# Laws of Exponents for Cosets - Fraleigh p. 142 14.22

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 $[Hg]^n = Hg^n$?

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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. –  user1729 Apr 28 '14 at 11:14
(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".) –  user1729 Apr 28 '14 at 11:16
I agree with user1729. Please open a new question. –  Adam Saltz Apr 28 '14 at 22:48

1. I think the easier way (thus also answering the question 2 about intuition) is to note that $x\mapsto xH$ is a group homomorphism $\phi\colon G\to G/H$. Therefore $x^m=1$ implies $\phi(x)^m=\phi(x^m)=\phi(1)=1$. At the basic level, even though $xh\ne hx$ in general, we are given that $H$ is normal, hence $xhx^{-1}\in H$ for all $h\in H$; which is equivalent ot $xH=Hx$.

2. see above

3. To make use of the definition of things involved should be among the first things to try. Actually, in these basic problems this method makes it often impossible to proceed in any direction other then towrds the goal.

4. Well, that's debatable. In answer 1., we only used that the elements of the group $\phi$ maps to are the elements $xH$ we are interested in. We do not actually make use of the fact that $\phi(x)=1$ iff $x\in H$. But of course this is exactly how $G/H$ is constructed ...

5. Note that $(Hg)^n=g^{-1}(gH)^ng$, so the situation is the same for left and right cosets. Normality is sufficient, as seen in 1. or other proofs of the original problem. (If $G$ is abelian, then all subgroups are normal, so $G$ being abelian is of course also sufficient). Normality is also necessary: Let $g\in G$. If $x\in gHg^{-1}$, then $xg^2\in (gH)^2=gHg^{-1}g^2H$, so for $(gH)^2=g^2H$ we need $xg^2\in g^2H$, i.e. $x\in g^2Hg^{-2}$. We conclude $gHg^{-1}\subseteq g^2Hg^{-2}$ for all $g\in G$. By conjugation $H\subseteq g H g^{-1}$ for all $g\in G$ and finally $H=gHg^{-1}$, i.e. $H$ is normal.

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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 $\phi: G\rightarrow K$ with $\ker\phi=H$. 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 $G$.

Comments: The OP asked me about the intuition here, and about what would happen if $H$ was abelian. 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 $n>2$.

Note also that in the question the OP claims that I am saying that $G$ is normal. No I am not. I am saying that $H$ is normal in $G$. The result holds for all $H$ when $G$ is an abelian subgroup, but this is a special case because all subgroups of an abelian group are normal.

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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 '14 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 Feb 21 '14 at 12:22

Original question (2.) Is this true on the whole: $\text{If H is any subgroup of any group G, then }[Hg]^n = Hg^n$

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 $(Ab)^2 \neq A^2b^2$.

Original question (1.) Can someone please flesh out $T(G)g^n = (T(G)g)^n$ from?

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 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 Feb 11 '14 at 16:46