Let $phi: S_n to G$ be a homomorphism, where $|G|$ is odd. Prove that $phi$ must be trivial.
$begingroup$
The Problem:
Let $n$ be a positive integer, let $G$ be a finite group with an odd number of elements, and let $phi: S_n to G$ be a homomorphism. Prove that $phi$ must be trivial, i.e., $phi(sigma) = e$ for all $sigma in S_n$.
My Approach:
I think I can go about it like this, but I keep getting stuck: We claim that, since $|G|$ is odd and $|S_n| = n!$ is even, it follows that $(|G|, |S_n|) = 1$; and so any homomorphism between them must be trivial (i.e., $|phi(S_n)| = 1$). So, we'll prove this more general statement, which follows from the First Isomorphism Theorem (and Lagrange's Theorem). Indeed, we have that
$$ S_n/kerphi cong phi(S_n) implies |S_n|/|kerphi| = |phi(S_n)|. $$
Now, since $phi(S_n) < G$, $|phi(S_n)|$ must divide $|G|$; and so $|phi(G)|$ cannot be even. But, since $|phi(S_n)| = |S_n / kerphi|$, this means that $|S_n / kerphi|$ cannot be even. Note that, since $ker phi < S_n$, $|kerphi| = k$ must divide $|S_n| = n(n-1)cdots2cdot1$; and so $k in {n, n-1, ..., 2, 1}$. Moreover, since $|S_n / kerphi|$ must be odd, $k$ must be even... and here's where I'm stuck...
abstract-algebra group-theory symmetric-groups
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
add a comment |
$begingroup$
The Problem:
Let $n$ be a positive integer, let $G$ be a finite group with an odd number of elements, and let $phi: S_n to G$ be a homomorphism. Prove that $phi$ must be trivial, i.e., $phi(sigma) = e$ for all $sigma in S_n$.
My Approach:
I think I can go about it like this, but I keep getting stuck: We claim that, since $|G|$ is odd and $|S_n| = n!$ is even, it follows that $(|G|, |S_n|) = 1$; and so any homomorphism between them must be trivial (i.e., $|phi(S_n)| = 1$). So, we'll prove this more general statement, which follows from the First Isomorphism Theorem (and Lagrange's Theorem). Indeed, we have that
$$ S_n/kerphi cong phi(S_n) implies |S_n|/|kerphi| = |phi(S_n)|. $$
Now, since $phi(S_n) < G$, $|phi(S_n)|$ must divide $|G|$; and so $|phi(G)|$ cannot be even. But, since $|phi(S_n)| = |S_n / kerphi|$, this means that $|S_n / kerphi|$ cannot be even. Note that, since $ker phi < S_n$, $|kerphi| = k$ must divide $|S_n| = n(n-1)cdots2cdot1$; and so $k in {n, n-1, ..., 2, 1}$. Moreover, since $|S_n / kerphi|$ must be odd, $k$ must be even... and here's where I'm stuck...
abstract-algebra group-theory symmetric-groups
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
4
$begingroup$
What is the image of a transposition?
$endgroup$
– Mindlack
Jan 16 at 16:34
1
$begingroup$
Note that $(|G|, |S_n|) = 1$ is false, in general, since you could simply take $n=|G|$.
$endgroup$
– Arnaud D.
Jan 16 at 16:35
add a comment |
$begingroup$
The Problem:
Let $n$ be a positive integer, let $G$ be a finite group with an odd number of elements, and let $phi: S_n to G$ be a homomorphism. Prove that $phi$ must be trivial, i.e., $phi(sigma) = e$ for all $sigma in S_n$.
My Approach:
I think I can go about it like this, but I keep getting stuck: We claim that, since $|G|$ is odd and $|S_n| = n!$ is even, it follows that $(|G|, |S_n|) = 1$; and so any homomorphism between them must be trivial (i.e., $|phi(S_n)| = 1$). So, we'll prove this more general statement, which follows from the First Isomorphism Theorem (and Lagrange's Theorem). Indeed, we have that
$$ S_n/kerphi cong phi(S_n) implies |S_n|/|kerphi| = |phi(S_n)|. $$
Now, since $phi(S_n) < G$, $|phi(S_n)|$ must divide $|G|$; and so $|phi(G)|$ cannot be even. But, since $|phi(S_n)| = |S_n / kerphi|$, this means that $|S_n / kerphi|$ cannot be even. Note that, since $ker phi < S_n$, $|kerphi| = k$ must divide $|S_n| = n(n-1)cdots2cdot1$; and so $k in {n, n-1, ..., 2, 1}$. Moreover, since $|S_n / kerphi|$ must be odd, $k$ must be even... and here's where I'm stuck...
abstract-algebra group-theory symmetric-groups
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
$endgroup$
The Problem:
Let $n$ be a positive integer, let $G$ be a finite group with an odd number of elements, and let $phi: S_n to G$ be a homomorphism. Prove that $phi$ must be trivial, i.e., $phi(sigma) = e$ for all $sigma in S_n$.
My Approach:
I think I can go about it like this, but I keep getting stuck: We claim that, since $|G|$ is odd and $|S_n| = n!$ is even, it follows that $(|G|, |S_n|) = 1$; and so any homomorphism between them must be trivial (i.e., $|phi(S_n)| = 1$). So, we'll prove this more general statement, which follows from the First Isomorphism Theorem (and Lagrange's Theorem). Indeed, we have that
$$ S_n/kerphi cong phi(S_n) implies |S_n|/|kerphi| = |phi(S_n)|. $$
Now, since $phi(S_n) < G$, $|phi(S_n)|$ must divide $|G|$; and so $|phi(G)|$ cannot be even. But, since $|phi(S_n)| = |S_n / kerphi|$, this means that $|S_n / kerphi|$ cannot be even. Note that, since $ker phi < S_n$, $|kerphi| = k$ must divide $|S_n| = n(n-1)cdots2cdot1$; and so $k in {n, n-1, ..., 2, 1}$. Moreover, since $|S_n / kerphi|$ must be odd, $k$ must be even... and here's where I'm stuck...
abstract-algebra group-theory symmetric-groups
abstract-algebra group-theory symmetric-groups
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
asked Jan 16 at 16:31
thisisourconcerndudethisisourconcerndude
1,1271022
1,1271022
4
$begingroup$
What is the image of a transposition?
$endgroup$
– Mindlack
Jan 16 at 16:34
1
$begingroup$
Note that $(|G|, |S_n|) = 1$ is false, in general, since you could simply take $n=|G|$.
$endgroup$
– Arnaud D.
Jan 16 at 16:35
add a comment |
4
$begingroup$
What is the image of a transposition?
$endgroup$
– Mindlack
Jan 16 at 16:34
1
$begingroup$
Note that $(|G|, |S_n|) = 1$ is false, in general, since you could simply take $n=|G|$.
$endgroup$
– Arnaud D.
Jan 16 at 16:35
4
4
$begingroup$
What is the image of a transposition?
$endgroup$
– Mindlack
Jan 16 at 16:34
$begingroup$
What is the image of a transposition?
$endgroup$
– Mindlack
Jan 16 at 16:34
1
1
$begingroup$
Note that $(|G|, |S_n|) = 1$ is false, in general, since you could simply take $n=|G|$.
$endgroup$
– Arnaud D.
Jan 16 at 16:35
$begingroup$
Note that $(|G|, |S_n|) = 1$ is false, in general, since you could simply take $n=|G|$.
$endgroup$
– Arnaud D.
Jan 16 at 16:35
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The order of the image of a transposition is $1$ or $2$, it cannot be $2$ since Lagrange implies $2$ divides the order of $G$ so the image of every transposition is trivial. Since $S_n$ is generated by transpositions, the morphism is trivial.
edited Jan 16 at 17:03
Shaun
9,083113683
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
add a comment |
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$begingroup$
The order of the image of a transposition is $1$ or $2$, it cannot be $2$ since Lagrange implies $2$ divides the order of $G$ so the image of every transposition is trivial. Since $S_n$ is generated by transpositions, the morphism is trivial.
edited Jan 16 at 17:03
Shaun
9,083113683
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
add a comment |
$begingroup$
The order of the image of a transposition is $1$ or $2$, it cannot be $2$ since Lagrange implies $2$ divides the order of $G$ so the image of every transposition is trivial. Since $S_n$ is generated by transpositions, the morphism is trivial.
edited Jan 16 at 17:03
Shaun
9,083113683
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
add a comment |
$begingroup$
The order of the image of a transposition is $1$ or $2$, it cannot be $2$ since Lagrange implies $2$ divides the order of $G$ so the image of every transposition is trivial. Since $S_n$ is generated by transpositions, the morphism is trivial.
edited Jan 16 at 17:03
Shaun
9,083113683
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
$endgroup$
The order of the image of a transposition is $1$ or $2$, it cannot be $2$ since Lagrange implies $2$ divides the order of $G$ so the image of every transposition is trivial. Since $S_n$ is generated by transpositions, the morphism is trivial.
edited Jan 16 at 17:03
Shaun
9,083113683
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
edited Jan 16 at 17:03
Shaun
9,083113683
edited Jan 16 at 17:03
Shaun
9,083113683
edited Jan 16 at 17:03
Shaun
9,083113683
9,083113683
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
answered Jan 16 at 16:42
Tsemo AristideTsemo Aristide
57.9k11445
57.9k11445
add a comment |
add a comment |
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$begingroup$
What is the image of a transposition?
$endgroup$
– Mindlack
Jan 16 at 16:34
1
$begingroup$
Note that $(|G|, |S_n|) = 1$ is false, in general, since you could simply take $n=|G|$.
$endgroup$
– Arnaud D.
Jan 16 at 16:35