Find a subgroup of $S_4$ that is isomorphic to V, the Klein group.












2












$begingroup$


So I know that the Klein group is the group with 4 elements that is not cyclic but I'm stuck from there onwards?






group-theory finite-groups







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    ${1,(12)(34),(14)(23),(13)(24)}$.
    $endgroup$
    – Pedro Tamaroff♦
    Jul 31 '14 at 3:59






  • 1




    $begingroup$
    The Klein group acts by left multiplication on itself.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:26
















2












$begingroup$


So I know that the Klein group is the group with 4 elements that is not cyclic but I'm stuck from there onwards?






group-theory finite-groups







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    ${1,(12)(34),(14)(23),(13)(24)}$.
    $endgroup$
    – Pedro Tamaroff♦
    Jul 31 '14 at 3:59






  • 1




    $begingroup$
    The Klein group acts by left multiplication on itself.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:26














2












2








2





$begingroup$


So I know that the Klein group is the group with 4 elements that is not cyclic but I'm stuck from there onwards?






group-theory finite-groups







share|cite|improve this question









$endgroup$




So I know that the Klein group is the group with 4 elements that is not cyclic but I'm stuck from there onwards?






group-theory finite-groups




group-theory finite-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 31 '14 at 3:56









user20391user20391

213




213








  • 6




    $begingroup$
    ${1,(12)(34),(14)(23),(13)(24)}$.
    $endgroup$
    – Pedro Tamaroff♦
    Jul 31 '14 at 3:59






  • 1




    $begingroup$
    The Klein group acts by left multiplication on itself.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:26














  • 6




    $begingroup$
    ${1,(12)(34),(14)(23),(13)(24)}$.
    $endgroup$
    – Pedro Tamaroff♦
    Jul 31 '14 at 3:59






  • 1




    $begingroup$
    The Klein group acts by left multiplication on itself.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:26








6




6




$begingroup$
${1,(12)(34),(14)(23),(13)(24)}$.
$endgroup$
– Pedro Tamaroff♦
Jul 31 '14 at 3:59




$begingroup$
${1,(12)(34),(14)(23),(13)(24)}$.
$endgroup$
– Pedro Tamaroff♦
Jul 31 '14 at 3:59




1




1




$begingroup$
The Klein group acts by left multiplication on itself.
$endgroup$
– Qiaochu Yuan
Jul 31 '14 at 4:26




$begingroup$
The Klein group acts by left multiplication on itself.
$endgroup$
– Qiaochu Yuan
Jul 31 '14 at 4:26










1 Answer
1






active

oldest

votes


















4












$begingroup$

Here's one way of going about this:



Every non-identity element in $V_4$ is of order $2$. Therefore, a good starting point would be to choose your favorite transposition in $S_4$. Call it $pi$. Next, you could choose another transposition and call it $sigma$. However, we want to ensure that the composition of $pi$ and $sigma$ also has order $2$. This will only happen if $pi$ and $sigma$ are disjoint transpositions (why?).



Finish up by considering the subgroup generated by $pi$ and $sigma$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    $V_4$? What is the subscript for? I thought the "V" stood for "4"?
    $endgroup$
    – bof
    Jul 31 '14 at 4:25






  • 1




    $begingroup$
    It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
    $endgroup$
    – user20391
    Jul 31 '14 at 4:56








  • 1




    $begingroup$
    I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 5:26














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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Here's one way of going about this:



Every non-identity element in $V_4$ is of order $2$. Therefore, a good starting point would be to choose your favorite transposition in $S_4$. Call it $pi$. Next, you could choose another transposition and call it $sigma$. However, we want to ensure that the composition of $pi$ and $sigma$ also has order $2$. This will only happen if $pi$ and $sigma$ are disjoint transpositions (why?).



Finish up by considering the subgroup generated by $pi$ and $sigma$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    $V_4$? What is the subscript for? I thought the "V" stood for "4"?
    $endgroup$
    – bof
    Jul 31 '14 at 4:25






  • 1




    $begingroup$
    It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
    $endgroup$
    – user20391
    Jul 31 '14 at 4:56








  • 1




    $begingroup$
    I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 5:26


















4












$begingroup$

Here's one way of going about this:



Every non-identity element in $V_4$ is of order $2$. Therefore, a good starting point would be to choose your favorite transposition in $S_4$. Call it $pi$. Next, you could choose another transposition and call it $sigma$. However, we want to ensure that the composition of $pi$ and $sigma$ also has order $2$. This will only happen if $pi$ and $sigma$ are disjoint transpositions (why?).



Finish up by considering the subgroup generated by $pi$ and $sigma$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    $V_4$? What is the subscript for? I thought the "V" stood for "4"?
    $endgroup$
    – bof
    Jul 31 '14 at 4:25






  • 1




    $begingroup$
    It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
    $endgroup$
    – user20391
    Jul 31 '14 at 4:56








  • 1




    $begingroup$
    I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 5:26
















4












4








4





$begingroup$

Here's one way of going about this:



Every non-identity element in $V_4$ is of order $2$. Therefore, a good starting point would be to choose your favorite transposition in $S_4$. Call it $pi$. Next, you could choose another transposition and call it $sigma$. However, we want to ensure that the composition of $pi$ and $sigma$ also has order $2$. This will only happen if $pi$ and $sigma$ are disjoint transpositions (why?).



Finish up by considering the subgroup generated by $pi$ and $sigma$.






share|cite|improve this answer









$endgroup$



Here's one way of going about this:



Every non-identity element in $V_4$ is of order $2$. Therefore, a good starting point would be to choose your favorite transposition in $S_4$. Call it $pi$. Next, you could choose another transposition and call it $sigma$. However, we want to ensure that the composition of $pi$ and $sigma$ also has order $2$. This will only happen if $pi$ and $sigma$ are disjoint transpositions (why?).



Finish up by considering the subgroup generated by $pi$ and $sigma$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jul 31 '14 at 4:09









Kaj HansenKaj Hansen

27.8k43980




27.8k43980








  • 2




    $begingroup$
    $V_4$? What is the subscript for? I thought the "V" stood for "4"?
    $endgroup$
    – bof
    Jul 31 '14 at 4:25






  • 1




    $begingroup$
    It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
    $endgroup$
    – user20391
    Jul 31 '14 at 4:56








  • 1




    $begingroup$
    I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 5:26
















  • 2




    $begingroup$
    $V_4$? What is the subscript for? I thought the "V" stood for "4"?
    $endgroup$
    – bof
    Jul 31 '14 at 4:25






  • 1




    $begingroup$
    It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
    $endgroup$
    – Qiaochu Yuan
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 4:27








  • 1




    $begingroup$
    Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
    $endgroup$
    – user20391
    Jul 31 '14 at 4:56








  • 1




    $begingroup$
    I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
    $endgroup$
    – Kaj Hansen
    Jul 31 '14 at 5:26










2




2




$begingroup$
$V_4$? What is the subscript for? I thought the "V" stood for "4"?
$endgroup$
– bof
Jul 31 '14 at 4:25




$begingroup$
$V_4$? What is the subscript for? I thought the "V" stood for "4"?
$endgroup$
– bof
Jul 31 '14 at 4:25




1




1




$begingroup$
It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
$endgroup$
– Qiaochu Yuan
Jul 31 '14 at 4:27






$begingroup$
It's the same $4$ as in $S_4$; in particular it makes the list of transitive subgroups of $S_4$ look more uniform. $C_4, V_4, D_4, A_4, S_4$.
$endgroup$
– Qiaochu Yuan
Jul 31 '14 at 4:27






1




1




$begingroup$
I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
$endgroup$
– Kaj Hansen
Jul 31 '14 at 4:27






$begingroup$
I think I've seen that notation for it before in textbooks. Personally, I prefer to just say $mathbb{Z}_2 times mathbb{Z}_2$ because it's much more descriptive. Edit: Qiaochu makes a reasonable point. In this context, adding the subscript would make sense.
$endgroup$
– Kaj Hansen
Jul 31 '14 at 4:27






1




1




$begingroup$
Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
$endgroup$
– user20391
Jul 31 '14 at 4:56






$begingroup$
Does this make sense? Consider the transpositions (1 3) and (2 4). As the subgroups are disjoint and normal to $S_4$, the product of the groups are commutative. Also they are of order two, therefore the product is isomorphic to $Z_2$ X $Z_2$ = V.
$endgroup$
– user20391
Jul 31 '14 at 4:56






1




1




$begingroup$
I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
$endgroup$
– Kaj Hansen
Jul 31 '14 at 5:26






$begingroup$
I'm not sure what you mean by normal. The argument you want to make is that $(1 2)$ and $(3 4)$ commute because they are disjoint. (All disjoint cycles commute). Can you explicitly list the four elements of our group?
$endgroup$
– Kaj Hansen
Jul 31 '14 at 5:26




















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