# Can anyone help me with these abstract algebra proofs?

1. Oct 30, 2012

### Homo Novus

1. The problem statement, all variables and given/known data

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.

2. Relevant 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 $a,b \in G$.

3. 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.

Last edited: Oct 30, 2012
2. Oct 30, 2012

### micromass

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

3. Oct 30, 2012

### Homo Novus

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

Last edited: Oct 30, 2012