Show that $tau_1tau_2$ of order 2 or 3, given that $tau_1$ and $tau_2$ are distinct transpositions.
Show that $tau_1tau_2$ of order $2$ or $3$, given that $tau_1$ and $tau_2$ are distinct transpositions.
I know that every cycle can be factored as a product of transpositions. But I'm not sure how to prove the order of $2$ or $3$.
abstract-algebra group-theory
edited Nov 26 at 16:35
Shaun
8,507113580
asked Dec 15 '11 at 19:14
user12691
2913
add a comment |
Show that $tau_1tau_2$ of order $2$ or $3$, given that $tau_1$ and $tau_2$ are distinct transpositions.
I know that every cycle can be factored as a product of transpositions. But I'm not sure how to prove the order of $2$ or $3$.
abstract-algebra group-theory
edited Nov 26 at 16:35
Shaun
8,507113580
asked Dec 15 '11 at 19:14
user12691
2913
add a comment |
Show that $tau_1tau_2$ of order $2$ or $3$, given that $tau_1$ and $tau_2$ are distinct transpositions.
I know that every cycle can be factored as a product of transpositions. But I'm not sure how to prove the order of $2$ or $3$.
abstract-algebra group-theory
edited Nov 26 at 16:35
Shaun
8,507113580
asked Dec 15 '11 at 19:14
user12691
2913
Show that $tau_1tau_2$ of order $2$ or $3$, given that $tau_1$ and $tau_2$ are distinct transpositions.
I know that every cycle can be factored as a product of transpositions. But I'm not sure how to prove the order of $2$ or $3$.
abstract-algebra group-theory
abstract-algebra group-theory
edited Nov 26 at 16:35
Shaun
8,507113580
asked Dec 15 '11 at 19:14
user12691
2913
edited Nov 26 at 16:35
Shaun
8,507113580
asked Dec 15 '11 at 19:14
user12691
2913
edited Nov 26 at 16:35
Shaun
8,507113580
edited Nov 26 at 16:35
Shaun
8,507113580
edited Nov 26 at 16:35
Shaun
8,507113580
8,507113580
asked Dec 15 '11 at 19:14
user12691
2913
asked Dec 15 '11 at 19:14
user12691
2913
asked Dec 15 '11 at 19:14
user12691
2913
2913
add a comment |
add a comment |
1 Answer
1
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Write $tau_1 = (a,b)$, $tau_2=(r,s)$.
We have two cases:
${a,b}cap{r,s} = emptyset$. What happens to the product then?
${a,b}cap{r,s}neqemptyset$. Then the two share exactly one element (Why?) so we can write $tau_1 = (a,b)$, $tau_2=(a,s)$. What is the product then?
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
|
show 2 more comments
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Write $tau_1 = (a,b)$, $tau_2=(r,s)$.
We have two cases:
${a,b}cap{r,s} = emptyset$. What happens to the product then?
${a,b}cap{r,s}neqemptyset$. Then the two share exactly one element (Why?) so we can write $tau_1 = (a,b)$, $tau_2=(a,s)$. What is the product then?
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
|
show 2 more comments
Write $tau_1 = (a,b)$, $tau_2=(r,s)$.
We have two cases:
${a,b}cap{r,s} = emptyset$. What happens to the product then?
${a,b}cap{r,s}neqemptyset$. Then the two share exactly one element (Why?) so we can write $tau_1 = (a,b)$, $tau_2=(a,s)$. What is the product then?
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
|
show 2 more comments
Write $tau_1 = (a,b)$, $tau_2=(r,s)$.
We have two cases:
${a,b}cap{r,s} = emptyset$. What happens to the product then?
${a,b}cap{r,s}neqemptyset$. Then the two share exactly one element (Why?) so we can write $tau_1 = (a,b)$, $tau_2=(a,s)$. What is the product then?
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
Write $tau_1 = (a,b)$, $tau_2=(r,s)$.
We have two cases:
${a,b}cap{r,s} = emptyset$. What happens to the product then?
${a,b}cap{r,s}neqemptyset$. Then the two share exactly one element (Why?) so we can write $tau_1 = (a,b)$, $tau_2=(a,s)$. What is the product then?
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
answered Dec 15 '11 at 19:17
Arturo Magidin
260k32584904
260k32584904
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
|
show 2 more comments
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
For Case 1, would the product just be (a b)(r s) because nothing is in common? And for Case 2, the product is (a b)(a s)(r) ?
– user12691
Dec 15 '11 at 19:27
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
@user12691: Yes; and what is the order of that?
– Arturo Magidin
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
The order of case one is 2 and the order of case two is 3.
– user12691
Dec 15 '11 at 19:28
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
And since those are the only two possible cases, that gives the answer you want.
– Arturo Magidin
Dec 15 '11 at 19:29
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
Oh okay. That was easier than I thought it would be. Thanks!
– user12691
Dec 15 '11 at 19:32
|
show 2 more comments
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