Mixing
Proving that if $(ab)^{p}=a^{p},b^{p}$, then the p-sylow subgroup is normal
$begingroup$
So, while studying Abstract Algebra, i ran into this problem (i.n. Herstein, second edition, chapter 2.12) and have been stuck since:
Given a group of finite order, and a prime $p$ that divides $o(G)$, and suppose that $forall,a,bin G$, $(ab)^{p}=a^{p},b^{p}$. Prove that the $p$-sylow subgroup is normal in $G$.
What I've tried: I defined a mapping $varphi:Gto H=lbrace x^{p}:,xin Grbrace;,,varphi(x)=x^{p}$, wich would be a surjective homomorphism. Then, I proved that $ker(varphi)subseteq P$, where $P$ is a $p$-sylow subgroup. If I could prove either that $Psubseteqker(varphi)$ or that $o(ker(varphi))=o(P)$, that would end it, because that would imply that $P=ker(varphi)$, and I know that $ker(varphi)unlhd G$.
One ideia to follow up on those would be to use the firs isomorphism theorem ($G/ker(varphi)simeq Im(varphi)$) to get $o(G)=o(ker(varphi))o(Im(varphi))$ and from there work something out about the orders, but I cannot think of how to do that.
I've also tried proving that $G$ only has one $p$-sylow subgroup, using Sylow's third theorem, but I believe it's a dead end.
Any ideas?
abstract-algebra finite-groups sylow-theory
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
$endgroup$
add a comment |
$begingroup$
So, while studying Abstract Algebra, i ran into this problem (i.n. Herstein, second edition, chapter 2.12) and have been stuck since:
Given a group of finite order, and a prime $p$ that divides $o(G)$, and suppose that $forall,a,bin G$, $(ab)^{p}=a^{p},b^{p}$. Prove that the $p$-sylow subgroup is normal in $G$.
What I've tried: I defined a mapping $varphi:Gto H=lbrace x^{p}:,xin Grbrace;,,varphi(x)=x^{p}$, wich would be a surjective homomorphism. Then, I proved that $ker(varphi)subseteq P$, where $P$ is a $p$-sylow subgroup. If I could prove either that $Psubseteqker(varphi)$ or that $o(ker(varphi))=o(P)$, that would end it, because that would imply that $P=ker(varphi)$, and I know that $ker(varphi)unlhd G$.
One ideia to follow up on those would be to use the firs isomorphism theorem ($G/ker(varphi)simeq Im(varphi)$) to get $o(G)=o(ker(varphi))o(Im(varphi))$ and from there work something out about the orders, but I cannot think of how to do that.
I've also tried proving that $G$ only has one $p$-sylow subgroup, using Sylow's third theorem, but I believe it's a dead end.
Any ideas?
abstract-algebra finite-groups sylow-theory
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
$endgroup$
$begingroup$
I think that you're on the right track. Suppose that $p^q$ is the highest power of $p$ dividing $o(G)$. Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Now, prove that $ker(varphi^q)subseteq P$ and $Psubseteqker(varphi^q)$.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:44
$begingroup$
But $ker(phi^q)$ contains all elements from any $p$-Sylow subgroup of $G$, so that only works if there is a unique $p$-Sylow subgroup $P$, which is equivalent to $P$ being normal.
$endgroup$
– Ethan Alwaise
Dec 1 '16 at 0:46
$begingroup$
@EthanAlwaise If there are multiple $p$-Sylow subgroups, then one can prove that the kernel of $varphi^q$ contains an element which is not of order $p^r$ for some $r$. This is impossible.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:48
add a comment |
$begingroup$
So, while studying Abstract Algebra, i ran into this problem (i.n. Herstein, second edition, chapter 2.12) and have been stuck since:
Given a group of finite order, and a prime $p$ that divides $o(G)$, and suppose that $forall,a,bin G$, $(ab)^{p}=a^{p},b^{p}$. Prove that the $p$-sylow subgroup is normal in $G$.
What I've tried: I defined a mapping $varphi:Gto H=lbrace x^{p}:,xin Grbrace;,,varphi(x)=x^{p}$, wich would be a surjective homomorphism. Then, I proved that $ker(varphi)subseteq P$, where $P$ is a $p$-sylow subgroup. If I could prove either that $Psubseteqker(varphi)$ or that $o(ker(varphi))=o(P)$, that would end it, because that would imply that $P=ker(varphi)$, and I know that $ker(varphi)unlhd G$.
One ideia to follow up on those would be to use the firs isomorphism theorem ($G/ker(varphi)simeq Im(varphi)$) to get $o(G)=o(ker(varphi))o(Im(varphi))$ and from there work something out about the orders, but I cannot think of how to do that.
I've also tried proving that $G$ only has one $p$-sylow subgroup, using Sylow's third theorem, but I believe it's a dead end.
Any ideas?
abstract-algebra finite-groups sylow-theory
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
$endgroup$
So, while studying Abstract Algebra, i ran into this problem (i.n. Herstein, second edition, chapter 2.12) and have been stuck since:
Given a group of finite order, and a prime $p$ that divides $o(G)$, and suppose that $forall,a,bin G$, $(ab)^{p}=a^{p},b^{p}$. Prove that the $p$-sylow subgroup is normal in $G$.
What I've tried: I defined a mapping $varphi:Gto H=lbrace x^{p}:,xin Grbrace;,,varphi(x)=x^{p}$, wich would be a surjective homomorphism. Then, I proved that $ker(varphi)subseteq P$, where $P$ is a $p$-sylow subgroup. If I could prove either that $Psubseteqker(varphi)$ or that $o(ker(varphi))=o(P)$, that would end it, because that would imply that $P=ker(varphi)$, and I know that $ker(varphi)unlhd G$.
One ideia to follow up on those would be to use the firs isomorphism theorem ($G/ker(varphi)simeq Im(varphi)$) to get $o(G)=o(ker(varphi))o(Im(varphi))$ and from there work something out about the orders, but I cannot think of how to do that.
I've also tried proving that $G$ only has one $p$-sylow subgroup, using Sylow's third theorem, but I believe it's a dead end.
Any ideas?
abstract-algebra finite-groups sylow-theory
abstract-algebra finite-groups sylow-theory
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
asked Dec 1 '16 at 0:34
João RuizJoão Ruiz
513
513
$begingroup$
I think that you're on the right track. Suppose that $p^q$ is the highest power of $p$ dividing $o(G)$. Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Now, prove that $ker(varphi^q)subseteq P$ and $Psubseteqker(varphi^q)$.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:44
$begingroup$
But $ker(phi^q)$ contains all elements from any $p$-Sylow subgroup of $G$, so that only works if there is a unique $p$-Sylow subgroup $P$, which is equivalent to $P$ being normal.
$endgroup$
– Ethan Alwaise
Dec 1 '16 at 0:46
$begingroup$
@EthanAlwaise If there are multiple $p$-Sylow subgroups, then one can prove that the kernel of $varphi^q$ contains an element which is not of order $p^r$ for some $r$. This is impossible.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:48
add a comment |
$begingroup$
I think that you're on the right track. Suppose that $p^q$ is the highest power of $p$ dividing $o(G)$. Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Now, prove that $ker(varphi^q)subseteq P$ and $Psubseteqker(varphi^q)$.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:44
$begingroup$
But $ker(phi^q)$ contains all elements from any $p$-Sylow subgroup of $G$, so that only works if there is a unique $p$-Sylow subgroup $P$, which is equivalent to $P$ being normal.
$endgroup$
– Ethan Alwaise
Dec 1 '16 at 0:46
$begingroup$
@EthanAlwaise If there are multiple $p$-Sylow subgroups, then one can prove that the kernel of $varphi^q$ contains an element which is not of order $p^r$ for some $r$. This is impossible.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:48
$begingroup$
I think that you're on the right track. Suppose that $p^q$ is the highest power of $p$ dividing $o(G)$. Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Now, prove that $ker(varphi^q)subseteq P$ and $Psubseteqker(varphi^q)$.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:44
$begingroup$
I think that you're on the right track. Suppose that $p^q$ is the highest power of $p$ dividing $o(G)$. Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Now, prove that $ker(varphi^q)subseteq P$ and $Psubseteqker(varphi^q)$.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:44
$begingroup$
But $ker(phi^q)$ contains all elements from any $p$-Sylow subgroup of $G$, so that only works if there is a unique $p$-Sylow subgroup $P$, which is equivalent to $P$ being normal.
$endgroup$
– Ethan Alwaise
Dec 1 '16 at 0:46
$begingroup$
But $ker(phi^q)$ contains all elements from any $p$-Sylow subgroup of $G$, so that only works if there is a unique $p$-Sylow subgroup $P$, which is equivalent to $P$ being normal.
$endgroup$
– Ethan Alwaise
Dec 1 '16 at 0:46
$begingroup$
@EthanAlwaise If there are multiple $p$-Sylow subgroups, then one can prove that the kernel of $varphi^q$ contains an element which is not of order $p^r$ for some $r$. This is impossible.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:48
$begingroup$
@EthanAlwaise If there are multiple $p$-Sylow subgroups, then one can prove that the kernel of $varphi^q$ contains an element which is not of order $p^r$ for some $r$. This is impossible.
$endgroup$
– Michael Burr
Dec 1 '16 at 0:48
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Let $varphi:Grightarrow G$ be defined by $varphi(x)=x^p$. $varphi(x)$ is a homomorphism because $varphi(xy)=(xy)^p=x^py^p$ by assumption. Suppose that $p^q$ is the largest power of $p$ which divides $o(G)$.
Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Since $varphi^q$ is a composition of homomorphisms, it is a homomorphism. Moreover, note that the kernel of $varphi^q$ consists of all elements of $G$ whose order is $p^r$ for some $r$.
Since every element of order $p^r$ is in some Sylow-$p$ subgroup, this means that $ker(varphi^q)$ is exactly the union of all of the Sylow-$p$ subgroups. Moreover, the order of $ker(varphi^q)$ is a power of $p$ (since otherwise, by Cauchy's theorem, there would be an element of order $p'$ where $p'$ is a prime not equal to $p$). Hence, $ker(varphi^q)$ is a $p$-subgroup of $G$ containing a Sylow-$p$ subgroup of $G$, so $ker(varphi^q)$ must be a Sylow-$p$ subgroup. Therefore, there is only one Sylow-$p$ subgroup and it is $ker(varphi^q)$ which is normal.
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
$endgroup$
add a comment |
$begingroup$
Let's try mathematical induction with respect to the group order $n = |G|$.
The claim is obiviously true for $n=1$. Let it be true for $n=N$. To show it is true for $n=N+1$, let $|G| = N+1$ with $p^alpha big | |G|$ and $alpha$ is the largest power of $p$ dividing $N+1$. And $(ab)^p = a^p b ^p$
Let $K = ker(phi)$
First notice that, $p big | |K|$. Consider the quotient group $G/K$, assuming $K$ is not the $p$ syllow s.g (otherwise we are done),
By induction $G/K$ {has order $< N+1$, and satisfies $(ab)^p = a^p b ^p$ } contains a p-syllow s.g $tilde{P}$ which is normal in $G/K$. Let $P = {xin G| xK in tilde{P}}$, which is a normal subgroup in $G$. And we can notice that $p^alpha big | |P|$ and hence $P$ contains a p-syllow s.g of $G$ which is normal in $P$ (by induction). Since $P$ is normal in $G$, the p-syllow s.g contained in $P$ is also normal in $G$.
answered Jan 21 at 7:08
pavanpavan
914
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $varphi:Grightarrow G$ be defined by $varphi(x)=x^p$. $varphi(x)$ is a homomorphism because $varphi(xy)=(xy)^p=x^py^p$ by assumption. Suppose that $p^q$ is the largest power of $p$ which divides $o(G)$.
Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Since $varphi^q$ is a composition of homomorphisms, it is a homomorphism. Moreover, note that the kernel of $varphi^q$ consists of all elements of $G$ whose order is $p^r$ for some $r$.
Since every element of order $p^r$ is in some Sylow-$p$ subgroup, this means that $ker(varphi^q)$ is exactly the union of all of the Sylow-$p$ subgroups. Moreover, the order of $ker(varphi^q)$ is a power of $p$ (since otherwise, by Cauchy's theorem, there would be an element of order $p'$ where $p'$ is a prime not equal to $p$). Hence, $ker(varphi^q)$ is a $p$-subgroup of $G$ containing a Sylow-$p$ subgroup of $G$, so $ker(varphi^q)$ must be a Sylow-$p$ subgroup. Therefore, there is only one Sylow-$p$ subgroup and it is $ker(varphi^q)$ which is normal.
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
$endgroup$
add a comment |
$begingroup$
Let $varphi:Grightarrow G$ be defined by $varphi(x)=x^p$. $varphi(x)$ is a homomorphism because $varphi(xy)=(xy)^p=x^py^p$ by assumption. Suppose that $p^q$ is the largest power of $p$ which divides $o(G)$.
Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Since $varphi^q$ is a composition of homomorphisms, it is a homomorphism. Moreover, note that the kernel of $varphi^q$ consists of all elements of $G$ whose order is $p^r$ for some $r$.
Since every element of order $p^r$ is in some Sylow-$p$ subgroup, this means that $ker(varphi^q)$ is exactly the union of all of the Sylow-$p$ subgroups. Moreover, the order of $ker(varphi^q)$ is a power of $p$ (since otherwise, by Cauchy's theorem, there would be an element of order $p'$ where $p'$ is a prime not equal to $p$). Hence, $ker(varphi^q)$ is a $p$-subgroup of $G$ containing a Sylow-$p$ subgroup of $G$, so $ker(varphi^q)$ must be a Sylow-$p$ subgroup. Therefore, there is only one Sylow-$p$ subgroup and it is $ker(varphi^q)$ which is normal.
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
$endgroup$
add a comment |
$begingroup$
Let $varphi:Grightarrow G$ be defined by $varphi(x)=x^p$. $varphi(x)$ is a homomorphism because $varphi(xy)=(xy)^p=x^py^p$ by assumption. Suppose that $p^q$ is the largest power of $p$ which divides $o(G)$.
Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Since $varphi^q$ is a composition of homomorphisms, it is a homomorphism. Moreover, note that the kernel of $varphi^q$ consists of all elements of $G$ whose order is $p^r$ for some $r$.
Since every element of order $p^r$ is in some Sylow-$p$ subgroup, this means that $ker(varphi^q)$ is exactly the union of all of the Sylow-$p$ subgroups. Moreover, the order of $ker(varphi^q)$ is a power of $p$ (since otherwise, by Cauchy's theorem, there would be an element of order $p'$ where $p'$ is a prime not equal to $p$). Hence, $ker(varphi^q)$ is a $p$-subgroup of $G$ containing a Sylow-$p$ subgroup of $G$, so $ker(varphi^q)$ must be a Sylow-$p$ subgroup. Therefore, there is only one Sylow-$p$ subgroup and it is $ker(varphi^q)$ which is normal.
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
$endgroup$
Let $varphi:Grightarrow G$ be defined by $varphi(x)=x^p$. $varphi(x)$ is a homomorphism because $varphi(xy)=(xy)^p=x^py^p$ by assumption. Suppose that $p^q$ is the largest power of $p$ which divides $o(G)$.
Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Since $varphi^q$ is a composition of homomorphisms, it is a homomorphism. Moreover, note that the kernel of $varphi^q$ consists of all elements of $G$ whose order is $p^r$ for some $r$.
Since every element of order $p^r$ is in some Sylow-$p$ subgroup, this means that $ker(varphi^q)$ is exactly the union of all of the Sylow-$p$ subgroups. Moreover, the order of $ker(varphi^q)$ is a power of $p$ (since otherwise, by Cauchy's theorem, there would be an element of order $p'$ where $p'$ is a prime not equal to $p$). Hence, $ker(varphi^q)$ is a $p$-subgroup of $G$ containing a Sylow-$p$ subgroup of $G$, so $ker(varphi^q)$ must be a Sylow-$p$ subgroup. Therefore, there is only one Sylow-$p$ subgroup and it is $ker(varphi^q)$ which is normal.
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
answered Dec 1 '16 at 0:56
Michael BurrMichael Burr
26.9k23262
26.9k23262
add a comment |
add a comment |
$begingroup$
Let's try mathematical induction with respect to the group order $n = |G|$.
The claim is obiviously true for $n=1$. Let it be true for $n=N$. To show it is true for $n=N+1$, let $|G| = N+1$ with $p^alpha big | |G|$ and $alpha$ is the largest power of $p$ dividing $N+1$. And $(ab)^p = a^p b ^p$
Let $K = ker(phi)$
First notice that, $p big | |K|$. Consider the quotient group $G/K$, assuming $K$ is not the $p$ syllow s.g (otherwise we are done),
By induction $G/K$ {has order $< N+1$, and satisfies $(ab)^p = a^p b ^p$ } contains a p-syllow s.g $tilde{P}$ which is normal in $G/K$. Let $P = {xin G| xK in tilde{P}}$, which is a normal subgroup in $G$. And we can notice that $p^alpha big | |P|$ and hence $P$ contains a p-syllow s.g of $G$ which is normal in $P$ (by induction). Since $P$ is normal in $G$, the p-syllow s.g contained in $P$ is also normal in $G$.
answered Jan 21 at 7:08
pavanpavan
914
$endgroup$
add a comment |
$begingroup$
Let's try mathematical induction with respect to the group order $n = |G|$.
The claim is obiviously true for $n=1$. Let it be true for $n=N$. To show it is true for $n=N+1$, let $|G| = N+1$ with $p^alpha big | |G|$ and $alpha$ is the largest power of $p$ dividing $N+1$. And $(ab)^p = a^p b ^p$
Let $K = ker(phi)$
First notice that, $p big | |K|$. Consider the quotient group $G/K$, assuming $K$ is not the $p$ syllow s.g (otherwise we are done),
By induction $G/K$ {has order $< N+1$, and satisfies $(ab)^p = a^p b ^p$ } contains a p-syllow s.g $tilde{P}$ which is normal in $G/K$. Let $P = {xin G| xK in tilde{P}}$, which is a normal subgroup in $G$. And we can notice that $p^alpha big | |P|$ and hence $P$ contains a p-syllow s.g of $G$ which is normal in $P$ (by induction). Since $P$ is normal in $G$, the p-syllow s.g contained in $P$ is also normal in $G$.
answered Jan 21 at 7:08
pavanpavan
914
$endgroup$
add a comment |
$begingroup$
Let's try mathematical induction with respect to the group order $n = |G|$.
The claim is obiviously true for $n=1$. Let it be true for $n=N$. To show it is true for $n=N+1$, let $|G| = N+1$ with $p^alpha big | |G|$ and $alpha$ is the largest power of $p$ dividing $N+1$. And $(ab)^p = a^p b ^p$
Let $K = ker(phi)$
First notice that, $p big | |K|$. Consider the quotient group $G/K$, assuming $K$ is not the $p$ syllow s.g (otherwise we are done),
By induction $G/K$ {has order $< N+1$, and satisfies $(ab)^p = a^p b ^p$ } contains a p-syllow s.g $tilde{P}$ which is normal in $G/K$. Let $P = {xin G| xK in tilde{P}}$, which is a normal subgroup in $G$. And we can notice that $p^alpha big | |P|$ and hence $P$ contains a p-syllow s.g of $G$ which is normal in $P$ (by induction). Since $P$ is normal in $G$, the p-syllow s.g contained in $P$ is also normal in $G$.
answered Jan 21 at 7:08
pavanpavan
914
$endgroup$
Let's try mathematical induction with respect to the group order $n = |G|$.
The claim is obiviously true for $n=1$. Let it be true for $n=N$. To show it is true for $n=N+1$, let $|G| = N+1$ with $p^alpha big | |G|$ and $alpha$ is the largest power of $p$ dividing $N+1$. And $(ab)^p = a^p b ^p$
Let $K = ker(phi)$
First notice that, $p big | |K|$. Consider the quotient group $G/K$, assuming $K$ is not the $p$ syllow s.g (otherwise we are done),
By induction $G/K$ {has order $< N+1$, and satisfies $(ab)^p = a^p b ^p$ } contains a p-syllow s.g $tilde{P}$ which is normal in $G/K$. Let $P = {xin G| xK in tilde{P}}$, which is a normal subgroup in $G$. And we can notice that $p^alpha big | |P|$ and hence $P$ contains a p-syllow s.g of $G$ which is normal in $P$ (by induction). Since $P$ is normal in $G$, the p-syllow s.g contained in $P$ is also normal in $G$.
answered Jan 21 at 7:08
pavanpavan
914
answered Jan 21 at 7:08
pavanpavan
914
answered Jan 21 at 7:08
pavanpavan
914
answered Jan 21 at 7:08
pavanpavan
914
914
add a comment |
add a comment |
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Required, but never shown
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I think that you're on the right track. Suppose that $p^q$ is the highest power of $p$ dividing $o(G)$. Consider $varphi^q$ so that $varphi^q(x)=x^{p^q}$. Now, prove that $ker(varphi^q)subseteq P$ and $Psubseteqker(varphi^q)$.
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– Michael Burr
Dec 1 '16 at 0:44
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But $ker(phi^q)$ contains all elements from any $p$-Sylow subgroup of $G$, so that only works if there is a unique $p$-Sylow subgroup $P$, which is equivalent to $P$ being normal.
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– Ethan Alwaise
Dec 1 '16 at 0:46
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@EthanAlwaise If there are multiple $p$-Sylow subgroups, then one can prove that the kernel of $varphi^q$ contains an element which is not of order $p^r$ for some $r$. This is impossible.
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– Michael Burr
Dec 1 '16 at 0:48