Let V= { (1),(12)(34) ,(13)(24),(14)(23) } subgroup . if H is subgroup such that |H|=6 , HV < A4 is absurde. Note that V is normal at A4. Calculating the HV norm, we arrive that is 12, which would imply that HV = A4, I need to find somebody of A4 that is not in HV, but I do not know much about arbitrary order H.
Asked
Active
Viewed 1,986 times
0
-
Didn't you fix $|H| = 6$? Also, please use MathJax to wrap expressions, makes them much clearer. – AspiringMathematician Apr 20 '17 at 03:08
1 Answers
1
Hint: $A_4$ has eight 3-cycles, so clearly there is one which is not in H. Can you work out a contradiction to existence of H by looking at its cosets now?
Maryam
- 316