A sequence of finite groups

Question: does there exist a strictly ascending sequence of finite groups $G_0<G_1<G_2<\dots$ such that for every $i \in \mathbb{N}$ there is $a_i \in G_{i+3}$ and the following two conditions are satisfied:

(i) $a_i$ centralizes $G_{i-1}$ in $G_{i+3}$;

(ii) $G_{i+3}=\langle G_i, a_i G_i a_i^{-1} \rangle$.

Remark: taking $G_i=Sym(d_i)$ for some increasing sequence of integers $(d_i)$ with the evident embedding $Sym(d_i) \hookrightarrow Sym(d_{i+1})$ does not seem to work.

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@Ashot: $i+1,i+2$ are missing on purpose? If you replace $i+3$ by $i+1$, the example exists? –  Mark Sapir Mar 25 '12 at 0:37
I think in that case, $G_{i}=S_{i}$ and $a_{i}=(i i+1)\in G_{i+1}$ would work. –  David Cohen Mar 25 '12 at 3:45
@Mark; yes, $i+3$ is important. As David Cohen notices symmetric groups give an example when $i+3$ is replaced by $i+1$. –  Ashot Minasyan Mar 25 '12 at 7:34
Just an idea: Try to use the regular embedding of $G$ into $\mathop{Sym}(G)$ given by the right multiplication of $G$ on itself. Given $U < H \le G$ with $U \cap \mathrm{Z}(H) = 1$, you need to show the existence of an element $a\in \mathop{Sym}(G)$ with $\langle H, H^a\rangle = \mathop{Sym}(G)$ that centralizes $U$. –  j.p. Mar 27 '12 at 13:19
@jp: in this case $U$ will act freely on G, so it may not be easy to see that an element centralizes $U$ just looking at the action... –  Ashot Minasyan Mar 27 '12 at 23:07

Apparently such sequences do exist. For example, one can take $G_i:=Alt(5^i)$ for $i=0,1,2,\dots$, where the embedding $\gamma_i: G_i \to G_{i+1}$ is the diagonal embedding of $Alt(5^i)$ into $Alt(5^{i+1})$ (i.e., if $\sigma \in Alt(5^i)$ then $\gamma_i(\sigma)(1)=\sigma(1)$, $\dots$, $\gamma_i(\sigma)(5^i)=\sigma(5^i)$, $\gamma_i(\sigma)(5^i+1)=\sigma(1)$, $\dots$, $\gamma_i(\sigma)(2\cdot 5^{i})=\sigma(5^i)$, $\gamma_i(\sigma)(2\cdot 5^i+1)=\sigma(1)$, $\dots$
For $i \in \mathbb{N}$, the element $a_i \in G_{i+3}$ can be chosen as a certain permutation of blocks of length $5^{i-1}$, which are the orbits of $G_{i-1}=Alt(5^{i-1})$ when it is considered as a subgroup of $G_{i+3}=Alt(5^{i+3})$. This clearly implies the condition (i). The argument that $G_i$ and $a_iG_ia_i^{-1}$ generate $G_{i+3}$ is somewhat technical. Hopefully the details will appear in the second version of http://arxiv.org/abs/1203.3317.