What Are the Proofs for Powers in Normal Subgroups and Orders in Homomorphisms?

[Homo Novus]

Homework Statement

a) Let H be a normal subgroup of G. If the index of H in G is n, show that y^n \in H for all y \in G.

b) Let \varphi : G \rightarrow G' be a homomorphism and suppose that x \in G has order n. Prove that the order of \varphi(x) (in the group G') divides n. (Suggestion: Use the Division Algorithm.)

c) Let \varphi : \mathbb{Z}_n \rightarrow \mathbb{Z}_m be a homomorphism. Show that \varphi has the form \varphi([x]) = [qx] for some 0 ≤ q ≤ m - 1. Then, by means of a counterexample, show that not every mapping from \mathbb{Z}_n to \mathbb{Z}_m of the form
\varphi([x]) = [qx] where 0 ≤ q ≤ m - 1 need be a homomorphism.

Homework Equations

For normal subset H:

yH=Hy (right coset = left coset) for all y \in G, and they partition G.
yhy^{-1} \in H for all h \in H, y \in G.

For homomorphism \varphi : G \rightarrow G':

\varphi(ab) = \varphi(a) \varphi(b) for all a,b \in G.

The Attempt at a Solution

b):

x^n = e; n \in \mathbb{P}
(\varphi(x))^{qn+r} = e; q,r \in \mathbb{Z}, 0≤ r < n.
(\varphi(x))^{qn}(\varphi(x))^{r}=e
\varphi(x^{qn})\varphi(x^r)=e
\varphi(e)\varphi(x^r)=e
\varphi(x^r)=e
...? Not sure where to go from here.

 

[micromass]

For (a), what can you say about the element yH of G/H ?

 

[Homo Novus]

Hmm... The order of yH = order of G divided by n...? That, and it contains y?