Mapping from a finite index subgroup onto the whole group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:16:01Z http://mathoverflow.net/feeds/question/90021 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90021/mapping-from-a-finite-index-subgroup-onto-the-whole-group Mapping from a finite index subgroup onto the whole group Victor 2012-03-02T08:34:29Z 2012-03-02T14:29:02Z <p>Dear All,</p> <p>here is the question:</p> <p>Does there exist a finitely generated group $G$ with a proper subgroup $H$ of finite index, and an (onto) homomorphism $\phi:G\to G$ such that $\phi(H)=G$?</p> <p>My guess is "no", for the following reason (and this is basically where the question came from): in Semigroup Theory there is a notion of Rees index -- for a subsemigroup $T$ in a semigroup $S$, the Rees index is just $|S\setminus T|$. The thing is that group index and Rees index share the same features: say for almost all classical finiteness conditions $\mathcal{P}$, which make sense both for groups and semigroups, the passage of $\mathcal{P}$ to sub- or supergroups of finite index holds if and only if this passage holds for sub- or supersemigroups of finite Rees index. There are also some other cases of analogy between the indices. Now, the question from the post is "no" for Rees index in the semigroup case, so I wonder if the same is true for the groups.</p> <p>Also, I beleive the answer to the question may shed some light on self-similar groups.</p> http://mathoverflow.net/questions/90021/mapping-from-a-finite-index-subgroup-onto-the-whole-group/90036#90036 Answer by HW for Mapping from a finite index subgroup onto the whole group HW 2012-03-02T13:19:09Z 2012-03-02T13:33:29Z <p>Here is a proof that there is no such finitely generated group. It's similar to Mal'cev's proof that finitely generated residually finite groups are non-Hopfian.</p> <p>First, note that $\ker\phi$ is not contained in $H$---otherwise, $|\phi(G):\phi(H)|=|G:H|$. Let $k\in\ker\phi\smallsetminus H$. Because $\phi$ is surjective, there are elements $k_n$ for each $n\in\mathbb{N}$ such that $\phi^n(k_n)=k$.</p> <p>Let $\eta:G\to\mathrm{Sym}(G/H)$ be the natural action by left translation. Then the homomorphisms $\eta\circ\phi^n$ are all distinct. Indeed,</p> <p>$\eta\circ\phi^n(k_n)=\eta(k)\neq 1$</p> <p>because $k\notin H$, whereas</p> <p>$\eta\circ\phi^{m}(k_n)=\eta(1)=1$</p> <p>for $m>n$. But there can only be finitely many distinct homomorphisms from a finitely generated group to a finite group.</p> http://mathoverflow.net/questions/90021/mapping-from-a-finite-index-subgroup-onto-the-whole-group/90043#90043 Answer by Benjamin Steinberg for Mapping from a finite index subgroup onto the whole group Benjamin Steinberg 2012-03-02T14:29:02Z 2012-03-02T14:29:02Z <p>Here is a variation on Henry's nice argument which uses Malcev's theorem. Let $N$ be the intersection of all finite index normal subgroups of $G$. Clearly $\phi(N)\subseteq N$ because a surjective endomorphism takes finite index normal subgroups to finite index normal subgroups. Thus $\phi$ induces a proper endomorphism of the finitely generated residually finite group $G/N$. By Malcev's theorem that f.g. residually finite groups are Hopfian, it follows $\phi$ induces an automorphism, which means $\ker \phi$ is contained in $N$. But then since each finite index subgroup contains a finite index normal subgroup, we have $\ker \phi\subseteq H$, which is a contradiction as Henry points out since in that case one would have $[G:H]=[\phi(G):\phi(H)]$.</p>