we can assume that $H<G$.
We can also assume that $p^t < |G|$.
Let $mathcal{X}$ and $mathcal{M}$, respectively, be the set of subgroups of order $p^t$ that contain $H$ and
the set of maximal subgroups that contain $H$.
We count in two ways the number $N$ of ordered pairs $(X,M)$ with
$X in mathcal{X}$, $M in mathcal{M}$ and $X leq M$.
First, $N$ is the sum over $X in mathcal{X}$ of the number of maximal subgroups that contain $X$.
Also, $N$ is the sum over $M in mathcal{M}$ of the number of members of $mathcal{X}$ that are contained in $M$.
It is easy to see that the number of maximal subgroups of $G$ that contain any given proper subgroup of $G$ is congruent to $1 bmod p$.
The first formula for $N$ gives $N equiv |mathcal{X}| bmod p$ because the number of $M in mathcal{M}$
containing any given member of $mathcal{X}$ is congruent to $1 bmod p$.
By the inductive hypothesis, the number of members of $mathcal{X}$ contained in any given member of $mathcal{M}$ is congruent to $1$, so the second formula for $N$ gives $$N equiv |mathcal{M}| equiv 1 bmod p.$$
Thus $$|mathcal{X}| equiv N equiv 1 bmod p,$$ as wanted.
Proof of Theorem. Let $mathcal{X}$ be the set of subgroups of order $p^t$ that contain $H$ and let $mathcal{P}$ be the set of $P in operatorname{Syl}_p(G)$ that contain $H$. We count the number $S$ of pairs $(X,P)$, where $X in mathcal{X}$, $P in mathcal{P}$ and $X leq P$.
First, $S$ is the sum over $X in mathcal{X}$ of the number of $P in mathcal{P}$ such that $X leq P$. Also, $S$ is the sum over $P in mathcal{P}$ of the number of $X in mathcal{X}$ such that $X leq P$. Since the number of $P$ containing a given $X$ is congruent to $1 bmod p$ by Lemma $mathbf{1}$, the first formula for $S$ yields $S equiv |mathcal{X}| bmod p$. The number of $X$ contained in a given $P$ is congruent to $1 bmod p$ by Lemma $mathbf{2}$, so the second formula for $S$ yields $S equiv |mathcal{P}| bmod p$.
Also, $$|mathcal{P}| equiv 1 bmod p$$ by Lemma $mathbf{1}$, so we get
$$|mathcal{X}| equiv S equiv 1 bmod p,$$
which concludes our proof.
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
add a comment |
I suspect that this is something well-known, but I am not aware
of any specific references. Anyway, here is a proof.
Theorem. Let $H$ be a $p$-subgroup of a group $G$, and suppose $p^t geq |H|$. Then the
number of subgroups $X leq G$ such that $H leq X$ and $|X|=p^t$ is congruent to $1 bmod p$.
Lemma 1. The theorem is true in the case where $p^t$ is the order of a Sylow $p$-subgroup of $G$.
Proof. Let $H leq P in operatorname{Syl}_p(G)$ and induct on $|P:H|$.
The result is trivial if $H=P$, so we can assume that $H < P$ and we let $N = N_P(H)$.
Conjugation by $N$ permutes the Sylow $p$-subgroups that contain $H$.
If $S$ is one of these and $N nleq S$, then $N$ does not normalise $S$,
and so the $N$-orbit of $S$ has size divisible by $p$.
Also, all members of this orbit fail to contain $N$,
so we need only count the Sylow $p$-subgroups of $G$ that contain $N$,
and since $N > H$, this number is congruent to $1 bmod p$ by the inductive hypothesis.
Lemma 2. The theorem is true in the case where $G$ is a $p$-group.
Proof. Induct on $|G:H|$. The result is trivial if $H=G$ so we can assume that $H<G$.
We can also assume that $p^t < |G|$.
Let $mathcal{X}$ and $mathcal{M}$, respectively, be the set of subgroups of order $p^t$ that contain $H$ and
the set of maximal subgroups that contain $H$.
We count in two ways the number $N$ of ordered pairs $(X,M)$ with
$X in mathcal{X}$, $M in mathcal{M}$ and $X leq M$.
First, $N$ is the sum over $X in mathcal{X}$ of the number of maximal subgroups that contain $X$.
Also, $N$ is the sum over $M in mathcal{M}$ of the number of members of $mathcal{X}$ that are contained in $M$.
It is easy to see that the number of maximal subgroups of $G$ that contain any given proper subgroup of $G$ is congruent to $1 bmod p$.
The first formula for $N$ gives $N equiv |mathcal{X}| bmod p$ because the number of $M in mathcal{M}$
containing any given member of $mathcal{X}$ is congruent to $1 bmod p$.
By the inductive hypothesis, the number of members of $mathcal{X}$ contained in any given member of $mathcal{M}$ is congruent to $1$, so the second formula for $N$ gives $$N equiv |mathcal{M}| equiv 1 bmod p.$$
Thus $$|mathcal{X}| equiv N equiv 1 bmod p,$$ as wanted.
Proof of Theorem. Let $mathcal{X}$ be the set of subgroups of order $p^t$ that contain $H$ and let $mathcal{P}$ be the set of $P in operatorname{Syl}_p(G)$ that contain $H$. We count the number $S$ of pairs $(X,P)$, where $X in mathcal{X}$, $P in mathcal{P}$ and $X leq P$.
First, $S$ is the sum over $X in mathcal{X}$ of the number of $P in mathcal{P}$ such that $X leq P$. Also, $S$ is the sum over $P in mathcal{P}$ of the number of $X in mathcal{X}$ such that $X leq P$. Since the number of $P$ containing a given $X$ is congruent to $1 bmod p$ by Lemma $mathbf{1}$, the first formula for $S$ yields $S equiv |mathcal{X}| bmod p$. The number of $X$ contained in a given $P$ is congruent to $1 bmod p$ by Lemma $mathbf{2}$, so the second formula for $S$ yields $S equiv |mathcal{P}| bmod p$.
Also, $$|mathcal{P}| equiv 1 bmod p$$ by Lemma $mathbf{1}$, so we get
$$|mathcal{X}| equiv S equiv 1 bmod p,$$
which concludes our proof.
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
add a comment |
I suspect that this is something well-known, but I am not aware
of any specific references. Anyway, here is a proof.
Theorem. Let $H$ be a $p$-subgroup of a group $G$, and suppose $p^t geq |H|$. Then the
number of subgroups $X leq G$ such that $H leq X$ and $|X|=p^t$ is congruent to $1 bmod p$.
Lemma 1. The theorem is true in the case where $p^t$ is the order of a Sylow $p$-subgroup of $G$.
Proof. Let $H leq P in operatorname{Syl}_p(G)$ and induct on $|P:H|$.
The result is trivial if $H=P$, so we can assume that $H < P$ and we let $N = N_P(H)$.
Conjugation by $N$ permutes the Sylow $p$-subgroups that contain $H$.
If $S$ is one of these and $N nleq S$, then $N$ does not normalise $S$,
and so the $N$-orbit of $S$ has size divisible by $p$.
Also, all members of this orbit fail to contain $N$,
so we need only count the Sylow $p$-subgroups of $G$ that contain $N$,
and since $N > H$, this number is congruent to $1 bmod p$ by the inductive hypothesis.
Lemma 2. The theorem is true in the case where $G$ is a $p$-group.
Proof. Induct on $|G:H|$. The result is trivial if $H=G$ so we can assume that $H<G$.
We can also assume that $p^t < |G|$.
Let $mathcal{X}$ and $mathcal{M}$, respectively, be the set of subgroups of order $p^t$ that contain $H$ and
the set of maximal subgroups that contain $H$.
We count in two ways the number $N$ of ordered pairs $(X,M)$ with
$X in mathcal{X}$, $M in mathcal{M}$ and $X leq M$.
First, $N$ is the sum over $X in mathcal{X}$ of the number of maximal subgroups that contain $X$.
Also, $N$ is the sum over $M in mathcal{M}$ of the number of members of $mathcal{X}$ that are contained in $M$.
It is easy to see that the number of maximal subgroups of $G$ that contain any given proper subgroup of $G$ is congruent to $1 bmod p$.
The first formula for $N$ gives $N equiv |mathcal{X}| bmod p$ because the number of $M in mathcal{M}$
containing any given member of $mathcal{X}$ is congruent to $1 bmod p$.
By the inductive hypothesis, the number of members of $mathcal{X}$ contained in any given member of $mathcal{M}$ is congruent to $1$, so the second formula for $N$ gives $$N equiv |mathcal{M}| equiv 1 bmod p.$$
Thus $$|mathcal{X}| equiv N equiv 1 bmod p,$$ as wanted.
Proof of Theorem. Let $mathcal{X}$ be the set of subgroups of order $p^t$ that contain $H$ and let $mathcal{P}$ be the set of $P in operatorname{Syl}_p(G)$ that contain $H$. We count the number $S$ of pairs $(X,P)$, where $X in mathcal{X}$, $P in mathcal{P}$ and $X leq P$.
First, $S$ is the sum over $X in mathcal{X}$ of the number of $P in mathcal{P}$ such that $X leq P$. Also, $S$ is the sum over $P in mathcal{P}$ of the number of $X in mathcal{X}$ such that $X leq P$. Since the number of $P$ containing a given $X$ is congruent to $1 bmod p$ by Lemma $mathbf{1}$, the first formula for $S$ yields $S equiv |mathcal{X}| bmod p$. The number of $X$ contained in a given $P$ is congruent to $1 bmod p$ by Lemma $mathbf{2}$, so the second formula for $S$ yields $S equiv |mathcal{P}| bmod p$.
Also, $$|mathcal{P}| equiv 1 bmod p$$ by Lemma $mathbf{1}$, so we get
$$|mathcal{X}| equiv S equiv 1 bmod p,$$
which concludes our proof.
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
I suspect that this is something well-known, but I am not aware
of any specific references. Anyway, here is a proof.
Theorem. Let $H$ be a $p$-subgroup of a group $G$, and suppose $p^t geq |H|$. Then the
number of subgroups $X leq G$ such that $H leq X$ and $|X|=p^t$ is congruent to $1 bmod p$.
Lemma 1. The theorem is true in the case where $p^t$ is the order of a Sylow $p$-subgroup of $G$.
Proof. Let $H leq P in operatorname{Syl}_p(G)$ and induct on $|P:H|$.
The result is trivial if $H=P$, so we can assume that $H < P$ and we let $N = N_P(H)$.
Conjugation by $N$ permutes the Sylow $p$-subgroups that contain $H$.
If $S$ is one of these and $N nleq S$, then $N$ does not normalise $S$,
and so the $N$-orbit of $S$ has size divisible by $p$.
Also, all members of this orbit fail to contain $N$,
so we need only count the Sylow $p$-subgroups of $G$ that contain $N$,
and since $N > H$, this number is congruent to $1 bmod p$ by the inductive hypothesis.
Lemma 2. The theorem is true in the case where $G$ is a $p$-group.
Proof. Induct on $|G:H|$. The result is trivial if $H=G$ so we can assume that $H<G$.
We can also assume that $p^t < |G|$.
Let $mathcal{X}$ and $mathcal{M}$, respectively, be the set of subgroups of order $p^t$ that contain $H$ and
the set of maximal subgroups that contain $H$.
We count in two ways the number $N$ of ordered pairs $(X,M)$ with
$X in mathcal{X}$, $M in mathcal{M}$ and $X leq M$.
First, $N$ is the sum over $X in mathcal{X}$ of the number of maximal subgroups that contain $X$.
Also, $N$ is the sum over $M in mathcal{M}$ of the number of members of $mathcal{X}$ that are contained in $M$.
It is easy to see that the number of maximal subgroups of $G$ that contain any given proper subgroup of $G$ is congruent to $1 bmod p$.
The first formula for $N$ gives $N equiv |mathcal{X}| bmod p$ because the number of $M in mathcal{M}$
containing any given member of $mathcal{X}$ is congruent to $1 bmod p$.
By the inductive hypothesis, the number of members of $mathcal{X}$ contained in any given member of $mathcal{M}$ is congruent to $1$, so the second formula for $N$ gives $$N equiv |mathcal{M}| equiv 1 bmod p.$$
Thus $$|mathcal{X}| equiv N equiv 1 bmod p,$$ as wanted.
Proof of Theorem. Let $mathcal{X}$ be the set of subgroups of order $p^t$ that contain $H$ and let $mathcal{P}$ be the set of $P in operatorname{Syl}_p(G)$ that contain $H$. We count the number $S$ of pairs $(X,P)$, where $X in mathcal{X}$, $P in mathcal{P}$ and $X leq P$.
First, $S$ is the sum over $X in mathcal{X}$ of the number of $P in mathcal{P}$ such that $X leq P$. Also, $S$ is the sum over $P in mathcal{P}$ of the number of $X in mathcal{X}$ such that $X leq P$. Since the number of $P$ containing a given $X$ is congruent to $1 bmod p$ by Lemma $mathbf{1}$, the first formula for $S$ yields $S equiv |mathcal{X}| bmod p$. The number of $X$ contained in a given $P$ is congruent to $1 bmod p$ by Lemma $mathbf{2}$, so the second formula for $S$ yields $S equiv |mathcal{P}| bmod p$.
Also, $$|mathcal{P}| equiv 1 bmod p$$ by Lemma $mathbf{1}$, so we get
$$|mathcal{X}| equiv S equiv 1 bmod p,$$
which concludes our proof.
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
edited Nov 22 '18 at 12:00
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
answered Nov 22 '18 at 11:54
the_foxthe_fox
2,47211431
2,47211431
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I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$.
– Derek Holt
Nov 8 '18 at 8:06
1
See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf.
– KCd
Nov 22 '18 at 11:28