oh ok, so what I said in my previous post means that a normal subgroup of order 5 exists but not a normal subgroup of order 4. However, both a subgroup of order 4 and 5 do exist?
If I can show that the group of rotational symmetries of a tetrahedron is A4 and the full group of symmetries of a tetrahedron is S4, then I can conclude that A4 is a subset of S4.
Does this approach satisfy the question?
Homework Statement
A question from artin 6.2:
Two tetrahedra can be inscribed into a cube C, each one using half the vertices. Relate this to
the inclusion A4 is a subset of S4.
The Attempt at a Solution
I can only think that the tetrahedral group is isomorphic to A4, and the cube is...
Yeah it is from artin, but I don't have my text with me.
So 20/4=5 so there is a subgroup of order 5, and because 20/5=4 there is a subgroup of order 4? Is this enough?
Thanks
Homework Statement
Given a class equation of a group H of 20=1+4+5+5+5, does H have a subgroup of order 5? Of order 4?
The Attempt at a Solution
Order 5 I can't get, but for order 4 I think I am correct in saying that H does have a subgroup of order 4 because each summand in the class...