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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?

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    You shouldn't edit questions once you have answers. This is bad form - I shouldn't have to up-date my answer sporadically to keep it in check with your changing question. Rather, you should ask a new question and link back to this old one.2014-04-28
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    (That said, I have edited my answer. I am not sure what I was meaning when I wrote the thing about the kernel which prompted (4), so I have removed it. You claim in (5) is wrong - I am saying that $H$ is normal in $G$ *not* "$G$ normal".)2014-04-28
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    I agree with user1729. Please open a new question.2014-04-28

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