# 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?

## 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!

• 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) – Paul Plummer Aug 26 '19 at 18:55
• Isn’t $S_3$, being the normal closure of $\langle (12)\rangle$ a counterexample? – Nicky Hekster Aug 26 '19 at 19:01
• 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 – Paul Plummer Aug 26 '19 at 19:02
• @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) – Paul Plummer Aug 28 '19 at 16:51
• 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). – Paul Plummer Aug 29 '19 at 23:43

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.

• I've edited in the name of the conjecture. I hope you don't mind. – user1729 Aug 27 '19 at 13:35
• @user1729 Thanks! I don't mind at all. – Paul Plummer Aug 27 '19 at 13:44

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.

• I'm insufficiently informed about knots to accept this answer. Thank you nonetheless. – Shaun Aug 26 '19 at 19:06
• May I have a reference for relatively hyperbolic groups being SQ-universal, please, @LeeMosher? – Shaun Sep 11 '19 at 22:09
• It's the second bullet of that SQ-universal link, or reference [3] at the bottom of the whole page. – Lee Mosher Sep 11 '19 at 23:51

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.

• For completeness, $A_5$ is a non-abelian simple group :p – diracdeltafunk Aug 26 '19 at 19:52