Mixing
Cardinality of Subgroup of Permutations
Clash Royale CLAN TAG#URR8PPP
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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
edited Jul 28 at 16:02
amWhy
189k25219431
asked Jul 15 at 4:00
Matt.P
768313
add a comment |Â
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
edited Jul 28 at 16:02
amWhy
189k25219431
asked Jul 15 at 4:00
Matt.P
768313
add a comment |Â
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
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
edited Jul 28 at 16:02
amWhy
189k25219431
asked Jul 15 at 4:00
Matt.P
768313
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 |Â
add a comment |Â
1 Answer
1
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 $S_n-1$.
answered Jul 15 at 5:13
Julian Benali
27212
add a comment |Â
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 $S_n-1$.
answered Jul 15 at 5:13
Julian Benali
27212
add a comment |Â
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 $S_n-1$.
answered Jul 15 at 5:13
Julian Benali
27212
add a comment |Â
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 $S_n-1$.
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 $S_n-1$.
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
answered Jul 15 at 5:13
Julian Benali
27212
27212
add a comment |Â
add a comment |Â
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