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
$(i)$ $Ha= Hb\quad\text{iff}\quad ab^{-1}\in H$
$(ii)$ $Ha = a\quad\text{iff}\quad a\in H$
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 $x\in G$.
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 $x^n\in H$.