We are working on the proof of showing G (the group of rigid motions of a regular dodecahedron) is isomorphic to the alternating group A_5.
Lemma: Let H be a normal subgroup of a finite group G, and let x be an element of G. If o(x) and [G:H] are relatively prime, then x is in H.
Theorem: Any 60 element group having 24 elements of order 5, 20 elements of order 3, and 15 elements of order 2 (which is exactly the group of rigid motions)is a simple group and is isomorphic to A_5.
Proof (parts of it): Suppose G satisfies the hypothesis and H is a nontrivial normal subgroup of G. By LaGrange, the possible orders for H are 2,3,4,5,6,10,12,15,20,30. If |H| = 3,6,12, or 15, then [G:H] is relatively prime to 3, so Lemma implies H contains all 20 elements of order 3, which is impossible.
(1) Use similar methods to prove that H cannot have order 4, 5, 10, 20, or 30.
(Omitting proof but assume phi: G --> S_5 and that phi is an isomorphism. So S_5 contains a subgroup phi(G) that is isomorphic to G. Then M is a simple group that has 60 elements. |MA_5| = (|M||A_5|)/|M and A_5|)
(2) Show that M = A_5. (use the above equation to get a good lower bound on |M and A_5| . You will then need to consider some cases, and use the fact that M is simple.)© BrainMass Inc. brainmass.com October 9, 2019, 5:12 pm ad1c9bdddf
(1) H is a nontrivial normal subgroup of G, |G|=60. If |H|=4, then [G:H|=60/4=15. For any x with o(x)=2, since gcd(2,15)=1, then x is in H. Thus H contains all 15 elements of order 2. This is impossible. If |H|=5, then [G:H]=60/5=12. For any x with order o(x)=5, since gcd(5,12)=1, then x is in H. So H contains all 24 elements of order 5. This is impossible. Similarly, If ...
This solution is comprised of a detailed explanation to prove isomorphism.