Transitive Subgroups of
S
5
T
Name(s)
|
G
|
Parity
|
C
S
5
(
G
)|
Other Representations
Resolvents
1
C(5) = 5
5
1
5
2
D(5) = 5:2
10 = 2 · 5
1
1
10T2
2:
2T1
3
F(5) = 5:4
20 = 2
2
· 5
-1
1
10T4
, 20T5
2:
2T1
4:
4T1
4
A5
60 = 2
2
· 3 · 5
1
1
6T12
,
10T7
,
12T33
, 15T5, 20T15
5
S5
120 = 2
3
· 3 · 5
-1
1
6T14
,
10T12
,
10T13
,
12T74
, 15T10, 20T30, 20T32, 20T35
2:
2T1