Given a Group, $G =A_5$.Then choose correct statement [closed]
$begingroup$
Given $G =A_5$.Then
choose correct statement
$a)$$ G$ one sylow $3$ subgroup
$b)$ $G$ has four sylow $3 $ subgroup
$c)$ $G$ has ten sylow $3$ subgroup
My attempt : In my View I thinks None of the option will correct ,
but My friends said in my hostel when i was taking lunch , that option $c)$ will correct.
my logics is that option a) will not correct because G is simple
option $b)$ will not correct because number of element of order $3$ in $S_4$ is not equal to number of element of oder $3$ in $A_5$
similarly same logic in option $c)$ that is number of element of order $3$ in $S_{10}$ is not equal to number of element of order $3$ in $A_5$
so, None of the option will correct..
Is its True ?
Any hints/solution will be apprecaited
thanks u
abstract-algebra
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
$endgroup$
closed as off-topic by Jyrki Lahtonen, Batominovski, Alexander Gruber♦ Dec 12 '18 at 5:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Batominovski, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given $G =A_5$.Then
choose correct statement
$a)$$ G$ one sylow $3$ subgroup
$b)$ $G$ has four sylow $3 $ subgroup
$c)$ $G$ has ten sylow $3$ subgroup
My attempt : In my View I thinks None of the option will correct ,
but My friends said in my hostel when i was taking lunch , that option $c)$ will correct.
my logics is that option a) will not correct because G is simple
option $b)$ will not correct because number of element of order $3$ in $S_4$ is not equal to number of element of oder $3$ in $A_5$
similarly same logic in option $c)$ that is number of element of order $3$ in $S_{10}$ is not equal to number of element of order $3$ in $A_5$
so, None of the option will correct..
Is its True ?
Any hints/solution will be apprecaited
thanks u
abstract-algebra
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
$endgroup$
closed as off-topic by Jyrki Lahtonen, Batominovski, Alexander Gruber♦ Dec 12 '18 at 5:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Batominovski, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I don't know what $S_4$ and $S_{10}$ have to do with anything.
$endgroup$
– Slade
Dec 10 '18 at 13:00
$begingroup$
@Slade $ 1+ 3k |60$ i put $k=0,1,3$
$endgroup$
– jasmine
Dec 10 '18 at 13:02
1
$begingroup$
I don't know what that means.
$endgroup$
– Slade
Dec 10 '18 at 13:03
$begingroup$
...@Slade see $1 + 3.1 = 4$ is that $n_3 = 4$, so i take $S_4$ similarly $1+3.3= 10,$ so i take $S_{10}$
$endgroup$
– jasmine
Dec 10 '18 at 13:04
1
$begingroup$
I see, you're constructing a map $A_5 to S_{10}$. But this map isn't an isomorphism, so the fact that $S_{10}$ has more elements of order $3$ than $A_5$ does is irrelevant.
$endgroup$
– Slade
Dec 10 '18 at 13:09
add a comment |
$begingroup$
Given $G =A_5$.Then
choose correct statement
$a)$$ G$ one sylow $3$ subgroup
$b)$ $G$ has four sylow $3 $ subgroup
$c)$ $G$ has ten sylow $3$ subgroup
My attempt : In my View I thinks None of the option will correct ,
but My friends said in my hostel when i was taking lunch , that option $c)$ will correct.
my logics is that option a) will not correct because G is simple
option $b)$ will not correct because number of element of order $3$ in $S_4$ is not equal to number of element of oder $3$ in $A_5$
similarly same logic in option $c)$ that is number of element of order $3$ in $S_{10}$ is not equal to number of element of order $3$ in $A_5$
so, None of the option will correct..
Is its True ?
Any hints/solution will be apprecaited
thanks u
abstract-algebra
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
$endgroup$
Given $G =A_5$.Then
choose correct statement
$a)$$ G$ one sylow $3$ subgroup
$b)$ $G$ has four sylow $3 $ subgroup
$c)$ $G$ has ten sylow $3$ subgroup
My attempt : In my View I thinks None of the option will correct ,
but My friends said in my hostel when i was taking lunch , that option $c)$ will correct.
my logics is that option a) will not correct because G is simple
option $b)$ will not correct because number of element of order $3$ in $S_4$ is not equal to number of element of oder $3$ in $A_5$
similarly same logic in option $c)$ that is number of element of order $3$ in $S_{10}$ is not equal to number of element of order $3$ in $A_5$
so, None of the option will correct..
Is its True ?
Any hints/solution will be apprecaited
thanks u
abstract-algebra
abstract-algebra
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
asked Dec 10 '18 at 12:58
jasminejasmine
1,781417
1,781417
closed as off-topic by Jyrki Lahtonen, Batominovski, Alexander Gruber♦ Dec 12 '18 at 5:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Batominovski, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, Batominovski, Alexander Gruber♦ Dec 12 '18 at 5:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, Batominovski, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
I don't know what $S_4$ and $S_{10}$ have to do with anything.
$endgroup$
– Slade
Dec 10 '18 at 13:00
$begingroup$
@Slade $ 1+ 3k |60$ i put $k=0,1,3$
$endgroup$
– jasmine
Dec 10 '18 at 13:02
1
$begingroup$
I don't know what that means.
$endgroup$
– Slade
Dec 10 '18 at 13:03
$begingroup$
...@Slade see $1 + 3.1 = 4$ is that $n_3 = 4$, so i take $S_4$ similarly $1+3.3= 10,$ so i take $S_{10}$
$endgroup$
– jasmine
Dec 10 '18 at 13:04
1
$begingroup$
I see, you're constructing a map $A_5 to S_{10}$. But this map isn't an isomorphism, so the fact that $S_{10}$ has more elements of order $3$ than $A_5$ does is irrelevant.
$endgroup$
– Slade
Dec 10 '18 at 13:09
add a comment |
2
$begingroup$
I don't know what $S_4$ and $S_{10}$ have to do with anything.
$endgroup$
– Slade
Dec 10 '18 at 13:00
$begingroup$
@Slade $ 1+ 3k |60$ i put $k=0,1,3$
$endgroup$
– jasmine
Dec 10 '18 at 13:02
1
$begingroup$
I don't know what that means.
$endgroup$
– Slade
Dec 10 '18 at 13:03
$begingroup$
...@Slade see $1 + 3.1 = 4$ is that $n_3 = 4$, so i take $S_4$ similarly $1+3.3= 10,$ so i take $S_{10}$
$endgroup$
– jasmine
Dec 10 '18 at 13:04
1
$begingroup$
I see, you're constructing a map $A_5 to S_{10}$. But this map isn't an isomorphism, so the fact that $S_{10}$ has more elements of order $3$ than $A_5$ does is irrelevant.
$endgroup$
– Slade
Dec 10 '18 at 13:09
2
2
$begingroup$
I don't know what $S_4$ and $S_{10}$ have to do with anything.
$endgroup$
– Slade
Dec 10 '18 at 13:00
$begingroup$
I don't know what $S_4$ and $S_{10}$ have to do with anything.
$endgroup$
– Slade
Dec 10 '18 at 13:00
$begingroup$
@Slade $ 1+ 3k |60$ i put $k=0,1,3$
$endgroup$
– jasmine
Dec 10 '18 at 13:02
$begingroup$
@Slade $ 1+ 3k |60$ i put $k=0,1,3$
$endgroup$
– jasmine
Dec 10 '18 at 13:02
1
1
$begingroup$
I don't know what that means.
$endgroup$
– Slade
Dec 10 '18 at 13:03
$begingroup$
I don't know what that means.
$endgroup$
– Slade
Dec 10 '18 at 13:03
$begingroup$
...@Slade see $1 + 3.1 = 4$ is that $n_3 = 4$, so i take $S_4$ similarly $1+3.3= 10,$ so i take $S_{10}$
$endgroup$
– jasmine
Dec 10 '18 at 13:04
$begingroup$
...@Slade see $1 + 3.1 = 4$ is that $n_3 = 4$, so i take $S_4$ similarly $1+3.3= 10,$ so i take $S_{10}$
$endgroup$
– jasmine
Dec 10 '18 at 13:04
1
1
$begingroup$
I see, you're constructing a map $A_5 to S_{10}$. But this map isn't an isomorphism, so the fact that $S_{10}$ has more elements of order $3$ than $A_5$ does is irrelevant.
$endgroup$
– Slade
Dec 10 '18 at 13:09
$begingroup$
I see, you're constructing a map $A_5 to S_{10}$. But this map isn't an isomorphism, so the fact that $S_{10}$ has more elements of order $3$ than $A_5$ does is irrelevant.
$endgroup$
– Slade
Dec 10 '18 at 13:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't really understand your explanation about number of elements in $S_4$ and $S_{10}$. Anyway, using Sylow theorems we can see that one of the three options must be correct. Option $a$ is not correct because $G$ is simple, you got it right. Option $b$ is not correct because of the following: suppose $G$ has four $3$-Sylow subgroups. We can define an action of $G$ on $Syl_3(G)$ by $g.P=gPg^{-1}$. I'll leave it to you to check that it is indeed an action. Now, an action gives a homomorphism $varphi:Gto S_{Syl_3(G)}$ by $varphi(g)(P)=g.P=gPg^{-1}$. The group $G$ is simple and hence the homomorphism must be either trivial or one to one. It is not trivial because Sylow subgroups are conjugate to each other. And it is obviously not one to one because there can't be a function from a set with $60$ elements to a set with $4!=24$ elements which is one to one. A contradiction.
answered Dec 10 '18 at 13:06
MarkMark
8,424521
$endgroup$
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
1
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
add a comment |
$begingroup$
$|G| = 60 = 2^2cdot 3cdot5$, so each Sylow 3-subgroup has order $3$.
Check that every set ${a,b,c}$ of three elements from ${1,2,3,4,5}$ uniquely determines a subgroup of order 3, namely ${(), (abc), (acb)}$. Check that every order 3 subgroup of $A_5$ is of this form.
Then the number of Sylow 3-subgroups is just ${5 choose 3} = 10$.
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
$endgroup$
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
1
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't really understand your explanation about number of elements in $S_4$ and $S_{10}$. Anyway, using Sylow theorems we can see that one of the three options must be correct. Option $a$ is not correct because $G$ is simple, you got it right. Option $b$ is not correct because of the following: suppose $G$ has four $3$-Sylow subgroups. We can define an action of $G$ on $Syl_3(G)$ by $g.P=gPg^{-1}$. I'll leave it to you to check that it is indeed an action. Now, an action gives a homomorphism $varphi:Gto S_{Syl_3(G)}$ by $varphi(g)(P)=g.P=gPg^{-1}$. The group $G$ is simple and hence the homomorphism must be either trivial or one to one. It is not trivial because Sylow subgroups are conjugate to each other. And it is obviously not one to one because there can't be a function from a set with $60$ elements to a set with $4!=24$ elements which is one to one. A contradiction.
answered Dec 10 '18 at 13:06
MarkMark
8,424521
$endgroup$
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
1
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
add a comment |
$begingroup$
I don't really understand your explanation about number of elements in $S_4$ and $S_{10}$. Anyway, using Sylow theorems we can see that one of the three options must be correct. Option $a$ is not correct because $G$ is simple, you got it right. Option $b$ is not correct because of the following: suppose $G$ has four $3$-Sylow subgroups. We can define an action of $G$ on $Syl_3(G)$ by $g.P=gPg^{-1}$. I'll leave it to you to check that it is indeed an action. Now, an action gives a homomorphism $varphi:Gto S_{Syl_3(G)}$ by $varphi(g)(P)=g.P=gPg^{-1}$. The group $G$ is simple and hence the homomorphism must be either trivial or one to one. It is not trivial because Sylow subgroups are conjugate to each other. And it is obviously not one to one because there can't be a function from a set with $60$ elements to a set with $4!=24$ elements which is one to one. A contradiction.
answered Dec 10 '18 at 13:06
MarkMark
8,424521
$endgroup$
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
1
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
add a comment |
$begingroup$
I don't really understand your explanation about number of elements in $S_4$ and $S_{10}$. Anyway, using Sylow theorems we can see that one of the three options must be correct. Option $a$ is not correct because $G$ is simple, you got it right. Option $b$ is not correct because of the following: suppose $G$ has four $3$-Sylow subgroups. We can define an action of $G$ on $Syl_3(G)$ by $g.P=gPg^{-1}$. I'll leave it to you to check that it is indeed an action. Now, an action gives a homomorphism $varphi:Gto S_{Syl_3(G)}$ by $varphi(g)(P)=g.P=gPg^{-1}$. The group $G$ is simple and hence the homomorphism must be either trivial or one to one. It is not trivial because Sylow subgroups are conjugate to each other. And it is obviously not one to one because there can't be a function from a set with $60$ elements to a set with $4!=24$ elements which is one to one. A contradiction.
answered Dec 10 '18 at 13:06
MarkMark
8,424521
$endgroup$
I don't really understand your explanation about number of elements in $S_4$ and $S_{10}$. Anyway, using Sylow theorems we can see that one of the three options must be correct. Option $a$ is not correct because $G$ is simple, you got it right. Option $b$ is not correct because of the following: suppose $G$ has four $3$-Sylow subgroups. We can define an action of $G$ on $Syl_3(G)$ by $g.P=gPg^{-1}$. I'll leave it to you to check that it is indeed an action. Now, an action gives a homomorphism $varphi:Gto S_{Syl_3(G)}$ by $varphi(g)(P)=g.P=gPg^{-1}$. The group $G$ is simple and hence the homomorphism must be either trivial or one to one. It is not trivial because Sylow subgroups are conjugate to each other. And it is obviously not one to one because there can't be a function from a set with $60$ elements to a set with $4!=24$ elements which is one to one. A contradiction.
answered Dec 10 '18 at 13:06
MarkMark
8,424521
answered Dec 10 '18 at 13:06
MarkMark
8,424521
answered Dec 10 '18 at 13:06
MarkMark
8,424521
answered Dec 10 '18 at 13:06
MarkMark
8,424521
8,424521
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
1
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
add a comment |
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
1
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
$begingroup$
@Marks from ur last sentence even 10 ! will also get contradiction ..can u elaborate that
$endgroup$
– jasmine
Dec 10 '18 at 13:17
1
1
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
Why will you get a contradiction? $10!$ is a number which is much bigger than $60$. There is no reason why there can't be an injective function from a set of $60$ elements to a set of $10!$ elements. In my answer I simply used the fact that $60>4!$.
$endgroup$
– Mark
Dec 10 '18 at 13:21
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
$begingroup$
okss ,thanks u @Marks i got its
$endgroup$
– jasmine
Dec 10 '18 at 13:25
add a comment |
$begingroup$
$|G| = 60 = 2^2cdot 3cdot5$, so each Sylow 3-subgroup has order $3$.
Check that every set ${a,b,c}$ of three elements from ${1,2,3,4,5}$ uniquely determines a subgroup of order 3, namely ${(), (abc), (acb)}$. Check that every order 3 subgroup of $A_5$ is of this form.
Then the number of Sylow 3-subgroups is just ${5 choose 3} = 10$.
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
$endgroup$
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
1
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
add a comment |
$begingroup$
$|G| = 60 = 2^2cdot 3cdot5$, so each Sylow 3-subgroup has order $3$.
Check that every set ${a,b,c}$ of three elements from ${1,2,3,4,5}$ uniquely determines a subgroup of order 3, namely ${(), (abc), (acb)}$. Check that every order 3 subgroup of $A_5$ is of this form.
Then the number of Sylow 3-subgroups is just ${5 choose 3} = 10$.
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
$endgroup$
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
1
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
add a comment |
$begingroup$
$|G| = 60 = 2^2cdot 3cdot5$, so each Sylow 3-subgroup has order $3$.
Check that every set ${a,b,c}$ of three elements from ${1,2,3,4,5}$ uniquely determines a subgroup of order 3, namely ${(), (abc), (acb)}$. Check that every order 3 subgroup of $A_5$ is of this form.
Then the number of Sylow 3-subgroups is just ${5 choose 3} = 10$.
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
$endgroup$
$|G| = 60 = 2^2cdot 3cdot5$, so each Sylow 3-subgroup has order $3$.
Check that every set ${a,b,c}$ of three elements from ${1,2,3,4,5}$ uniquely determines a subgroup of order 3, namely ${(), (abc), (acb)}$. Check that every order 3 subgroup of $A_5$ is of this form.
Then the number of Sylow 3-subgroups is just ${5 choose 3} = 10$.
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
answered Dec 10 '18 at 13:07
SladeSlade
25.1k12665
25.1k12665
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
1
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
add a comment |
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
1
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
$begingroup$
..@Slade is their any formula to find the number of sylow p-subgroup ??
$endgroup$
– jasmine
Dec 10 '18 at 13:14
1
1
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
@jasmine For symmetric groups in general? No, this is a difficult research problem.
$endgroup$
– Slade
Dec 10 '18 at 13:51
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
$begingroup$
okss @Slade....
$endgroup$
– jasmine
Dec 10 '18 at 13:54
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I don't know what $S_4$ and $S_{10}$ have to do with anything.
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– Slade
Dec 10 '18 at 13:00
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@Slade $ 1+ 3k |60$ i put $k=0,1,3$
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– jasmine
Dec 10 '18 at 13:02
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I don't know what that means.
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– Slade
Dec 10 '18 at 13:03
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...@Slade see $1 + 3.1 = 4$ is that $n_3 = 4$, so i take $S_4$ similarly $1+3.3= 10,$ so i take $S_{10}$
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– jasmine
Dec 10 '18 at 13:04
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I see, you're constructing a map $A_5 to S_{10}$. But this map isn't an isomorphism, so the fact that $S_{10}$ has more elements of order $3$ than $A_5$ does is irrelevant.
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– Slade
Dec 10 '18 at 13:09