Let $phi: S_n to G$ be a homomorphism, where $|G|$ is odd. Prove that $phi$ must be trivial.












1












$begingroup$


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...










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$endgroup$








  • 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
















1












$begingroup$


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...










share|cite|improve this question









$endgroup$








  • 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














1












1








1


1



$begingroup$


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...










share|cite|improve this question









$endgroup$




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






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asked Jan 16 at 16:31









thisisourconcerndudethisisourconcerndude

1,1271123




1,1271123








  • 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




    $begingroup$
    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










1 Answer
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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.






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    4












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 at 17:03









        Shaun

        10.3k113686




        10.3k113686










        answered Jan 16 at 16:42









        Tsemo AristideTsemo Aristide

        60.2k11446




        60.2k11446






























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