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