Cardinality of Subgroup of Permutations

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Say we have the symmetric group, $S_n$, and we take a subgroup, $H$, specified by the rule that $sigma(5) = 5$. The question is: what is the cardinality of $H$?



After working out some examples, it seems logical that the answer is $(n-1)!$. I believe the argument is that we're fixing a single element, $5$, by mapping it to itself, but are permuting each of the remaining elements, i.e., we're permutting $n-1$ elements. Similarly, we could say that there are $n!$ total permutations in $S_n$, and we want to discard all those permutations that send $sigma(5)$ to something else. So, there are $n-1$ possibilities for $sigma(5)$ and $(n-2)!$ combinations for all of the remaining elements, as we've already fixed some value to send $5$ to, so the result is $(n-1)(n-2)! = (n-1)!$.



How does this logic sound?




group-theory symmetric-groups




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    Say we have the symmetric group, $S_n$, and we take a subgroup, $H$, specified by the rule that $sigma(5) = 5$. The question is: what is the cardinality of $H$?



    After working out some examples, it seems logical that the answer is $(n-1)!$. I believe the argument is that we're fixing a single element, $5$, by mapping it to itself, but are permuting each of the remaining elements, i.e., we're permutting $n-1$ elements. Similarly, we could say that there are $n!$ total permutations in $S_n$, and we want to discard all those permutations that send $sigma(5)$ to something else. So, there are $n-1$ possibilities for $sigma(5)$ and $(n-2)!$ combinations for all of the remaining elements, as we've already fixed some value to send $5$ to, so the result is $(n-1)(n-2)! = (n-1)!$.



    How does this logic sound?




    group-theory symmetric-groups




    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Say we have the symmetric group, $S_n$, and we take a subgroup, $H$, specified by the rule that $sigma(5) = 5$. The question is: what is the cardinality of $H$?



      After working out some examples, it seems logical that the answer is $(n-1)!$. I believe the argument is that we're fixing a single element, $5$, by mapping it to itself, but are permuting each of the remaining elements, i.e., we're permutting $n-1$ elements. Similarly, we could say that there are $n!$ total permutations in $S_n$, and we want to discard all those permutations that send $sigma(5)$ to something else. So, there are $n-1$ possibilities for $sigma(5)$ and $(n-2)!$ combinations for all of the remaining elements, as we've already fixed some value to send $5$ to, so the result is $(n-1)(n-2)! = (n-1)!$.



      How does this logic sound?




      group-theory symmetric-groups




      share|cite|improve this question













      Say we have the symmetric group, $S_n$, and we take a subgroup, $H$, specified by the rule that $sigma(5) = 5$. The question is: what is the cardinality of $H$?



      After working out some examples, it seems logical that the answer is $(n-1)!$. I believe the argument is that we're fixing a single element, $5$, by mapping it to itself, but are permuting each of the remaining elements, i.e., we're permutting $n-1$ elements. Similarly, we could say that there are $n!$ total permutations in $S_n$, and we want to discard all those permutations that send $sigma(5)$ to something else. So, there are $n-1$ possibilities for $sigma(5)$ and $(n-2)!$ combinations for all of the remaining elements, as we've already fixed some value to send $5$ to, so the result is $(n-1)(n-2)! = (n-1)!$.



      How does this logic sound?





      group-theory symmetric-groups





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      share|cite|improve this question




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      edited Jul 28 at 16:02









      amWhy

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      asked Jul 15 at 4:00









      Matt.P

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          This sounds exactly right. In fact, if for any given $x,yin1,...,n$ the subset $H=sigmain S_n:sigma(x)=y$ will have cardinality $(n-1)!$ using pretty much the same argument you mentioned above. In the case $x=y$, $H$ will be a subgroup of $S_n$ isomorphic to $S_n-1$.






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            up vote
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            accepted










            This sounds exactly right. In fact, if for any given $x,yin1,...,n$ the subset $H=sigmain S_n:sigma(x)=y$ will have cardinality $(n-1)!$ using pretty much the same argument you mentioned above. In the case $x=y$, $H$ will be a subgroup of $S_n$ isomorphic to $S_n-1$.






            share|cite|improve this answer

























              up vote
              4
              down vote



              accepted










              This sounds exactly right. In fact, if for any given $x,yin1,...,n$ the subset $H=sigmain S_n:sigma(x)=y$ will have cardinality $(n-1)!$ using pretty much the same argument you mentioned above. In the case $x=y$, $H$ will be a subgroup of $S_n$ isomorphic to $S_n-1$.






              share|cite|improve this answer























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                This sounds exactly right. In fact, if for any given $x,yin1,...,n$ the subset $H=sigmain S_n:sigma(x)=y$ will have cardinality $(n-1)!$ using pretty much the same argument you mentioned above. In the case $x=y$, $H$ will be a subgroup of $S_n$ isomorphic to $S_n-1$.






                share|cite|improve this answer













                This sounds exactly right. In fact, if for any given $x,yin1,...,n$ the subset $H=sigmain S_n:sigma(x)=y$ will have cardinality $(n-1)!$ using pretty much the same argument you mentioned above. In the case $x=y$, $H$ will be a subgroup of $S_n$ isomorphic to $S_n-1$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 15 at 5:13









                Julian Benali

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