Suppose we have a finitely presented group $G$ with decidable word problem. Is it decidable to check whether a given element $x\in G$ has finite order or infinite?
Group Theory – Decidability of Finite Order Elements
computability-theorydecidabilitygr.group-theory
Related Solutions
I think the answer to your final question is no. Given a finite set of generators of a subgroup $H$ of a Tarski Monster $G$, systematically evaluate words in those generators. You can use your solution to the word problem to check equality in $G$ between words. Either you will eventually produce a finite set of group elements which is closed under multiplication by the generators of $H$ and their inverses, in which case $H$ is finite, or else you will eventually find all of the generators of $G$, in which case $H=G$.
This is just an expanded version of Igor's comment. Indeed this is an open problem. The common opinion (I believe) is that such groups do exist, but the best result in this direction so far is the Olshanskii-Sapir group, which is finitely presented and (infinite torsion)-by-cyclic.
There is a general idea, commonly attributed to Rips, which shows that such groups should exist. The idea is the following. Take the free Burnside group $B(m,n)$ on $m\ge 2$ generators and of exponent $n>>1$, so that $B(m,n)$ is infinite. Embed it into a finitely presented group $G$ by using a version of the Higman embedding theorem. Let $G=\langle x_1, \ldots , x_n\rangle$. Then take the quotient group $Q$ of $G$ by the relations
(*) $x_1=b_1, \ldots , x_n=b_n$, where $b_1, \ldots , b_n\in B(m,n)$.
Clearly $Q$ is torsion being a quotient of $B(m,n)$ and is finitely presented. And it is believable that if the embedding of $B(m,n)$ in $G$ is "reasonably hyperbolic" and elements $b_1, \ldots , b_n$ are chosen randomly, then $Q$ is infinite with probability close to $1$. This idea was essentially implemented by Olshanskii and Sapir, but they managed to impose all relations of type (*) but one. (This is in fact a very rough interpretation of what they did, so Mark may correct me here.) There are even very particular choices of $b_1, \ldots , b_n$ (e.g., long aperiodic words with small cancellation), for which the group $Q$ should be infinite. So there are even particular finite presentations which should represent infinite torsion groups, but nobody knows how to prove that.
Best Answer
A finitely presented group with decidable word problem and undecidable order problem is in McCool, James Unsolvable problems in groups with solvable word problem. Canad. J. Math. 22 1970 836–838.