Mixing
Group of order 60
$begingroup$
Let $G$ be a simple group of order $60$:
i) Find the number of Sylow $3$- and $5$-subgroups of $G$.
ii) Show that the alternating group $A_5$ has a subgroup of order $12$.
iii)Show that $G$ is isomorphic to $A_5$ if it has a subgroup of order $12$.
iv) $G$ is isomorphic to $A_5$.
I'm stuck at i), iii) and iv):
Let $s_p$ denote the number of Sylow p-subgroups of $G$:
Because of $(60=2^2cdot3cdot5)$ I already know that $G$ must have Sylow $3$- and $5$-subgroups and that $s_3 in {1;4;10}$ and $s_5 in {1;6}$ because of $(s_3 mid text{ord}(G))$ and $(s_5 mid text{ord}(G))$ aswell as $(s_3 equiv 1 text{ mod }3)$ and $(s_5 equiv 1 text{ mod }5)$.
Is the following sufficient to exclude $1$ out of each set:
If $s_3=1$ or $s_5=1$ there is only one Sylow 3- or 5-subgroup. Because Sylow subgroups are conjugate to only one-another these Sylow 3- and 5-subgroups have to be normal in $G$. Contradiction: $G$ is simple. It follows that there are more than one Sylow 3- and 5-subgroups.
How would I go about narrowing down these even further?
The fact that $A_5$ has a subgroup of order $12$ I have already shown.
As for iii) and iv) I don't even know where to begin with. As other proofs of $G cong A_5$ on Stackexchange use "Extended Sylow Theorem", "Burnside Transfer Theorem" and "Strong Cayley Theorem" (probably the same as the first one) I'm quite curious how to prove this without those theorems.
Thanks for checking in!
~Cedric
group-theory finite-groups sylow-theory
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
$endgroup$
add a comment |
$begingroup$
Let $G$ be a simple group of order $60$:
i) Find the number of Sylow $3$- and $5$-subgroups of $G$.
ii) Show that the alternating group $A_5$ has a subgroup of order $12$.
iii)Show that $G$ is isomorphic to $A_5$ if it has a subgroup of order $12$.
iv) $G$ is isomorphic to $A_5$.
I'm stuck at i), iii) and iv):
Let $s_p$ denote the number of Sylow p-subgroups of $G$:
Because of $(60=2^2cdot3cdot5)$ I already know that $G$ must have Sylow $3$- and $5$-subgroups and that $s_3 in {1;4;10}$ and $s_5 in {1;6}$ because of $(s_3 mid text{ord}(G))$ and $(s_5 mid text{ord}(G))$ aswell as $(s_3 equiv 1 text{ mod }3)$ and $(s_5 equiv 1 text{ mod }5)$.
Is the following sufficient to exclude $1$ out of each set:
If $s_3=1$ or $s_5=1$ there is only one Sylow 3- or 5-subgroup. Because Sylow subgroups are conjugate to only one-another these Sylow 3- and 5-subgroups have to be normal in $G$. Contradiction: $G$ is simple. It follows that there are more than one Sylow 3- and 5-subgroups.
How would I go about narrowing down these even further?
The fact that $A_5$ has a subgroup of order $12$ I have already shown.
As for iii) and iv) I don't even know where to begin with. As other proofs of $G cong A_5$ on Stackexchange use "Extended Sylow Theorem", "Burnside Transfer Theorem" and "Strong Cayley Theorem" (probably the same as the first one) I'm quite curious how to prove this without those theorems.
Thanks for checking in!
~Cedric
group-theory finite-groups sylow-theory
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
$endgroup$
add a comment |
$begingroup$
Let $G$ be a simple group of order $60$:
i) Find the number of Sylow $3$- and $5$-subgroups of $G$.
ii) Show that the alternating group $A_5$ has a subgroup of order $12$.
iii)Show that $G$ is isomorphic to $A_5$ if it has a subgroup of order $12$.
iv) $G$ is isomorphic to $A_5$.
I'm stuck at i), iii) and iv):
Let $s_p$ denote the number of Sylow p-subgroups of $G$:
Because of $(60=2^2cdot3cdot5)$ I already know that $G$ must have Sylow $3$- and $5$-subgroups and that $s_3 in {1;4;10}$ and $s_5 in {1;6}$ because of $(s_3 mid text{ord}(G))$ and $(s_5 mid text{ord}(G))$ aswell as $(s_3 equiv 1 text{ mod }3)$ and $(s_5 equiv 1 text{ mod }5)$.
Is the following sufficient to exclude $1$ out of each set:
If $s_3=1$ or $s_5=1$ there is only one Sylow 3- or 5-subgroup. Because Sylow subgroups are conjugate to only one-another these Sylow 3- and 5-subgroups have to be normal in $G$. Contradiction: $G$ is simple. It follows that there are more than one Sylow 3- and 5-subgroups.
How would I go about narrowing down these even further?
The fact that $A_5$ has a subgroup of order $12$ I have already shown.
As for iii) and iv) I don't even know where to begin with. As other proofs of $G cong A_5$ on Stackexchange use "Extended Sylow Theorem", "Burnside Transfer Theorem" and "Strong Cayley Theorem" (probably the same as the first one) I'm quite curious how to prove this without those theorems.
Thanks for checking in!
~Cedric
group-theory finite-groups sylow-theory
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
$endgroup$
Let $G$ be a simple group of order $60$:
i) Find the number of Sylow $3$- and $5$-subgroups of $G$.
ii) Show that the alternating group $A_5$ has a subgroup of order $12$.
iii)Show that $G$ is isomorphic to $A_5$ if it has a subgroup of order $12$.
iv) $G$ is isomorphic to $A_5$.
I'm stuck at i), iii) and iv):
Let $s_p$ denote the number of Sylow p-subgroups of $G$:
Because of $(60=2^2cdot3cdot5)$ I already know that $G$ must have Sylow $3$- and $5$-subgroups and that $s_3 in {1;4;10}$ and $s_5 in {1;6}$ because of $(s_3 mid text{ord}(G))$ and $(s_5 mid text{ord}(G))$ aswell as $(s_3 equiv 1 text{ mod }3)$ and $(s_5 equiv 1 text{ mod }5)$.
Is the following sufficient to exclude $1$ out of each set:
If $s_3=1$ or $s_5=1$ there is only one Sylow 3- or 5-subgroup. Because Sylow subgroups are conjugate to only one-another these Sylow 3- and 5-subgroups have to be normal in $G$. Contradiction: $G$ is simple. It follows that there are more than one Sylow 3- and 5-subgroups.
How would I go about narrowing down these even further?
The fact that $A_5$ has a subgroup of order $12$ I have already shown.
As for iii) and iv) I don't even know where to begin with. As other proofs of $G cong A_5$ on Stackexchange use "Extended Sylow Theorem", "Burnside Transfer Theorem" and "Strong Cayley Theorem" (probably the same as the first one) I'm quite curious how to prove this without those theorems.
Thanks for checking in!
~Cedric
group-theory finite-groups sylow-theory
group-theory finite-groups sylow-theory
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
asked Jan 10 at 16:55
C. BrendelC. Brendel
538
538
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Yes that is sufficient to rule out $s_3,s_5 = 1$. Here's some advice on the points you asked about.
Advice on (i). I'll help you get started on $s_3$. First of all, you know $A_5$ has ${5 choose 3} = 10$ subgroups of order 3, which must be the 3-sylow subgroups, so you probably shouldn't try to prove $s_3 = 4$. To prove $s_3 neq 4$, use that $G$ acts by conjugation the set $X$ of $p$-sylow subgroups, which induces a homomorphism $G to S_n$ where $n = |X|$. Prove there is no such injection in the case where $X$ is the set of 3-sylow subgroups if $s_3 = 4$.
Advice on (iii). Try to show that $G$ acts transitively on a set of order 5. Where can you find such a set? Hint: the assumption provides one for you!
Advice on (iv). This part is more tricky. At least its clear the problem is to show that $G$ has a subgroup of order 12 :) Hint: consider the intersection $P_2cap P_2'$ of a pair of 2-sylow subgroups. What can you say about its normalizer? You should be able to reduce to the case it has size 12 or 20, and then rule out 20 for reasons similar to (i).
You can let me know with a comment if you think a while and are still stick.
answered Jan 11 at 9:09
BenBen
3,736616
$endgroup$
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
add a comment |
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$begingroup$
Yes that is sufficient to rule out $s_3,s_5 = 1$. Here's some advice on the points you asked about.
Advice on (i). I'll help you get started on $s_3$. First of all, you know $A_5$ has ${5 choose 3} = 10$ subgroups of order 3, which must be the 3-sylow subgroups, so you probably shouldn't try to prove $s_3 = 4$. To prove $s_3 neq 4$, use that $G$ acts by conjugation the set $X$ of $p$-sylow subgroups, which induces a homomorphism $G to S_n$ where $n = |X|$. Prove there is no such injection in the case where $X$ is the set of 3-sylow subgroups if $s_3 = 4$.
Advice on (iii). Try to show that $G$ acts transitively on a set of order 5. Where can you find such a set? Hint: the assumption provides one for you!
Advice on (iv). This part is more tricky. At least its clear the problem is to show that $G$ has a subgroup of order 12 :) Hint: consider the intersection $P_2cap P_2'$ of a pair of 2-sylow subgroups. What can you say about its normalizer? You should be able to reduce to the case it has size 12 or 20, and then rule out 20 for reasons similar to (i).
You can let me know with a comment if you think a while and are still stick.
answered Jan 11 at 9:09
BenBen
3,736616
$endgroup$
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
add a comment |
$begingroup$
Yes that is sufficient to rule out $s_3,s_5 = 1$. Here's some advice on the points you asked about.
Advice on (i). I'll help you get started on $s_3$. First of all, you know $A_5$ has ${5 choose 3} = 10$ subgroups of order 3, which must be the 3-sylow subgroups, so you probably shouldn't try to prove $s_3 = 4$. To prove $s_3 neq 4$, use that $G$ acts by conjugation the set $X$ of $p$-sylow subgroups, which induces a homomorphism $G to S_n$ where $n = |X|$. Prove there is no such injection in the case where $X$ is the set of 3-sylow subgroups if $s_3 = 4$.
Advice on (iii). Try to show that $G$ acts transitively on a set of order 5. Where can you find such a set? Hint: the assumption provides one for you!
Advice on (iv). This part is more tricky. At least its clear the problem is to show that $G$ has a subgroup of order 12 :) Hint: consider the intersection $P_2cap P_2'$ of a pair of 2-sylow subgroups. What can you say about its normalizer? You should be able to reduce to the case it has size 12 or 20, and then rule out 20 for reasons similar to (i).
You can let me know with a comment if you think a while and are still stick.
answered Jan 11 at 9:09
BenBen
3,736616
$endgroup$
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
add a comment |
$begingroup$
Yes that is sufficient to rule out $s_3,s_5 = 1$. Here's some advice on the points you asked about.
Advice on (i). I'll help you get started on $s_3$. First of all, you know $A_5$ has ${5 choose 3} = 10$ subgroups of order 3, which must be the 3-sylow subgroups, so you probably shouldn't try to prove $s_3 = 4$. To prove $s_3 neq 4$, use that $G$ acts by conjugation the set $X$ of $p$-sylow subgroups, which induces a homomorphism $G to S_n$ where $n = |X|$. Prove there is no such injection in the case where $X$ is the set of 3-sylow subgroups if $s_3 = 4$.
Advice on (iii). Try to show that $G$ acts transitively on a set of order 5. Where can you find such a set? Hint: the assumption provides one for you!
Advice on (iv). This part is more tricky. At least its clear the problem is to show that $G$ has a subgroup of order 12 :) Hint: consider the intersection $P_2cap P_2'$ of a pair of 2-sylow subgroups. What can you say about its normalizer? You should be able to reduce to the case it has size 12 or 20, and then rule out 20 for reasons similar to (i).
You can let me know with a comment if you think a while and are still stick.
answered Jan 11 at 9:09
BenBen
3,736616
$endgroup$
Yes that is sufficient to rule out $s_3,s_5 = 1$. Here's some advice on the points you asked about.
Advice on (i). I'll help you get started on $s_3$. First of all, you know $A_5$ has ${5 choose 3} = 10$ subgroups of order 3, which must be the 3-sylow subgroups, so you probably shouldn't try to prove $s_3 = 4$. To prove $s_3 neq 4$, use that $G$ acts by conjugation the set $X$ of $p$-sylow subgroups, which induces a homomorphism $G to S_n$ where $n = |X|$. Prove there is no such injection in the case where $X$ is the set of 3-sylow subgroups if $s_3 = 4$.
Advice on (iii). Try to show that $G$ acts transitively on a set of order 5. Where can you find such a set? Hint: the assumption provides one for you!
Advice on (iv). This part is more tricky. At least its clear the problem is to show that $G$ has a subgroup of order 12 :) Hint: consider the intersection $P_2cap P_2'$ of a pair of 2-sylow subgroups. What can you say about its normalizer? You should be able to reduce to the case it has size 12 or 20, and then rule out 20 for reasons similar to (i).
You can let me know with a comment if you think a while and are still stick.
answered Jan 11 at 9:09
BenBen
3,736616
edited Jan 11 at 9:32
answered Jan 11 at 9:09
BenBen
3,736616
answered Jan 11 at 9:09
BenBen
3,736616
answered Jan 11 at 9:09
BenBen
3,736616
3,736616
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
add a comment |
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Thanks for your Answer! It's very helpful :) iii) is now done! As for i): I understand that there can't be an injection from $G$ to $S_4$ as $vert{G}vert=60 > vert{S_4}vert=4!=24$, but how exactly does the conjugation induce this homomorphism and why is it sufficient to show it isn't injective (I guess the last part will follow from the first :P). iv) I'll try next!
$endgroup$
– C. Brendel
Jan 11 at 17:35
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
$begingroup$
Nevermind, got it all! Thanks :)
$endgroup$
– C. Brendel
Jan 12 at 9:17
add a comment |
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