C In parts 1-5 below, $G$ is a group and $H$ is a normal subgroup of $G$. Prove the following (Theorem 5 will play a crucial role)

Theorem 5- Let G be a group and H be a subgroup of G. Then

$(i)$ $Ha= Hb\quad\text{iff}\quad ab^{-1}\in H$

$(ii)$ $Ha = a\quad\text{iff}\quad a\in H$

1. if $x^2\in H$ for every $x\in G$, then every element of $G/H$ is its own inverse. Conversely, if every element of $G/H$ is its own inverse, then $x^2\in H$ for all $x\in G$.

2. Suppose that for every $x\in G$, there is an integer $n$, such that $x^n\in H$; then every element of $G/H$ has finite order. Conversely, if every element of $G/H$ has finite order, then for every $x\in G$ there is an integer $n$, such that $x^n\in H$.

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This should be fairly straightforward. Just use the definition of a quotient group. –  hasnohat Nov 15 '12 at 2:41
First, you should really go to the FAQ and read how to properly type mathematics in this site using LaTeX. Second, in your (ii) I'm guessing you actually meant $\,Ha=H\,$ . Third, try at least to introduce more order in your post using freely spaces, lines, etc. between statements. Fourth and last, and perhaps more important than the above: give us some of your ideas, insights, own effort on the above. This is a very basic question in group theory and it's important you make your own work. After that we can focus on your main problems –  DonAntonio Nov 15 '12 at 2:42
Im not sure how to start either problem? How do i use the definition to prove this –  Jeremy Nov 15 '12 at 3:40
For $(ii)$, do you mean $Ha=H$? –  robjohn Nov 15 '12 at 9:29

You must understand how the quotient group $\,G/H\,$ is defined when $\,H\triangleleft G\,$ , though you don't need normality to prove (i)-(ii): this is true for any subgroup.
Now, if $\,x\in G\,$ then, by definition, in the quotient $\,G/H\,$ we have
$$(xH)^n:=x^nH\Longrightarrow (xH)^2=x^2H=H\Longleftrightarrow x^2\in H$$