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Motivation:

In playing around with some rather complicated groups in GAP during my PhD, I found that, with some of them, if I set some element $g$ of such a group $G$ equal to the identity by quotienting out by the normal subgroup $\langle\langle g\rangle\rangle$ it generates, so that, as they say, I "kill" $g$, the resulting group $G/\langle\langle g\rangle\rangle$ is the trivial group, just like it is when $G$ is cyclic.

The Question:

If I kill any element $g$ of a group $G$ and as a result $G/\langle\langle g\rangle\rangle$ is killed, is $G$ then cyclic?

Further Details:

The motivating groups were sufficiently "big" and their relations sufficiently convoluted, that neither GAP nor inspection of the groups (well . . . presentations) could tell me whether these groups were cyclic. Moreover, this was before I got into the habit of using LogTo("name-of-document"); (and I think the simple definitions are "scoop sensitive"), so I can't share them here.

Thoughts:

Suppose $G\cong\langle X\cup\{g\}\mid R\rangle$ such that $G/\langle\langle g\rangle\rangle\cong E,$ where $E$ is the trivial group up to isomorphism (so I don't get bogged down by what element $e$ is for $\{e\}\cong E$).

Now what? I don't know.

I guess if I were to build a group $\mathcal{G}$ to the contrary, any presentation $\mathcal{P}$ would have at least two generators $a, g$. A hunch of mine would be to try, say, $\mathcal{P}$ with relators approximately some "conjugation" $g^a=aga^{-1}$ of $g$ by $a$ such that the relation(s) are somehow cyclicaly-reduced but not $g$; something like $g^ag$. That seems like a contradiction, and what's more is that the motivating presentations did not appear (as far as I can remember) to be like that.

They, of course, had more than one generator though!

Please help :)

Shaun
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  • Isn't a group with only one generator always cyclic? – Andrew Tindall Aug 26 '19 at 18:35
  • I don't follow, @AndrewTindall. Did I say that they weren't? – Shaun Aug 26 '19 at 18:36
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    Oh, I misunderstood the question. You're asking, if you can kill the group by killing an element, was that element the only generator? – Andrew Tindall Aug 26 '19 at 18:43
  • Yes, @AndrewTindall; well put! – Shaun Aug 26 '19 at 18:43
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    Ok, so if you can kill the group this way, you've essentially made an exact sequence $0 \to \langle a \rangle \to G \to 0$, where $0 = G / \langle a \rangle$. That exact sequence is also a presentation of $G$ by one generator. Does that sound right? – Andrew Tindall Aug 26 '19 at 18:46
  • Again, I don't follow. Perhaps my intuition with exact sequences is underdeveloped. If you could explain the details in an answer, @AndrewTindall, please, then that'd help me a great deal. – Shaun Aug 26 '19 at 18:49
  • Let us continue this discussion in chat. – Andrew Tindall Aug 26 '19 at 18:52
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    What do you mean kill a generator? Do you just mean kill an element? If that is the case the group doesn't have to be cyclic(for example simple groups) –  Aug 26 '19 at 18:55
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    There are also non simple examples too. –  Aug 26 '19 at 18:56
  • My first impression is that your two questions equivalent, @PaulPlummer. Are they? I mean: a follow-up would be: if you kill a group by killing an element, is that element then a generator? – Shaun Aug 26 '19 at 18:58
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    Isn’t $S_3$, being the normal closure of $\langle (12)\rangle$ a counterexample? – Nicky Hekster Aug 26 '19 at 19:01
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    I guess I am wondering if by generator you mean part of a "free basis" or nice generating set. Of course simple groups still provide an example and any element can be added to a generating set –  Aug 26 '19 at 19:02
  • I think you're right, @NickyHekster. Why not promote that comment to an answer? – Shaun Aug 26 '19 at 19:10
  • Those are worthy of being an answer, too, @PaulPlummer! – Shaun Aug 26 '19 at 19:12
  • Why is this is getting downvotes? – Shaun Aug 27 '19 at 13:49
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    What's your double langle rangle notation mean here? – Alexander Gruber Aug 28 '19 at 03:02
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    @AlexanderGruber It means the normal subgroup generated by the element $g$. – user1729 Aug 28 '19 at 07:55
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    @Shaun I am guessing it is getting downvotes for "low research effort" reasons, as there are simple examples to try like $S_3$, the definition of simple groups etc. Perhaps unclear because of early comments confusion.(I am guessing you where focused on presentations to much where it can be a little trickier or trying to prove it true which can happen to everyone) –  Aug 28 '19 at 16:51
  • That makes sense, @PaulPlummer. I was looking at the problem in terms of presentations, as you said just now. (Hence the use of "generator" rather than simply "element".) It didn't occur to me that perhaps the answer is easy from a more traditional perspective. Besides, it had been years since I had worked with simple groups (believe it or not). – Shaun Aug 28 '19 at 17:46
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    @Shaun Its worth pointing out that this question is close to some interesting questions. One related theory with lots of interesting problems is that of "one relator products" (this is implicit in Paul Plummer's answer, and is a speciality of a chap called Jim Howie, who once had a PhD student called Gerald Williams). Another interesting question is to ask if it is decidable if a group in this class if cyclic: "If $G/\langle\langle g\rangle\rangle$ is trivial for all $g\in G\setminus{1}$, can I decide if $G$ is cyclic?". I suspect the answer is "no", but can't see how to prove this...……... – user1729 Aug 29 '19 at 09:58
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    Another interesting question is: given a group $G$, can we decide if $G/\langle\langle g\rangle\rangle$ is trivial for all $g\in G\setminus{1}$? (If you could prove that there exists some group $H$ which never embeds as a subgroup of such a group $G$ then this would be a Markov property and hence undecidable. For example, being a simple group is a Markov property as they have decidable word problem, so any group with undecidable word problem can be taken to be your "poison" group $H$. After a glance, this proof seems hard to mimic here.) – user1729 Aug 29 '19 at 10:18
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    Along the same lines of these recent comments by user1729: Algebraic Generalizations of Discrete Groups: A Path to Combinatorial Group Theory Through One-Relator Products by Fine and Rosenberger is a nice book on the topic(definitely on one relator products). –  Aug 29 '19 at 23:43

3 Answers3

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This is very far from true. For example any simple group, no matter the cardinality are counter examples.

There are also examples which are far from simple, for example $\langle x,y \mid x^2,y^3 \rangle$(isomorphic to $\mathrm{PSL}(2, \mathbb{Z})$) and quotienting out by $xy^{-1}$ gives the trivial group since this is like setting an order two and order three equal, which can only happen if both generators are trivial. This generalizes easily to where the powers are coprime.

To justify the above example being far from simple first $\mathrm{PSL}(2,\mathbb{Z})$ has many natural quotients coming from reducing modulo $n$. Further the group belongs to a class of groups called non elementary hyperbolic groups, which in some sense any "complicated enough" word will give a nontrivial quotient group which is still hyperbolic(a sort of generalization of small cancellation theory).

There was a conjecture called the Scott-Wiegold conjecture which asked if you can ever get the trivial group when killing an element in a free product of three cyclic groups. This was solved fairly recently in a paper A proof of the Scott–Wiegold conjecture on free products of cyclic groups by James Howie where he shows that you can never get the trivial group in the above set up.

Also there is a very important question in topology (combinatorial group theory) of a similar flavour: The Kervaire–Laudenbach conjecture conjectures that if $G$ is non-trivial then $G * \mathbb{Z} / \langle \langle w \rangle \rangle $ is non-trivial for all $w\in G * \mathbb{Z}$. This is still open.

user1729
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The answer is "no". The issue is that the normal subgroup $\langle\langle g \rangle \rangle < G$ that is generated by $g$ can be much larger and more complicated than the ordinary subgroup $\langle g \rangle < G$ that is generated by $g$.

For an example, if $K \subset S^3$ is any knot, if $\gamma : [0,1] \to S^3 - K$ is a small loop that goes once around the boundary circle of a small disc which pierces the knot at one point, with $\gamma$ based at a point $p$, and if $g = [\gamma] \in \pi_1(K,p)$ is the corresponding group element, then $\pi_1(K,p) / \langle\langle g \rangle\rangle$ is trivial, despite that $\pi_1(K,p)$ can be a very complicated group.

In addition, unlike the other answers, knot groups can be very far from simple, in fact most of them are relatively hyperbolic groups hence SQ-universal, which roughly means that it has zillions of weird quotient groups.

Lee Mosher
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The answer is no. For example, if $G$ is any non-abelian simple group, then $G/\langle\langle g\rangle\rangle$ is trivial for every $g \neq e$, but $G$ is not abelian.

Shaun
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N. S.
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