Mixing
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
asked Jul 31 '14 at 3:56
user20391user20391
213
$endgroup$
add a comment |
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
asked Jul 31 '14 at 3:56
user20391user20391
213
$endgroup$
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
add a comment |
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
asked Jul 31 '14 at 3:56
user20391user20391
213
$endgroup$
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
asked Jul 31 '14 at 3:56
user20391user20391
213
asked Jul 31 '14 at 3:56
user20391user20391
213
asked Jul 31 '14 at 3:56
user20391user20391
213
asked Jul 31 '14 at 3:56
user20391user20391
213
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
add a comment |
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
add a comment |
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 $sigma$.
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
$endgroup$
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
|
show 4 more comments
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1 Answer
1
active
oldest
votes
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 $sigma$.
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
$endgroup$
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
|
show 4 more comments
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 $sigma$.
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
$endgroup$
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
|
show 4 more comments
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 $sigma$.
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
$endgroup$
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 $sigma$.
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
answered Jul 31 '14 at 4:09
Kaj HansenKaj Hansen
27.2k43779
27.2k43779
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
|
show 4 more comments
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
|
show 4 more comments
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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