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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edited Jul 28 at 16:02

amWhy

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asked Jul 15 at 4:00

Matt.P

768313

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up vote
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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

edited Jul 28 at 16:02

amWhy

189k25219431

asked Jul 15 at 4:00

Matt.P

768313

add a comment | 

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

edited Jul 28 at 16:02

amWhy

189k25219431

asked Jul 15 at 4:00

Matt.P

768313

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

edited Jul 28 at 16:02

amWhy

189k25219431

asked Jul 15 at 4:00

Matt.P

768313

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 16:02

amWhy

189k25219431

edited Jul 28 at 16:02

amWhy

189k25219431

edited Jul 28 at 16:02

amWhy

189k25219431

189k25219431

asked Jul 15 at 4:00

Matt.P

768313

asked Jul 15 at 4:00

Matt.P

768313

asked Jul 15 at 4:00

Matt.P

768313

768313

add a comment | 

add a comment | 

1 Answer
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4
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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

answered Jul 15 at 5:13

Julian Benali

27212

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

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

answered Jul 15 at 5:13

Julian Benali

27212

add a comment | 

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

answered Jul 15 at 5:13

Julian Benali

27212

add a comment | 

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

answered Jul 15 at 5:13

Julian Benali

27212

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

answered Jul 15 at 5:13

Julian Benali

27212

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 5:13

Julian Benali

27212

answered Jul 15 at 5:13

Julian Benali

27212

answered Jul 15 at 5:13

Julian Benali

27212

27212

add a comment | 

add a comment | 

 

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