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?