find all $x in S_5$ such that $x^3 = (12)$
So I have to find all $x in S_5$ such that $x^3 = (12)$. For example, one solution would be $(12)$ itself, because its order is $2$. How can I find all of the solutions though? Is it just trial and error?
(also, the multiplication is defined from left to right, just fyi)
permutations symmetric-groups permutation-cycles
asked Dec 1 '18 at 13:30
mandella
717521
add a comment |
So I have to find all $x in S_5$ such that $x^3 = (12)$. For example, one solution would be $(12)$ itself, because its order is $2$. How can I find all of the solutions though? Is it just trial and error?
(also, the multiplication is defined from left to right, just fyi)
permutations symmetric-groups permutation-cycles
asked Dec 1 '18 at 13:30
mandella
717521
1
Well, $;x^3=(12)implies x^6=(x^3)^2=(12)^2=1implies o(x);$ is a divisor of $;6;$ , so you already eliminate lots of elements of $;S_5;$ ...
– DonAntonio
Dec 1 '18 at 13:35
ok, makes sense. still, I need to find the solutions.
– mandella
Dec 1 '18 at 13:36
add a comment |
So I have to find all $x in S_5$ such that $x^3 = (12)$. For example, one solution would be $(12)$ itself, because its order is $2$. How can I find all of the solutions though? Is it just trial and error?
(also, the multiplication is defined from left to right, just fyi)
permutations symmetric-groups permutation-cycles
asked Dec 1 '18 at 13:30
mandella
717521
So I have to find all $x in S_5$ such that $x^3 = (12)$. For example, one solution would be $(12)$ itself, because its order is $2$. How can I find all of the solutions though? Is it just trial and error?
(also, the multiplication is defined from left to right, just fyi)
permutations symmetric-groups permutation-cycles
permutations symmetric-groups permutation-cycles
asked Dec 1 '18 at 13:30
mandella
717521
asked Dec 1 '18 at 13:30
mandella
717521
asked Dec 1 '18 at 13:30
mandella
717521
asked Dec 1 '18 at 13:30
mandella
717521
asked Dec 1 '18 at 13:30
mandella
717521
717521
1
Well, $;x^3=(12)implies x^6=(x^3)^2=(12)^2=1implies o(x);$ is a divisor of $;6;$ , so you already eliminate lots of elements of $;S_5;$ ...
– DonAntonio
Dec 1 '18 at 13:35
ok, makes sense. still, I need to find the solutions.
– mandella
Dec 1 '18 at 13:36
add a comment |
1
Well, $;x^3=(12)implies x^6=(x^3)^2=(12)^2=1implies o(x);$ is a divisor of $;6;$ , so you already eliminate lots of elements of $;S_5;$ ...
– DonAntonio
Dec 1 '18 at 13:35
ok, makes sense. still, I need to find the solutions.
– mandella
Dec 1 '18 at 13:36
1
1
Well, $;x^3=(12)implies x^6=(x^3)^2=(12)^2=1implies o(x);$ is a divisor of $;6;$ , so you already eliminate lots of elements of $;S_5;$ ...
– DonAntonio
Dec 1 '18 at 13:35
Well, $;x^3=(12)implies x^6=(x^3)^2=(12)^2=1implies o(x);$ is a divisor of $;6;$ , so you already eliminate lots of elements of $;S_5;$ ...
– DonAntonio
Dec 1 '18 at 13:35
ok, makes sense. still, I need to find the solutions.
– mandella
Dec 1 '18 at 13:36
ok, makes sense. still, I need to find the solutions.
– mandella
Dec 1 '18 at 13:36
add a comment |
1 Answer
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$o(x)$ is a divisor of $6$ means $o(x)=1,2,3$ 0r $6$. $o(x)=1,3$ are automatically excluded. For $o(x)=2$, $x=(12)$ is the only solution. For $x=6$, $x=(12)(abc)$ is a solution, where $a,b,cin {3,4,5}$.
answered Dec 1 '18 at 13:50
Anupam
2,3721823
add a comment |
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$o(x)$ is a divisor of $6$ means $o(x)=1,2,3$ 0r $6$. $o(x)=1,3$ are automatically excluded. For $o(x)=2$, $x=(12)$ is the only solution. For $x=6$, $x=(12)(abc)$ is a solution, where $a,b,cin {3,4,5}$.
answered Dec 1 '18 at 13:50
Anupam
2,3721823
add a comment |
$o(x)$ is a divisor of $6$ means $o(x)=1,2,3$ 0r $6$. $o(x)=1,3$ are automatically excluded. For $o(x)=2$, $x=(12)$ is the only solution. For $x=6$, $x=(12)(abc)$ is a solution, where $a,b,cin {3,4,5}$.
answered Dec 1 '18 at 13:50
Anupam
2,3721823
add a comment |
$o(x)$ is a divisor of $6$ means $o(x)=1,2,3$ 0r $6$. $o(x)=1,3$ are automatically excluded. For $o(x)=2$, $x=(12)$ is the only solution. For $x=6$, $x=(12)(abc)$ is a solution, where $a,b,cin {3,4,5}$.
answered Dec 1 '18 at 13:50
Anupam
2,3721823
$o(x)$ is a divisor of $6$ means $o(x)=1,2,3$ 0r $6$. $o(x)=1,3$ are automatically excluded. For $o(x)=2$, $x=(12)$ is the only solution. For $x=6$, $x=(12)(abc)$ is a solution, where $a,b,cin {3,4,5}$.
answered Dec 1 '18 at 13:50
Anupam
2,3721823
answered Dec 1 '18 at 13:50
Anupam
2,3721823
answered Dec 1 '18 at 13:50
Anupam
2,3721823
answered Dec 1 '18 at 13:50
Anupam
2,3721823
2,3721823
add a comment |
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Well, $;x^3=(12)implies x^6=(x^3)^2=(12)^2=1implies o(x);$ is a divisor of $;6;$ , so you already eliminate lots of elements of $;S_5;$ ...
– DonAntonio
Dec 1 '18 at 13:35
ok, makes sense. still, I need to find the solutions.
– mandella
Dec 1 '18 at 13:36