Mixing
Explicit exceptional Embedding of $S_5$ in $S_6$
We know $S_5$ acting on its Sylow 5-groups gives an embedding $S_5 rightarrow S_6$. Now $S_6$ acts on ${(x_i) in V=mathbb C^6| sum_{i=1}^{6}x_i=0}$ by permuting the coordinates. $V$ has a basis $v_i=e_{i}-e_{i+1}$ $i=1,dots,5$. Restricted on $S_5$, this is one $5$ dimensional irreducible representation of $S_5$.
My question is, how to write the embedding $S_5$ explicitly with respect to $v_i$ ? In particular, for different $v_i, v_j$ could we always find $sigma in S_5$ such that $sigma(v_i)=v_j$?
group-theory representation-theory
asked Nov 20 '18 at 9:26
zzy
2,3621419
add a comment |
We know $S_5$ acting on its Sylow 5-groups gives an embedding $S_5 rightarrow S_6$. Now $S_6$ acts on ${(x_i) in V=mathbb C^6| sum_{i=1}^{6}x_i=0}$ by permuting the coordinates. $V$ has a basis $v_i=e_{i}-e_{i+1}$ $i=1,dots,5$. Restricted on $S_5$, this is one $5$ dimensional irreducible representation of $S_5$.
My question is, how to write the embedding $S_5$ explicitly with respect to $v_i$ ? In particular, for different $v_i, v_j$ could we always find $sigma in S_5$ such that $sigma(v_i)=v_j$?
group-theory representation-theory
asked Nov 20 '18 at 9:26
zzy
2,3621419
$S_6$ has $36$ Sylow $5$-subgroups, so I do not understand your first sentence.
– Derek Holt
Nov 20 '18 at 10:17
@DerekHolt Do you know the exceptional embedding from $S_5$ to $S_6$?
– zzy
Nov 20 '18 at 16:00
You could deduce it from the isomorphism $S_5 cong {rm PGL}(2,5)$. Then, since ${rm PGL}(2,5)$ acts $2$-transitively (in fact $3$-transitively) on the six points, the restriction of the representation you describe is irreducible. To get it explicitly, I guess you would need a specific realization of ${rm PGL}(2,5)$, but I don't believe that there is a single natural copy of this kind. For example, in GAP it is the group $langle (3,6,5,4), (1,2,5)(3,4,6) rangle$.
– Derek Holt
Nov 20 '18 at 16:37
add a comment |
We know $S_5$ acting on its Sylow 5-groups gives an embedding $S_5 rightarrow S_6$. Now $S_6$ acts on ${(x_i) in V=mathbb C^6| sum_{i=1}^{6}x_i=0}$ by permuting the coordinates. $V$ has a basis $v_i=e_{i}-e_{i+1}$ $i=1,dots,5$. Restricted on $S_5$, this is one $5$ dimensional irreducible representation of $S_5$.
My question is, how to write the embedding $S_5$ explicitly with respect to $v_i$ ? In particular, for different $v_i, v_j$ could we always find $sigma in S_5$ such that $sigma(v_i)=v_j$?
group-theory representation-theory
asked Nov 20 '18 at 9:26
zzy
2,3621419
We know $S_5$ acting on its Sylow 5-groups gives an embedding $S_5 rightarrow S_6$. Now $S_6$ acts on ${(x_i) in V=mathbb C^6| sum_{i=1}^{6}x_i=0}$ by permuting the coordinates. $V$ has a basis $v_i=e_{i}-e_{i+1}$ $i=1,dots,5$. Restricted on $S_5$, this is one $5$ dimensional irreducible representation of $S_5$.
My question is, how to write the embedding $S_5$ explicitly with respect to $v_i$ ? In particular, for different $v_i, v_j$ could we always find $sigma in S_5$ such that $sigma(v_i)=v_j$?
group-theory representation-theory
group-theory representation-theory
asked Nov 20 '18 at 9:26
zzy
2,3621419
asked Nov 20 '18 at 9:26
zzy
2,3621419
asked Nov 20 '18 at 9:26
zzy
2,3621419
asked Nov 20 '18 at 9:26
zzy
2,3621419
asked Nov 20 '18 at 9:26
zzy
2,3621419
2,3621419
$S_6$ has $36$ Sylow $5$-subgroups, so I do not understand your first sentence.
– Derek Holt
Nov 20 '18 at 10:17
@DerekHolt Do you know the exceptional embedding from $S_5$ to $S_6$?
– zzy
Nov 20 '18 at 16:00
You could deduce it from the isomorphism $S_5 cong {rm PGL}(2,5)$. Then, since ${rm PGL}(2,5)$ acts $2$-transitively (in fact $3$-transitively) on the six points, the restriction of the representation you describe is irreducible. To get it explicitly, I guess you would need a specific realization of ${rm PGL}(2,5)$, but I don't believe that there is a single natural copy of this kind. For example, in GAP it is the group $langle (3,6,5,4), (1,2,5)(3,4,6) rangle$.
– Derek Holt
Nov 20 '18 at 16:37
add a comment |
$S_6$ has $36$ Sylow $5$-subgroups, so I do not understand your first sentence.
– Derek Holt
Nov 20 '18 at 10:17
@DerekHolt Do you know the exceptional embedding from $S_5$ to $S_6$?
– zzy
Nov 20 '18 at 16:00
You could deduce it from the isomorphism $S_5 cong {rm PGL}(2,5)$. Then, since ${rm PGL}(2,5)$ acts $2$-transitively (in fact $3$-transitively) on the six points, the restriction of the representation you describe is irreducible. To get it explicitly, I guess you would need a specific realization of ${rm PGL}(2,5)$, but I don't believe that there is a single natural copy of this kind. For example, in GAP it is the group $langle (3,6,5,4), (1,2,5)(3,4,6) rangle$.
– Derek Holt
Nov 20 '18 at 16:37
$S_6$ has $36$ Sylow $5$-subgroups, so I do not understand your first sentence.
– Derek Holt
Nov 20 '18 at 10:17
$S_6$ has $36$ Sylow $5$-subgroups, so I do not understand your first sentence.
– Derek Holt
Nov 20 '18 at 10:17
@DerekHolt Do you know the exceptional embedding from $S_5$ to $S_6$?
– zzy
Nov 20 '18 at 16:00
@DerekHolt Do you know the exceptional embedding from $S_5$ to $S_6$?
– zzy
Nov 20 '18 at 16:00
You could deduce it from the isomorphism $S_5 cong {rm PGL}(2,5)$. Then, since ${rm PGL}(2,5)$ acts $2$-transitively (in fact $3$-transitively) on the six points, the restriction of the representation you describe is irreducible. To get it explicitly, I guess you would need a specific realization of ${rm PGL}(2,5)$, but I don't believe that there is a single natural copy of this kind. For example, in GAP it is the group $langle (3,6,5,4), (1,2,5)(3,4,6) rangle$.
– Derek Holt
Nov 20 '18 at 16:37
You could deduce it from the isomorphism $S_5 cong {rm PGL}(2,5)$. Then, since ${rm PGL}(2,5)$ acts $2$-transitively (in fact $3$-transitively) on the six points, the restriction of the representation you describe is irreducible. To get it explicitly, I guess you would need a specific realization of ${rm PGL}(2,5)$, but I don't believe that there is a single natural copy of this kind. For example, in GAP it is the group $langle (3,6,5,4), (1,2,5)(3,4,6) rangle$.
– Derek Holt
Nov 20 '18 at 16:37
add a comment |
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$S_6$ has $36$ Sylow $5$-subgroups, so I do not understand your first sentence.
– Derek Holt
Nov 20 '18 at 10:17
@DerekHolt Do you know the exceptional embedding from $S_5$ to $S_6$?
– zzy
Nov 20 '18 at 16:00
You could deduce it from the isomorphism $S_5 cong {rm PGL}(2,5)$. Then, since ${rm PGL}(2,5)$ acts $2$-transitively (in fact $3$-transitively) on the six points, the restriction of the representation you describe is irreducible. To get it explicitly, I guess you would need a specific realization of ${rm PGL}(2,5)$, but I don't believe that there is a single natural copy of this kind. For example, in GAP it is the group $langle (3,6,5,4), (1,2,5)(3,4,6) rangle$.
– Derek Holt
Nov 20 '18 at 16:37