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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 that 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. Computing $(xH)^m$ in G/H using the representative $x$, we have $\color{magenta}{(xH)^m = x^mH} = eH = H$,
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 we can't do $(xH)...(xH) = \underbrace{x...x}_{m \;times}\underbrace{H...H}_{m \;times} = x^mH^m$ ?

(2.) What's the intuition for the result here?

Update (3.) user1729's Method 2 is $\color{brown}{\text{definition of normality :}} g\color{brown}{H}hH=g\color{brown}{(h \quad Hh^{-1})}hH=gh \; H$.
How do you envisage and envision to use this magical trick, to substitute H out?

Update. (4.) What are the sufficient conditions for $[Hg]^n = Hg^n$? H G Abelian or normal?

I'm filling in this old question because I didn't know I don't understand it perfectly.

First post - Can someone please flesh out $T(G)g^n = (T(G)g)^n$ from?

Also, is this true on the whole: $\text{If H is any subgroup of any group G, then }[Hg]^n = Hg^n$

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2 Answers

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.

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Thank you Adam Saltz. So $H \text{subgroup of}$ Abelian $G and $g \in G$ is a sufficient condition. What's the necessary and sufficient condition though for this "exponent rule"? –  Frank Muer Dec 23 '12 at 21:57
I think it is enough for the group to be normal so that the quotient group is well-defined. –  Adam Saltz Dec 24 '12 at 4:35
Apologies. I shirked perfect understanding hence I filled in my original question. Upvoted. –  Frank Muer Feb 11 at 16:46
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Method 1: The best way, I think, to approach this problem is to forget about cosets and instead think about homomorphic images. We can do this because of the first isomorphism theorem. So, $G/H$ is just some group, $K$ say, and we have a homomorphism $\phi: G\rightarrow K$. The kernel of $\phi$ is $H$, but this doesn't matter. All your question is asking is the following.

Is it true that $(\phi(g))^n=\phi(g^n)$?

And, of course, this is true because $\phi(g)\phi(h)=\phi(gh)$ for all $g, h\in G$ (as $\phi$ is a homomorphism).

Method 2: If you do still want to think about cosets then induct on $n$ using the following trick. $$gHhH=g(hHh^{-1})hH=ghH$$ This holds because $H$ is normal in $G$.

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Thanks a lot. Upvoted. Can you please notify me on my updated questions? And do you have an answer for (2) ? –  Frank Muer Feb 21 at 9:23
The intuition is the homomorphism. The trick I talk about holds because $H$ is normal. Indeed, it holds precisely when $h$ is contained in the normaliser of $H$ (the normaliser is the largest subgroup of $G$ in which $H$ forms a normal subgroup). It is possible for this trick to not hold when $H$ is abelian - the structure of $H$ is irrelevant, rather, it is how it lies in $G$ which is important. For example, consider a dihedral group $\langle a, b; a^2, b^n, aba=b^{-1}\rangle$. Then $\langle a\rangle$ is abelian but not normal for $n>2$. –  user1729 Feb 21 at 10:20
Thanks a lot. Can you please move your comment into your answer? Forgot to ask this. Still perplexed hence want to clarify. (3.) I meant how you envisaged/envisioned to use this trick to induce this result. The trick does operate but where does it spring from? It feels magical and I want to remove the magic. (4.) Hence are the sufficient conditions $G$ is Abelian or normal? If not, what are they? Can you please answer me in your answer? –  Frank Muer Feb 21 at 12:22
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