A torsion group is a group all of whose elements have finite order. A group is torsion free if the identity is the only element of finite order. Prove: if G is a torsion group, then so is G/H for every normal subgroup H of G.
Proof: Let $xH ∈ G/H$. Because G is a torsion group, $x^m = e$ in G for some positive integer m. Compute $(xH)^m$ in G/H using the representative $x$: $\color{magenta}{(xH)^m = x^mH} = eH = H$,
so $xH$ is of finite order. Because xH can be any element of G/H, we see G/H is a torsion group.
(1.) Can someone please flesh out $\color{magenta}{(xH)^m = x^mH}$? I understand $(xH)^m = (xH)...(xH)$ (m times). But the question doesn't presuppose $G,H$ are commutative, hence this is impossible: $(xH)...(xH) = \underbrace{x...x}_{m \;times}\underbrace{H...H}_{m \;times} = x^mH^m$ ?
(2.) What's the intuition?
(3.) user1729's Method 2 is $\color{brown}{\text{definition of normality :}} g\color{brown}{H}hH=g\color{brown}{(h \quad Hh^{-1})}hH=gh \; H$.
Where does this trick of substituting H spring from? How do you predestine to do this?
(4.) user1729 says it doesn't matter that $\ker \phi = H$? Why not? Doesn't Fundamental Homomorphism Theorem require this?
(5.) What are the sufficient conditions for $[Hg]^n = Hg^n$?
Adam Saltz's answer says G H Abelian. But user1729's answer says G normal?
Let me answer your second question first. That statement is false. Take the free product on generators $a, b$. Let $A$ be the subgroup generated by $a$. Then $(Ab)^2 \neq A^2b^2$.
Make sure you read the original question: the group $G$ is assumed to be abelian. It is true that $(Hg)^n = Hg^n$ for an abelian group $G$ with subgroup $H$ and $g \in G$. To see this, note that $(Hg)^n$ is the set of elements of form $h_1gh_2g\cdots h_k g$ with $h_i \in H$ for all $i \in \{1, k\}$. As $G$ is abelian, we can rearrange this to $h_1h_2\cdots h_kg^n$, and $h_1h_2\cdots h_k \in H$ because $H$ is a subgroup.