Mixing
Is $n_p(G)$ unique for different groups of size $p$?
If $G$ has that $|G| = p^am$ where $p$ is prime and $gcd(p,m) = 1$, we have that $n_p(G)$ counts of Sylow $p$-subgroups in $G$. If $P$ is a Sylow $p$-subgroup, by the third Sylow theorem
begin{align*}
n_p(G) &equiv 1text{ mod } p\
n_p(G) &| m\
n_p(G) &= [G:N_G(P)]
end{align*}
$n_p(G)$ seems to depend on $G$, and these properties seem to only give possible candidates for what $n_p(G)$ will be. If I am given $G$ and $H$ with $|G| = |H| = p^am$, do we necessarily have that $n_p(G) = n_p(H)$? It feels sketchy to say so, but I can't come up with any counterexamples off the top of my head.
abstract-algebra group-theory sylow-theory
asked Nov 20 '18 at 19:30
J. Pistachio
477212
add a comment |
If $G$ has that $|G| = p^am$ where $p$ is prime and $gcd(p,m) = 1$, we have that $n_p(G)$ counts of Sylow $p$-subgroups in $G$. If $P$ is a Sylow $p$-subgroup, by the third Sylow theorem
begin{align*}
n_p(G) &equiv 1text{ mod } p\
n_p(G) &| m\
n_p(G) &= [G:N_G(P)]
end{align*}
$n_p(G)$ seems to depend on $G$, and these properties seem to only give possible candidates for what $n_p(G)$ will be. If I am given $G$ and $H$ with $|G| = |H| = p^am$, do we necessarily have that $n_p(G) = n_p(H)$? It feels sketchy to say so, but I can't come up with any counterexamples off the top of my head.
abstract-algebra group-theory sylow-theory
asked Nov 20 '18 at 19:30
J. Pistachio
477212
add a comment |
If $G$ has that $|G| = p^am$ where $p$ is prime and $gcd(p,m) = 1$, we have that $n_p(G)$ counts of Sylow $p$-subgroups in $G$. If $P$ is a Sylow $p$-subgroup, by the third Sylow theorem
begin{align*}
n_p(G) &equiv 1text{ mod } p\
n_p(G) &| m\
n_p(G) &= [G:N_G(P)]
end{align*}
$n_p(G)$ seems to depend on $G$, and these properties seem to only give possible candidates for what $n_p(G)$ will be. If I am given $G$ and $H$ with $|G| = |H| = p^am$, do we necessarily have that $n_p(G) = n_p(H)$? It feels sketchy to say so, but I can't come up with any counterexamples off the top of my head.
abstract-algebra group-theory sylow-theory
asked Nov 20 '18 at 19:30
J. Pistachio
477212
If $G$ has that $|G| = p^am$ where $p$ is prime and $gcd(p,m) = 1$, we have that $n_p(G)$ counts of Sylow $p$-subgroups in $G$. If $P$ is a Sylow $p$-subgroup, by the third Sylow theorem
begin{align*}
n_p(G) &equiv 1text{ mod } p\
n_p(G) &| m\
n_p(G) &= [G:N_G(P)]
end{align*}
$n_p(G)$ seems to depend on $G$, and these properties seem to only give possible candidates for what $n_p(G)$ will be. If I am given $G$ and $H$ with $|G| = |H| = p^am$, do we necessarily have that $n_p(G) = n_p(H)$? It feels sketchy to say so, but I can't come up with any counterexamples off the top of my head.
abstract-algebra group-theory sylow-theory
abstract-algebra group-theory sylow-theory
asked Nov 20 '18 at 19:30
J. Pistachio
477212
asked Nov 20 '18 at 19:30
J. Pistachio
477212
asked Nov 20 '18 at 19:30
J. Pistachio
477212
asked Nov 20 '18 at 19:30
J. Pistachio
477212
asked Nov 20 '18 at 19:30
J. Pistachio
477212
477212
add a comment |
add a comment |
2 Answers
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Hint: take $S_3$ and $C_6$ Both groups of order $6$. And look at the prime $2$.
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
add a comment |
By the way, suppose you have two finite groups $G$ and $H$ of the same order and for every prime $p$ dividing the order of $G$, $n_p(G)=n_p(H)$. Then $G$ and $H$ do not have to be isomorphic! (Example, take two non-isomorphic (non-abelian) groups $P_1$ and $P_2$ of order $p^3$ (see here), $p$ an odd prime and look at $G=C_2 times P_1$ and $H=C_2 times P_2$)
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
add a comment |
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2 Answers
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2 Answers
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Hint: take $S_3$ and $C_6$ Both groups of order $6$. And look at the prime $2$.
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
add a comment |
Hint: take $S_3$ and $C_6$ Both groups of order $6$. And look at the prime $2$.
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
add a comment |
Hint: take $S_3$ and $C_6$ Both groups of order $6$. And look at the prime $2$.
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
Hint: take $S_3$ and $C_6$ Both groups of order $6$. And look at the prime $2$.
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
answered Nov 20 '18 at 19:40
Nicky Hekster
28.2k53456
28.2k53456
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
add a comment |
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
If I am correct, $C_6$ has only one Sylow $2$-subgroup. Namely,if looking at $mathbb{Z}/6mathbb{Z}$, we have that ${0,3}$ is the only subgroup of size $2$. On the other hand, $S_3$ has $langle(1 2)rangle$, $langle(1 3)rangle$, and $langle(2 3)rangle$. Therefore, $n_2(C_6) = 1$ and $n_2(S_3) = 3$.
– J. Pistachio
Nov 20 '18 at 19:47
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
Yes correct! Well done! Perhaps you can look for other examples and other primes now.
– Nicky Hekster
Nov 20 '18 at 19:52
add a comment |
By the way, suppose you have two finite groups $G$ and $H$ of the same order and for every prime $p$ dividing the order of $G$, $n_p(G)=n_p(H)$. Then $G$ and $H$ do not have to be isomorphic! (Example, take two non-isomorphic (non-abelian) groups $P_1$ and $P_2$ of order $p^3$ (see here), $p$ an odd prime and look at $G=C_2 times P_1$ and $H=C_2 times P_2$)
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
add a comment |
By the way, suppose you have two finite groups $G$ and $H$ of the same order and for every prime $p$ dividing the order of $G$, $n_p(G)=n_p(H)$. Then $G$ and $H$ do not have to be isomorphic! (Example, take two non-isomorphic (non-abelian) groups $P_1$ and $P_2$ of order $p^3$ (see here), $p$ an odd prime and look at $G=C_2 times P_1$ and $H=C_2 times P_2$)
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
add a comment |
By the way, suppose you have two finite groups $G$ and $H$ of the same order and for every prime $p$ dividing the order of $G$, $n_p(G)=n_p(H)$. Then $G$ and $H$ do not have to be isomorphic! (Example, take two non-isomorphic (non-abelian) groups $P_1$ and $P_2$ of order $p^3$ (see here), $p$ an odd prime and look at $G=C_2 times P_1$ and $H=C_2 times P_2$)
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
By the way, suppose you have two finite groups $G$ and $H$ of the same order and for every prime $p$ dividing the order of $G$, $n_p(G)=n_p(H)$. Then $G$ and $H$ do not have to be isomorphic! (Example, take two non-isomorphic (non-abelian) groups $P_1$ and $P_2$ of order $p^3$ (see here), $p$ an odd prime and look at $G=C_2 times P_1$ and $H=C_2 times P_2$)
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
edited Nov 21 '18 at 7:04
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
answered Nov 20 '18 at 20:01
Nicky Hekster
28.2k53456
28.2k53456
add a comment |
add a comment |
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