Let $G = <a_1, . . . , a_n|r_1, . . . , r_k>$ and $H = <b_1, . . . , b_s|u_1, . . . , u_t>$ be finitely presented groups. Let $i: H → G$ be an injective homomorphism given by $i(b_j ) = w_j$ for $(j = 1, . . . , s)$, where $w_1, . . . , w_s$ are words in $a_1, . . . , a_n$.

Suppose that $H$ has solvable word problem and that $i(H)$ is a finite index subgroup of $G$. I am trying to show first that, given a word $w$ in the $a_1, . . . , a_n$, we can decide whether $w$ lies in $i(H)$. Then using this I wish to find an algorithm for solving the word problem in $G$.

So far I have the following:

Since $i(H)$ has finite index, there exists a normal subgroup $N$ of finite index in $G$, contained in $i(H)$. Then $w$ is contained in $i(H)$ iff $w$ is trivial in $G/N$. We can get a finite presentation of $G/N$ in terms of the $a_i$, and since the word problem is solvable for a finitely presented finite group, we are done. Does this work?

Then for WP in G:

for $w$ a word in the $a_i$, run the following in parallel:

  1. Enumerate elements of $<<{r_1...r_k}>>$. If $w=1$, then $w$ will appear in this list.

  2. Use the above to decide if $w$ lies in $i(H)$.

(a) If yes, and if $w\ne1$ in $G$ then also in $H$, which has solvable word problem... but $w\ne1$ in $H$ doesn't necessarily imply $w\ne1$ in $G$. Here I am stuck.

(b) If no, ... I am not sure how to progress.

Any help would be very much appreciated.

  • 1
    $\begingroup$ Your statement "Then $w$ is contained in $i(H)$ iff $w$ is trivial in $G/N$" is only true when $H=N$, which is not true in general. But that is not a serious problem, since $G/N$ is finite and has solvable subgroup membership probelm. $\endgroup$ – Derek Holt Apr 18 '17 at 11:44
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    $\begingroup$ Also it is not clear what you mean by "$w \ne 1$ in $H$" because $w$ is a word in the generators of $G$ not $H$. If $w \not\in i(H)$ then clearly $w \ne 1$ in $G$, so you can assume that $w \in i(H)$. What you then need to do is to express $w$ as a word in the generators $i(b_j)$ of $i(H)$. Then you can use the word problem in $H$ to decide whether $w = 1$ in $G$. $\endgroup$ – Derek Holt Apr 18 '17 at 11:51
  • $\begingroup$ @DerekHolt Thank you, I understand and can finish using this. I was given another hint, that for the case $w$ in $G$ \ $H$ (with $H$ identified with $i(H)$ and considered as a subgroup of $G$) I should consider homomorphisms from $G$ to finite groups. I'm not sure how I could have used this. $\endgroup$ – ech-93 Apr 18 '17 at 14:16

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