Decide how many elements who commutate with this symmetric group?
Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$
Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.
And then I'm insecure. What shall I do next?
permutations
edited Nov 25 at 20:11
amWhy
191k28224439
asked Nov 25 at 19:58
soetirl13
114
add a comment |
Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$
Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.
And then I'm insecure. What shall I do next?
permutations
edited Nov 25 at 20:11
amWhy
191k28224439
asked Nov 25 at 19:58
soetirl13
114
add a comment |
Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$
Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.
And then I'm insecure. What shall I do next?
permutations
edited Nov 25 at 20:11
amWhy
191k28224439
asked Nov 25 at 19:58
soetirl13
114
Let $S_3$ be the symmetric group on $lbrace 1,2,3rbrace.$ Decide how many elements who commutate with $(23)$
Permutation naturally commutes itself, with it's inverse and with the identity permutation. So that's 3.
And then I'm insecure. What shall I do next?
permutations
permutations
edited Nov 25 at 20:11
amWhy
191k28224439
asked Nov 25 at 19:58
soetirl13
114
edited Nov 25 at 20:11
amWhy
191k28224439
asked Nov 25 at 19:58
soetirl13
114
edited Nov 25 at 20:11
amWhy
191k28224439
edited Nov 25 at 20:11
amWhy
191k28224439
edited Nov 25 at 20:11
amWhy
191k28224439
191k28224439
asked Nov 25 at 19:58
soetirl13
114
asked Nov 25 at 19:58
soetirl13
114
asked Nov 25 at 19:58
soetirl13
114
114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.
answered Nov 25 at 20:04
Mark
5,955415
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
add a comment |
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1 Answer
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First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.
answered Nov 25 at 20:04
Mark
5,955415
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
add a comment |
First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.
answered Nov 25 at 20:04
Mark
5,955415
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
add a comment |
First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.
answered Nov 25 at 20:04
Mark
5,955415
First of all, the permutation $(23)$ is the inverse of itself so you actually found only $2$ permutations that commute with it so far, not $3$. Next, $S_3$ is a very small group, it has only $6$ elements. So you can just check every element in a direct way.
answered Nov 25 at 20:04
Mark
5,955415
answered Nov 25 at 20:04
Mark
5,955415
answered Nov 25 at 20:04
Mark
5,955415
answered Nov 25 at 20:04
Mark
5,955415
5,955415
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
add a comment |
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
How is it a inverse of itself? the inverse is (32) ?
– soetirl13
Nov 25 at 20:08
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
$(23)$ and $(32)$ is the same permutation. It maps $2$ to $3$, $3$ to $2$, and $1$ to itself. It's just two different ways to write the same permutation.
– Mark
Nov 25 at 20:09
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
Cool. Ok, so far we have 2. And then I look at {1,2,3} and therefore get, (12)(21)(23) that is 3.. So 5?
– soetirl13
Nov 25 at 20:13
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
It depends what is your knowledge in group theory. The set of permutations that commute with $(23)$ is actually a subgroup of $S_3$, so its number of elements must divide the order of $S_3$ by Lagrange's theorem. So it can't be $5$ elements. Just check all the permutations and find what the answer will be.
– Mark
Nov 25 at 20:23
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
Ahh 2. :))))) thank you
– soetirl13
Nov 25 at 21:06
add a comment |
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