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发帖数: 115 | 1 Is it a problem of a take-home test? If yes, stop here.
If |G|=72=2^3*3^2, consider its sylow 3 subgroups.
G has either 1 or 4 sylow 3 subgroups.
If 1, this sylow subgroup is normal.
If 4, consider the map f:G->S(4), that maps each g in
G to a permutation of the set of this sylow subgroups
by conjugation. Since |S(4)|=24, and f is not constant,
3<=ker(f)<24, which is a notrivial normal subgroup. |
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