It is well known that in category of groups there are Push-outs so it is possible to realize amalgamation in some kind of most free way. My question is what about category of locally free groups? I think there are not Push-outs so the question is can we realize every amalgamation? (i.e. for every situation $f\colon H \to G_1$, $g\colon H\to G_2$ in locally finite groups there is locally finite group $G$ and monomorphisms $f′ \colon G_1\to G$, $g′\colon G_2\to G$ such that $f′f=g′g$)
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If I am understanding the question correctly, then I think the answer is no. (Did you mean "what about the category of locally finite group" rather than locally free groups?)
Let $G$ and $H$ be defined by the presentations
$$G = \langle x_i\ (i \in {\mathbb Z}), t \mid x_i^2=[x_i,x_j]=t^2=1, x_i^t=x_{-i}\ (i,j \in {\mathbb Z})\,\rangle,$$
$$H = \langle x_i\ (i \in {\mathbb Z}), u \mid x_i^2=[x_i,x_j]=u^2=1, x_{i+1}^u=x_{-i}\ (i,j \in {\mathbb Z})\,\rangle.$$
Then $G$ and $H$ are locally finite, with the common elementary abelian normal subgroup $\langle x_i \mid i \in {\mathbb Z} \rangle$.
But in the amalgamated product $G*_NH$, $tu$ induces an autmorphism of $N$ of infinite order, and any factor group of $G*_N H$ that reduces $tu$ to finite order will cause collapse in $N$.
answered Apr 20, 2015 at 7:44
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3$\begingroup$ I'll just translate for those who like, so as to avoid group presentations. If $F$ is a group, write $F^{(\mathbf{Z})}$ the finitely supported elements in $F^\mathbf{Z}$, which we view as functions $\mathbf{Z}\to F$. We have two automorphisms $u,v$ of order 2 of these products, defined as $u(f)(n)=f(-n)$ and $v(f)(n)=f(1-n)$. Now let $C$ be cyclic of order 2, $F$ any nontrivial finite group (of order 2 in your example) and consider the groups $G_u,G_v$ semidirect products $F^{(\mathbf{Z})}\rtimes C$, where the element of order 2 in $C$ acts by $u$, respectively $v$. (...) $\endgroup$– YCorApr 20, 2015 at 8:19
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3$\begingroup$ These groups are both locally finite. Then $F^{(\mathbf{Z})}$ is normal in both $G_u$ and $G_v$ and the amalgam $G_u\ast_{F^{(\mathbf{Z})}}G_v$ is the semidirect product $F^{(\mathbf{Z})}\rtimes D$, where $D$ is the infinite dihedral group, whose two generators of order 2 act by $u$ and $v$. This action is faithful because $uv$ acts as an element of infinite order (since $uv(f)(n)=vf(-n)=f(n+1)$). Hence the normal subgroup generated by any nontrivial power of $uv$ intersects $F^{(\mathbf{Z})}$ non-trivially. So no "amalgam" of $G_u$ and $G_v$ is torsion. $\endgroup$– YCorApr 20, 2015 at 8:24
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2$\begingroup$ Yes, this also proves that we cannot define amalgams in the class of torsion groups. $\endgroup$ Apr 20, 2015 at 9:25