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I have in fact become stuck by the very first problem in Serre's book on Trees. It is a little bit embarrassing but ho-hum. I start with Serre's definition of direct limits.

Let $(G_i)_{i \in I}$ be a family of groups and for each pair $(i,j)$, let $F_{ij} \subset \mathrm{Hom}(G_i,G_j)$. There exists a unique group $G$ and a family of homomorphisms $f_i \colon G_i \to G$ such that $f_j \circ f = f_i$ for all $f \in F_{ij}$ that has the following universal property: if $H$ is a group and if $h_i \colon G_i \to H$ is a family of homomorphisms such that $h_j \circ f = h_i$ for all $f \in F_{ij}$, then there exists exactly one homomorphism $h \colon G \to H$ such that $h_i = h \circ f_i$. We say $G$ is the direct limit of $G_i$ relative to $F_{ij}$.

Take three groups $A$, $G_1$, $G_2$ and two homomorphisms $f_1 \colon A \to G_1$, $f_1 \colon A \to G_2$. The amalgamated free product $G_1 \ast_A G_2$ is the direct limit of $A, G_1, G_2$ relative to $f_1, f_2$.

Usually one defines the amalgamated free product with monomorphisms, however the next exercise shows that there isn't any difference between Serre's definition and the definition with monomorphisms.

Take $G =G_1 \ast_A G_2$ relative to the above maps. Define subgroups $A^n$, $G_1^n$, $G_2^n$ recursively by the following conditions: $$ A^1 = \{ 1 \}, \quad G_1^1 = \{1 \}, \quad G_2^1= \{ 1 \} $$ $$ A^n = \text{subgroup generated by } f_1^{-1}(G_1^{n-1}) \text{ and } f_2^{-1}(G_2^{n-1}) $$ $$ G^n_i = \text{subgroup generated by } f_i(A^{n}) $$ Let $A^{\infty}$ and $G_i^{\infty}$ be the unions of $A^n$ and $G_i^n$ respectively. Now the exercise says to show that $f_i$ defines an injection $A / A^{\infty} \to G_i / G^{\infty}_i$ and that $G$ may be identified with the amalgam $G_1 / G_1^{\infty} \ast_{A / A^{\infty}} G_2 / G_2^{\infty}$.

The question I have is: why are $A^{\infty}$ and $G_i^{\infty}$ normal inside $A$ and $G_i$ respectively?

EDIT:

So after discussing this with user10193, here is another reason why we think $G_i^n$ are meant to be the normaliser of $f_i(A^n)$.

First note that whenever $\varphi \colon G \to H$ is a homomorphism and $S$ is a subgroup of $G$ then $\varphi^{-1}(\varphi(S)) = SK$ where $K$ is the kernel of $\varphi$.

Set $K_1 = \mbox{ker}(f_1)$ and $K_2 = \mbox{ker}(f_2)$. Then $A^2 = \langle K_1, K_2 \rangle = K_1 K_2$ as $K_1$ and $K_2$ are normal. Then if $G^n_i$ are meant to be defined as it is, it follows that $G_1^2 = f_1(K_2)$ and $G_2^2 = f_2(K_1)$. Hence $$ A^3 = \langle f_1^{-1}(f_1(K_2)), f_2^{-1}(f_2(K_1)) \rangle = \langle K_1 K_2, K_1 K_2 \rangle = K_1 K_2. $$ Therefore $A^n$ and $G_i^n$ stabilise. This just seems nonsense. So my question just reduces to: why is $A^{\infty}$ normal inside $A$?

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    $\begingroup$ I wonder whether $G_i^n$ should be the normal subgroup if $G_i$ generated by $f_i(A^n)$ i.e. its normal closure. It's strange to write the subgroup generated by $f_i(A^n)$, because $f_i(A^n)$ is already a subgroup. $\endgroup$ – Derek Holt Nov 5 '14 at 11:21
  • $\begingroup$ I agree because $A^n$ is defined to be the subgroup generated by these pre-images. Do you think it should be the same condition for $A^n$? $\endgroup$ – Chris Cave Nov 5 '14 at 11:31
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Well, to answer your reduced question: the inverse image of a normal subgroup is a normal subgroup. Thus, each $A^i$ is normal in $A$ and hence their union is.

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  • $\begingroup$ Yep, that was a dumb question to ask. $\endgroup$ – Chris Cave Nov 5 '14 at 17:14

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