Another Group Proof

Need a breather from PHL, so here's another group theory proof.
Properties of Isomorphisms Acting on Elements and Subgroups
Let G and G' be groups, and let P map G to G'. Let H be a subgroup G.

1. Claim: P maps the identity of G to the identity of G'
   Pf: Because e=ee, P(e)=P(ee)=P(e)P(e). Also, because P(e) is in G', P(e)=e'P(e). Then, e'P(e)=P(e)P(e) --> e'=P(e). qed
2. Claim: For every integer n and for every group element a in G, P(a^n)=[P(a)]^n.
   Pf: By definition of isomorphisms and by mathematical induction, if n is positive, then the claim is true. If n=0, P(a^n)=P(a^0)=P(e) and [P(a)]^n=[P(a)]^0=e. If n is negative, then -n is positive, and we have from the first claim and mathematical induction that e=P(e)=P([g^n][g^-n])=P(g^n)P(g^-n)=P(g^n)[P(g)]^-n =e. Then, P(g^n)=[P(g)]^-(-n)=P(g)^n. qed
3. Claim: For any elements a and b in G, a and b commute iff P(a) and P(b) commute.
   Pf: I will prove the claim "P(a) and P(b) commute if a and b commute" first. Assume that a and b commute. Then, ab=ba. P(ab)=P(a)P(b) and P(ab)=P(ba)=P(b)P(a), so P(a)P(b)=P(b)P(a). Now I will prove the converse. Assume that P(a) and P(b) commute. Then, P(a)P(b)=P(b)P(a). Then, P(a)P(b)=P(ab)=P(b)P(a)=P(ba), so P(ab)=P(ba). Because P is one-to-one, ab=ba. qed
4. Claim: G='a' iff G'='P(a)'.
   Pf: I will prove the claim "G'='P(a)'" first. Let G='a'; then, by closure, 'P(a)' is a subset of G'. Because P is onto, for any element b in G', there is an element a^k in G such that P(a^k)=b. Thus, b=[P(a)]^k, so b is in 'P(a)'. Then, G'='P(a)'. Now I will prove the converse. Suppose that G'='P(a)'. 'a' is a subset of G (by closure, etc.). For any b in G, we have P(b) is an element of 'P(a)'. Then, for some integer k, we have P(b)=[P(a)]^k=P(a^k). Because P is one-to-one, b=a^k. Then, G='a'. qed
5. Claim: Isomorphisms preserve orders.
   Pf: Let g be in G, where |g| is n. Then, P(g^n)=P(e)=e and P(g^n)=[P(g)]^n. So, [P(g)]^n=e. So, |P(g)|=|g|. qed
6. Claim: If G is finite, then G and G' have exactly the same number of elements of every order.
   Pf: By property 5, elements of order n map to elements of order n. Then, G and G' must have the same number of elements of each order. qed
7. Claim: G is Abelian iff G' is Abelian.
   Pf: By property 3, this property follows.
   
8. Claim: G is cyclic iff G' is cyclic.
   Pf: By property 4, this property follows.
   
9. If K is a subgroup of G, then P(K) is a subgroup of G'.
   Pf: Let a,b be in K, and P(a)=c and P(b)=d. I need to show that cd^-1 is in P(K). By definition, c and d are in P(K). P(ab^-1)=P(a)P(b^-1)=P(a)[P(b)]^-1=cd^-1, which is in P(K) by closure.
   

That was fun. Now, back to PHL.