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.)

Solution Preview

Proof:
(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 ...

Solution Summary

This solution is comprised of a detailed explanation to prove isomorphism.

Note: ~~ means an isomorphism exists. Moreover,if an isomorphism existed from G to G1 I would say G ~~ G1
Questions: If G is an infinite cyclic group, show that G ~~ Z (Z is the set of integers)

Note:
S4 means symmetric group of degree 4
A4 means alternating group of degree 4
e is the identity
Is there a group homomorphism $:S4 -> A4, with
kernel $ = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}?

I have some trouble understanding the solution to the attached problem (solution included). Could you please provide some clarification of the solution. I have indicated what my points of concern are.
Show that if M1 and M2 are irreducible R modules, then any nonzero R-module homomorphism from
M1 to M2 is an isomorphism. De

(a) The kernel of this homomorphism is the principal ideal (x-1). Therefore, Z[x]/(x-1) is isomorphic to Z. According to the correspondence theorem, ideals of Z[x]/(x-1) are in one-to-one correspondence with ideals of Z[x] containing (x-1). Taking into account the above-mentioned isomorphism, we obtain that ideals of Z are in

What does it mean for two graphs to be the same? Let G and H be graphs. We say that G is isomorphic to H provided that there is a bijection f:V(G) -> V(H) so that for all a, b, in V(G) there is an edge connecting a and b (in G) if and only if there is an edge connecting f(a) and f(b) (in H). The function f is called an isomorphi