Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
@YCor Sure, although I'm not sure why you mention simple groups. My point is that any "criterion for determining the simplicity" of finitely presented groups cannot be algorithmically checked on general examples (unless the criterion implies, but is not equivalent to, simplicity - the word "determines" makes the precise situation unclear to me, but that's maybe just me).
Finitely presented simple groups have decidable word problem, and so being simple is a Markov property, and so it is undecidable if a given presentation defines a simple group. (In his survey article "Decision problems for groups: survey and reflections", Miller attributes this to Kuznetsov. He also provides an easy, half-page proof.)
@YCor Thanks. Yes, the other questions make sense and I would be interested to know their answer also. I only asked one question as I didn't want to make the question too convoluted (and this specific question cropped up in my work).
@HJRW Yes, my mistake. It should read "... Then $G$ has more than one end if and only if there exists a generator which does not occur in $R$." The first sentence in my second comment ("If $|\mathbf{x}|=2$ and...") is also incorrect. More can be said regarding this test for endedness, but I guess here isn't the place!
(If $|\mathbf{x}|=2$ and $R$ is primitive then $G$ is two-ended, otherwise $G$ is one-ended. Note that your relator satisfies this theorem, which can be seen by using the Whitehead automorphisms to verify minimality of its length. So $G$ is not infinitely-ended, and hence has to be one-ended (erm......there is presumably an easy way of showing non-0 and non-2 endedness.....but for your group you can use the Baumslag-Pride result to see that it is large and so not 0- or 2-ended.).)
Nice example! I just wanted to point out that your group is one-ended, but not for the reasons you state. For example, the following group is not one-ended even though the relator is not primitive and every generator occurs in the relator: $\langle a,b,c,d\mid (ad)b^2(ad)c^{12}\rangle$. The theorem you want is as follows: Suppose $R$ has shortest possible length in its $Aut(F(\mathbf{x}))$-orbit. Then $G=\langle \mathbf{x}\mid R\rangle$ is infinitely-ended if and only if there exists a generator which does not occur in $R$.
@HJRW Fischer, Karrass and Solitar proved (On one-relator groups having elements of finite order, PAMS 1972) that $G=\langle X; R^n\rangle$, $n>1$, is virtually free if and only if $G$ is the free product of a free group and a finite cyclic group. In fact, they prove that if $G$ has more than one end then it has infinitely many and is a free product of a nontrivial free group and an indecomposable one-relator group (which is decidable). It follows (perhaps not "clearly", but it does follow!) that if $G=\langle X; R\rangle$, $R$ not a proper power, is virtually free then it is free.
