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

[Homo Novus]

Homework Statement

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

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

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

Homework Equations

For normal subset H:

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

For homomorphism [itex]\varphi : G \rightarrow G'[/itex]:

[itex]\varphi(ab) = \varphi(a) \varphi(b)[/itex] for all [itex]a,b \in G[/itex].

The Attempt at a Solution

b):

[itex]x^n = e; n \in \mathbb{P}[/itex]
[itex](\varphi(x))^{qn+r} = e; q,r \in \mathbb{Z},[/itex] 0≤ r < n.
[itex](\varphi(x))^{qn}(\varphi(x))^{r}=e[/itex]
[itex]\varphi(x^{qn})\varphi(x^r)=e[/itex]
[itex]\varphi(e)\varphi(x^r)=e[/itex]
[itex]\varphi(x^r)=e[/itex]
...? 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?