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Let $N=\langle x_1, \ldots, x_m; \mathbf{r}\rangle$ and $H=\langle y_1, \ldots, y_n; \mathbf{s}\rangle$ ($m$ and $n$ can be infinite). Then if we define,

$G=\langle x_1, \ldots, y_1, \ldots; \mathbf{r}, \mathbf{s}, x_i^{y_j}=x_i\phi_j\rangle$

where $\phi_j\in Aut(N)$ for all $j=1, \ldots, n$ we get a short exact sequence, $$1\rightarrow \overline{N}\rightarrow G\rightarrow H\rightarrow 1$$ where $\overline{N}=\langle x_1, \ldots, x_m\rangle$.

(Basically, pin two groups together via an automorphism of one of them.)

My questions are these:

-Is it true that $\overline{N}\cong N$ ?

-If $\overline{N}\cong N$, does it hold that $G=N\rtimes H$ ?

-Do the above two questions hold if we stipulate that the $\phi_j$ arise in the natural way from a map $\langle y_1, \ldots, y_n\rangle\rightarrow Aut(N)?$

If these do not hold, does anyone know of a reference which explores when they do or do not?

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It only works if the $\phi_j$ are "functorial"; that is, you need $\phi_i\circ\phi_j$ to reflect the product $y_iy_j$ (composition of functions left to right). Otherwise, you don't. E.g., if $y_i^n=e$, say, and $\phi_i^n\neq \mathrm{id}$, then you get $x_j = (x_j)\phi^n_j$ in $\overline{N}$, and you know that those equalities cannot all hold in $N$ if $\phi_i^n\neq\mathrm{id}$. –  Arturo Magidin Aug 22 '11 at 16:17
Arturo: I suppose that is conceivable that the answer to the first question could be yes, even if the $\phi_j$ are not functorial, if $N$ was a nonHopfian group? –  Derek Holt Aug 22 '11 at 16:31
@Swlabr: In that case, the answer to both is "yes"; essentially, an internal semidirect product is equivalent to an external one, and an external one is equivalent to a homomorphism $H\to\mathrm{Aut}(N)$. –  Arturo Magidin Aug 22 '11 at 16:34
Okay, thanks, I'll re-edit my question to remove my previous edit, so your answer makes sense. Alternatively, this will just confuse everyone, but hey... –  user1729 Aug 22 '11 at 16:36
@Swlabr: How about adding the condition below the original question? –  Arturo Magidin Aug 22 '11 at 16:40
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2 Answers

up vote 3 down vote accepted

Here's a simple counterexample, given the requirements specified. Take $N=C_3 = \langle x|x^3\rangle$, the cyclic group of order $3$, and let $H=C_5 = \langle y|y^5\rangle$, the cyclic group of order $5$. Let $\phi\in\mathrm{Aut}(H)$ be the nontrivial automorphism. You then have $$G = \langle x,y\mid x^3,y^5, x^y=x^2\rangle.$$

We then have $x^{y^2} = x$, so $x^{y^4}=x$, hence $x = x^1 = x^{y^5} = (x^{y^4})^y = x^y = x^2$. Thus, $x=1$ in $G$, So $\overline{N}$ is trivial.

The problem is that the assignment $\{y_1,\ldots,y_m\}\to \mathrm{Aut}(N)$ is not being asked to "reflect" the relations $\mathbf{s}$, so that forces new relation on the $x_i$. In order to get the isomorphisms, you need the assignment $y_i\mapsto \phi_i$ (which induces a group homomorphism from the free group in the $y_i$ into $\mathrm{Aut}(N)$) to factor through $H$; i.e., you really need a group homomorphism $H\to \mathrm{Aut}(N)$, not just an assignment of the generators to homomorphisms.

If you have a homomorphism $H\to\mathrm{Aut}(N)$, then the answer to both question is "yes". An "external semidirect product of $N$ by $H$" is equivalent to a homomorphism $H\to\mathrm{Aut}(N)$ (see, e.g., Exercise I.12.12 in Lang's Algebra, or the section on semidirect products (pages 167-171, especially Lemmas 7.20 and 7.21, and Theorems 7.22 and 7.23 in Rotman's Introduction to the Theory of Groups, 4th Edition), and the presentation of the semidirect product is the one you give.

If, as it happens, you have that $\overline{N}\cong N$, then the answer to your second question is "yes". Note that we certainly have $\overline{N}\triangleleft G$, there is a generating set $S$ such that $x^{-1}Nx = N$ for all $x\in S$, namely $S=\{x_1,\ldots,x_n,y_1,\ldots y_m\}$. Also, $G/\overline{N} \cong \langle y_1,\ldots,y_m,\mathbf{s}\rangle\cong H$, as that's the presentation we get if we add the conditions $x_i=1$, $i=1,\ldots,n$. So $\overline{N}\triangleleft G$, and $G/\overline{N}\cong H$ with $\langle y_1,\ldots,y_m\rangle\cong H$ in $G$, so $G\cong \overline{N}\rtimes H$. Since we are assuming that $\overline{N}\cong N$, you certainly get a semidirect product $G\cong N\rtimes H$. However, it need not be given by the action of $H$ on $N$ given by the $\phi_i$: you would need to conjugate that action with the isomorphism $\psi\colon\overline{N}\to N$.

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Hi - yes, sorry I had meant to stipulate this when asking the question. I'll edit it just now. –  user1729 Aug 22 '11 at 16:31
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Just to turn Arturo's comment into an answer to the question, if the assignment $y_j \mapsto \phi_j$ extends to a homomorphism $\phi: H \rightarrow {\rm Aut}(N)$ (which is the case if and only if the $\phi_j$ satisfy the relations of $H$ when substituted for $y_j$), then the answer to both questions is yes. (I think the answer to the second question is always yes.)

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