Is there exist a group homomorphism from the symmetric group $S_n$ to $S_{n-1}$ for $n ge 5?$
$begingroup$
Does there exist a group homomorphism from the symmetric group $S_n$ to $S_{n-1}$ for $n ge 5?$
My attempt: I think not, because for $n ge 5$ , $A_n$ is the only normal subgroup of $S_n$.
Is it true?
group-theory symmetric-groups normal-subgroups group-homomorphism
edited Dec 29 '18 at 10:58
Shaun
9,585113684
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
$endgroup$
add a comment |
$begingroup$
Does there exist a group homomorphism from the symmetric group $S_n$ to $S_{n-1}$ for $n ge 5?$
My attempt: I think not, because for $n ge 5$ , $A_n$ is the only normal subgroup of $S_n$.
Is it true?
group-theory symmetric-groups normal-subgroups group-homomorphism
edited Dec 29 '18 at 10:58
Shaun
9,585113684
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
$endgroup$
7
$begingroup$
The trivial homomorphism?
$endgroup$
– Randall
Dec 29 '18 at 1:35
add a comment |
$begingroup$
Does there exist a group homomorphism from the symmetric group $S_n$ to $S_{n-1}$ for $n ge 5?$
My attempt: I think not, because for $n ge 5$ , $A_n$ is the only normal subgroup of $S_n$.
Is it true?
group-theory symmetric-groups normal-subgroups group-homomorphism
edited Dec 29 '18 at 10:58
Shaun
9,585113684
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
$endgroup$
Does there exist a group homomorphism from the symmetric group $S_n$ to $S_{n-1}$ for $n ge 5?$
My attempt: I think not, because for $n ge 5$ , $A_n$ is the only normal subgroup of $S_n$.
Is it true?
group-theory symmetric-groups normal-subgroups group-homomorphism
group-theory symmetric-groups normal-subgroups group-homomorphism
edited Dec 29 '18 at 10:58
Shaun
9,585113684
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
edited Dec 29 '18 at 10:58
Shaun
9,585113684
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
edited Dec 29 '18 at 10:58
Shaun
9,585113684
edited Dec 29 '18 at 10:58
Shaun
9,585113684
edited Dec 29 '18 at 10:58
Shaun
9,585113684
9,585113684
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
asked Dec 29 '18 at 1:33
jasminejasmine
1,888418
1,888418
7
$begingroup$
The trivial homomorphism?
$endgroup$
– Randall
Dec 29 '18 at 1:35
add a comment |
7
$begingroup$
The trivial homomorphism?
$endgroup$
– Randall
Dec 29 '18 at 1:35
7
7
$begingroup$
The trivial homomorphism?
$endgroup$
– Randall
Dec 29 '18 at 1:35
$begingroup$
The trivial homomorphism?
$endgroup$
– Randall
Dec 29 '18 at 1:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you are aware that $A_n$ is a normal subgroup how about trying for a homomorphism that will have $A_n$ as its kernel.
Take any permutation $tauin S_{n-1}$ of order 2. For example $tau =(17)(38)(26)in S_8$. (cycle notation).
Define for $sigmain A_n$, $phi(sigma) =id$, and $phi(sigma)=tau$ for $sigmain S_n-A_n$. This is a nontrivial homomorphism with image a subgroup of order $2$. This shows that in fact there are as many homomorphims as there are elemnts of order 2 in $S_{n-1}$. The same argument works by replacing the codomain $S_{n-1}$ by any group of even order (they always have elements of order 2).
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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votes
$begingroup$
As you are aware that $A_n$ is a normal subgroup how about trying for a homomorphism that will have $A_n$ as its kernel.
Take any permutation $tauin S_{n-1}$ of order 2. For example $tau =(17)(38)(26)in S_8$. (cycle notation).
Define for $sigmain A_n$, $phi(sigma) =id$, and $phi(sigma)=tau$ for $sigmain S_n-A_n$. This is a nontrivial homomorphism with image a subgroup of order $2$. This shows that in fact there are as many homomorphims as there are elemnts of order 2 in $S_{n-1}$. The same argument works by replacing the codomain $S_{n-1}$ by any group of even order (they always have elements of order 2).
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
add a comment |
$begingroup$
As you are aware that $A_n$ is a normal subgroup how about trying for a homomorphism that will have $A_n$ as its kernel.
Take any permutation $tauin S_{n-1}$ of order 2. For example $tau =(17)(38)(26)in S_8$. (cycle notation).
Define for $sigmain A_n$, $phi(sigma) =id$, and $phi(sigma)=tau$ for $sigmain S_n-A_n$. This is a nontrivial homomorphism with image a subgroup of order $2$. This shows that in fact there are as many homomorphims as there are elemnts of order 2 in $S_{n-1}$. The same argument works by replacing the codomain $S_{n-1}$ by any group of even order (they always have elements of order 2).
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
add a comment |
$begingroup$
As you are aware that $A_n$ is a normal subgroup how about trying for a homomorphism that will have $A_n$ as its kernel.
Take any permutation $tauin S_{n-1}$ of order 2. For example $tau =(17)(38)(26)in S_8$. (cycle notation).
Define for $sigmain A_n$, $phi(sigma) =id$, and $phi(sigma)=tau$ for $sigmain S_n-A_n$. This is a nontrivial homomorphism with image a subgroup of order $2$. This shows that in fact there are as many homomorphims as there are elemnts of order 2 in $S_{n-1}$. The same argument works by replacing the codomain $S_{n-1}$ by any group of even order (they always have elements of order 2).
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
As you are aware that $A_n$ is a normal subgroup how about trying for a homomorphism that will have $A_n$ as its kernel.
Take any permutation $tauin S_{n-1}$ of order 2. For example $tau =(17)(38)(26)in S_8$. (cycle notation).
Define for $sigmain A_n$, $phi(sigma) =id$, and $phi(sigma)=tau$ for $sigmain S_n-A_n$. This is a nontrivial homomorphism with image a subgroup of order $2$. This shows that in fact there are as many homomorphims as there are elemnts of order 2 in $S_{n-1}$. The same argument works by replacing the codomain $S_{n-1}$ by any group of even order (they always have elements of order 2).
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
answered Dec 29 '18 at 1:50
P VanchinathanP Vanchinathan
15.5k12136
15.5k12136
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
add a comment |
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
$begingroup$
Good approach. The way the question is phrased with the restriction on $n$ makes me think something is missing.
$endgroup$
– Randall
Dec 29 '18 at 1:57
add a comment |
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7
$begingroup$
The trivial homomorphism?
$endgroup$
– Randall
Dec 29 '18 at 1:35