Good Morning, Proof!

Here is the third proof (or, rather, set of proofs) I'll be expected to know.
Properties of Elements and Subgroups Under Homomorphisms
Let G and G' be groups, where P maps G into G', and H is a subgroup of G.

1. Claim: P maps the identity of G to the identity of G'.
   Pf: This proof is the same as the proof for property 1 of the previous set of proofs. But I'll do it again for practice. P(e)=P(ee)=P(e)P(e), and e'P(e)=P(e). So, e'P(e)=P(e)P(e), implying that e'=P(e). qed
2. Claim: P(g^n)=[P(g)]^n.
   Pf: Again this is the same as property 2 of the previous set of properties. If n is positive, then the claim follows by mathematical induction and by the definition of homomorphisms. If n=0, then the claim follows by the previous property. If n is negative, then -n is positive. Then, P(e)=P((g^n)(g^-n))=P(g^n)P(g^-n)=P(g^n)[P(g)]^-n. Then, P(g^n)=[P(g)]^n. qed
3. Claim: If |g| is finite, then |P(g)| divides |g|.
   Pf: Let |g|=n. Then, g^n=e. Then, e=P(e)=P(g^n)=[P(g)]^n. Then, by Thm 4.2, |P(g)| divides n=|g|. qed
4. Claim: KerP is a subgroup of G.
   Pf: By property 1, KerP is nonempty. Let a,b be in KerP. Then, P(a)=P(b)=e. Then, P(ab^-1)=P(a)P(b^-1)=P(a)[P(b)]^-1=ee^-1=e. Then, ab^-1 is in KerP, so KerP is a subgroup of G. qed
5. Claim: P(a)=P(b) iff aKerP=bKerP.
   Pf: I will prove the statement "If P(a)=P(b), then aKerP=bKerP" first. Assume that P(a)=P(b). Then, e=P(a)[P(b)]^-1=P(a)P(b^-1)=P(ab^-1), which is in KerP. By property 5 of the Lemma of Cosets, iff ab^-1 is an element of a subgroup H, then aH=bH. Then, the claim follows. Now, conversely, assume that aKerP=bKerP. Then, by the same property of the Lemma of Cosets, P(ab^-1) is in KerP, and so on backwards to P(a)=P(b). qed

There is a bunch more, but I think I get the general idea. Besides, I'll be asked to prove one of several, and I doubt I'd choose this one. I'll probably choose the set of properties in the last posting. Ok, back to PHL.