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)

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Let R1 and R2 be integral domains with quotient fields F1 and F2 respectively. If phi: R1 -> R2 is a ring isomorphism, show that phi extends to an isomorphism phi hat : F1 -> F2. Here extends means that phi hat(a) = phi (a) for all a in R1 (Hint: under the givien assumptions, there is really only one way to

Show that a homomorphism from a field onto a ring with more than one element must be an isomorphism.
Recall The the function f is an isomorphism if and only if f is onto and Kernel ={ 0}.
Please explain step by step with reasons in every step.

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)}?

a) Determine the structure of the ring R obtained from Z by adjoining an element w satisfying each set of relations: (i) 2w-6=0, w-10=0 and (ii) w3+w2+1=0, w2+w=0.
b) Let f=x4+x3+x2+x+1 and let y denote the residue of x in the ring R=Z[x]/(f). Express (y3+y2+y)(y5+1) in terms of the basis (1,y,y2,y3) of R.

Let M be a two by two matrix for which each element is a real number, and let S be the set of all such matrices. Consider the mapping f of S to the real numbers R defined by the relation f(M) = determinant of M.
a. Is this mapping onto? Why or why not?
b. Is this mapping one to one? Why or why not?
c. Is this mapping a h

f(x)=x^3+x+1 and g(x)=x^3+x^2+1 are irreducible over F_2. K is the field extension obtained by adjoining a root of f and L is the extension obtained by adjoining a root of g. Determine the number of isomorphisms from K to L. (It is not necessary to explicitly describe such an isomorphism)