# writing a subgroup of the affine group as a semidirect product

The group $\mathbb Z +...+\mathbb Z=\mathbb Z^n$ of covering translations and $S_n$ acting on $\mathbb R^n$ by permuting coordinates, both lie in $aff(\mathbb R^n)$ the group of affine motions of $\mathbb R^n$. Let $G$ be the subgroup of $aff(\mathbb R^n)$ generated by $\mathbb Z^n$ and $\mathbb S_n$. then $\mathbb Z^n$ is normal in $G$ with quotient $S_n$. Hence $\mathbb R^n/G$ is the quotient first by $\mathbb Z^n$ then by $S_n$.

I understand that he means $G=\mathbb Z^n\ltimes_\phi\ S_n$ but what is $\phi$ here? i have two guesses:

i) my first guess is that he identifies the group $$\mathbb Z^n=\{(m_1,...,m_n)\in Z^n\subset \mathbb R^n\}$$ with the group $$T=\{t:\mathbb R^n\rightarrow \mathbb R^n;\;(x_1,...,x_n)\mapsto (x_1+m_1,\cdots,x_n+m_n \} \subset Aff(\mathbb R^n)$$ and now we have $T$ is normalized by $S_n$ and $T\cap S_n=\{1\}$ and so $G$ is THE internal semidirect product $G=TS_n$ or equivalently this is The external semidirect product $G=T\ltimes_\phi S_n$ with $\phi$ is the action of conjugation of $S_n$ on $T$.

ii)My second guess is that since there is an isomorphism $$Aff(\mathbb R^n)\cong \mathbb R^n \ltimes_\psi Gl(\mathbb R^n)$$ where $$\psi:Gl(\mathbb R^n)\rightarrow Aut(\mathbb R^n);\; f\mapsto \psi_f(x)=f(x)$$

The isomorphism above maps an invertible affine motion $a:\mathbb R\rightarrow \mathbb R;\;x\mapsto f(x)+y$ to the couple $(y,f)$.

now since $\mathbb Z^n\subset \mathbb R^n$ and $S_n\subset Gl(\mathbb R^n)$ then $G$ is just the restriction of $Aff(\mathbb R^n)\cong \mathbb R^n \ltimes_\psi Gl(\mathbb R^n)$ which means $G= \mathbb Z^n \ltimes_\psi S_n$ where $$\psi: S_n\rightarrow Aut(\mathbb Z^n);\; \sigma \rightarrow \psi_\sigma(m_1,...,m_n)=\sigma(m_1,...,m_n)=(m_{\sigma(1)},...,m_{\sigma(n)})$$

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@Dylan: Yes, but that's true in both constructions; $T$ and $\mathbb R^n$ are the normal factors, respectively. –  joriki Aug 6 '11 at 17:54
@Dylan: I don't understand. Then you'd be putting the normal factor second, but you rightly said that it's the first? –  joriki Aug 7 '11 at 4:54
@joriki Er, but $N \rtimes H$ means the same thing as $H \ltimes N$, no? Anyway, as long as there's some convention that makes us all understandable I'm quite happy. –  Dylan Moreland Aug 7 '11 at 19:58
Also, I think the last display needs a small tweak: if you want your automorphisms to act on the left, this won't be a group homomorphism. (What is the effect of $\psi_\tau \circ \psi_\sigma$?) –  Dylan Moreland Aug 7 '11 at 20:08
@Dylan: Sorry, I must have not noticed that you flipped the product sign :-) Can't tell what was what now since you've deleted the comments, but I guess this is all resolved now then? –  joriki Aug 8 '11 at 8:21

Literally speaking, he's doing the former, since he speaks of $\mathbb Z^n$ as a group of translations, but on a more abstract level it doesn't matter, since those two constructions, if I understand them correctly, are basically saying the same thing. The map inducing the semidirect product is always conjugation on the first factor by the second; there's no difference in that respect between your two constructions; the $\psi$ in your second construction is also conjugation of $S_n$ on $\mathbb Z^n$; you're just describing it internally in $\mathrm{Aff}(\mathbb R^n)$ in one case and externally as the semidirect product of two abstract groups in the other case.
are you saying that $T\ltimes_\phi S_n$ is isomorphic to $\mathbb Z\ltimes_\psi S_n$ where $T$, $\phi$ and $\psi$ are as defined above? –  palio Aug 6 '11 at 21:35
@palio: If you mean $\mathbb Z^n$ instead of $\mathbb Z$, yes, that's basically what I'm saying. Don't you think so? –  joriki Aug 6 '11 at 21:41
@ joriki: i'm not sure that $T\ltimes_\phi S_n$ is isomorphic to $\mathbb Z^n\ltimes_\psi S_n$ because i know that the semidirect product is not unique; two actions $\phi$ and $\psi$ give non isomorphic groups, and in fact these give different extensions to $S_n$ by $T\cong \mathbb Z^n$ –  palio Aug 7 '11 at 2:40
@palio: (You don't need to ping me in comments under my answer, I get notified automatically; and when you do ping someone, it doesn't work with a space between the '@' and the name.) We agree that the semidirect product is not unique and different actions $\phi$ and $\psi$ give non-isomorphic groups; the question is whether your $\phi$ and $\psi$ are different. As I wrote, as far as I can tell these are both $S_n$ acting by conjugation on a group isomorphic to $\mathbb Z^n$, so I think they're just the same thing expressed slightly differently. How do you think they differ? –  joriki Aug 7 '11 at 4:52