Let $G = \mathbb{Z}/4\mathbb{Z} \ltimes H_4$, where $H_4$ is the Higman group and $\mathbb{Z}/4\mathbb{Z}$ acts on $H_4$ in the obvious way (permuting the four standard generators cyclicly). The group $G$ is generated by two elements, $a$ and $t$: here $t$ is a generator of $\mathbb{Z}/4\mathbb{Z}$, and $a$ is such that $t a t^{-1} \cdot a \cdot t a^{-1} t^{-1} = a^2$.

As is well-known, $H_4$ has plenty of normal subgroups (though none of finite index). My question is about normal subgroups of $G$ other than $\{e\}$, $H_4$, $G$ and (thanks to a commenter for reminding me of this last one) $2\mathbb{Z}/4\mathbb{Z} \ltimes H_4$. They do seem to exist, though this is non-obvious (to me). One may wonder how complicated they need to be.

(a) Is there any word of the form $a^{k_1} t a^{k_2} t a^{k_3}$ whose normal closure in $G$ is neither $\{e\}$, $H_4$ nor $G$? My guess is that there isn't one, but how does one prove this?

(b) Is there any word of the form $a^{k_1} t a^{k_2} t a^{k_3} t a^{k_4}$ whose normal closure in $G$ is neither $\{e\}$, $H_4$ nor $G$? Can there be two such words $w_1$, $w_2$ such that the normal closure of $\langle w_1, w_2\rangle$ is still neither $\{e\}$, $H_4$ nor $G$?

(Related comments (on $a^{k_1} t a^{k_2} t^{-1} a^{k_3} t a^{k_4}$, say) are of course also welcome.)

Note: the answers below (as of 24/10/15 at noon) address (a) and also clarify why $G$ has uncountably many normal subgroups. I am still keenly interested in (b).

Note 2: Thank you for all your answers.

Part (c) of the question: is there any word of the form $a^{k_1} t a^{k_2} t a^{k_3} t a^{k_4} t$ whose normal closure in $G$ is neither $\{e\}$, $H_4$, $2\mathbb{Z}/4\mathbb{Z} \ltimes H_4$ nor $G$? Can there be two such words $w_1$, $w_2$ such that the normal closure of $\langle w_1, w_2\rangle$ is still none of the above? Is there any bound on the number of words $w_1, w_2,\dotsc,w_k$ of this form such that the normal closure is still none of the above?

Note: part (c) has become its own question at Quotients of an extension of the Higman group . Please take all discussion there.

  • $\begingroup$ Presumably you exclude $k_1=k_2=k_3=0$? $\endgroup$ Oct 23, 2015 at 8:13
  • $\begingroup$ Are you using the notation $H\triangleright K$ to denote the semi-direct product of $H$ acting on $K$? If so, then the following paper of Usenko might be relevant: link.springer.com/article/10.1007%2FBF01058705 $\endgroup$
    – Mark Grant
    Oct 23, 2015 at 9:21
  • $\begingroup$ Yes, I mean a semi-direct product. I still don't see quite how that (very general) article helps. And yes, I exclude $k_1=k_2=k_3=0$, but the normal closure of that is all of $G$, simply because the Higman-like group with 3 instead of 4 in the definition is trivial. $\endgroup$ Oct 23, 2015 at 16:01
  • $\begingroup$ Sorry for confusion in comments, let me start over again. The kernel of the composite homomorphism $G\twoheadrightarrow\mathbb Z/4\mathbb Z\twoheadrightarrow\mathbb Z/2\mathbb Z$ is the normal subgroup $2\mathbb Z/4\mathbb Z\ltimes H_4$ containing $t^2$, so it contains the normal closure of the latter. Whether it coincides with that normal closure I don't see, but in any case the normal closure of $t^2$ is neither $\{e\}$ nor $H_4$ nor $G$. $\endgroup$ Oct 23, 2015 at 20:46
  • 2
    $\begingroup$ Thanks! Well, neither $\{e\}$ nor a group containing $H_4$, then. (I suspect your normal closure does coincide with $2\math{Z}/4\mathbb{Z} $\endgroup$ Oct 24, 2015 at 10:56

4 Answers 4


If $G$ is a group $H\subset G$ a subgroup of finite index and $H$ is SQ-universal (this means that every countable group embeds into a quotient of $H$) then so is $G$ (the easier converse also holds). This is due to P. Neumann (The SQ-universality of some finitely presented groups. Collection of articles dedicated to the memory of Hanna Neumann, I. J. Austral. Math. Soc. 16 (1973), 1–6.)

Consequence: since Higman's group is SQ-universal, your group $G$ is also SQ-universal; in particular it admits $2^{\aleph_0}$ distinct normal subgroups.

Variant: $H$ has the property that any pair of nontrivial normal subgroups has a nontrivial intersection. This is because it has a faithful action of general type (i.e., unbounded and not fixing any endpoint or pair of endpoints) on a tree, so any nontrivial normal subgroup also has an action of general type and has a trivial centralizer, while any pair of normal subgroups with trivial intersection should centralize each other. So picking any nontrivial normal subgroup $N$, intersecting its four $G$-conjugates provides a nontrivial normal subgroup of $G$.


The answer to part (a) of your question is that the normal closure of $a^{k_1}\cdot t\cdot a^{k_2} \cdot t \cdot a^{k_3}$ is always equal to the normal closure of $a$ and $t^2$. This is because the quotient of $G$ obtained by adding a relation of the type written in part (a) always gives $a=1$!

Let me continue to work from now on in the quotient (and by abuse of notation with the same letters $a$ and $t$.) The added relation can be written in the form $t\cdot a^m\cdot t \cdot a^n=1$, so $t\cdot a^m = a^{-n} \cdot t^{-1}$. Then $b=t\cdot a\cdot t^{-1}=(t\cdot a^m)\cdot a\cdot (t\cdot a^m)^{-1}$, so also $b=(a^{-n} \cdot t^{-1})\cdot a\cdot (t\cdot a^n)$ and hence $t^{-1}\cdot a \cdot t = a^n \cdot b\cdot a^{-n}$.

It follows that not only does one have $b\cdot a\cdot b^{-1}=a^2$, but (after conjugating all letters in the last written relation through by $a^{-n}$) we can hence deduce $(t^{-1}\cdot a\cdot t)\cdot a\cdot (t^{-1}\cdot a\cdot t)^{-1}=a^2$. Now conjugating (every letter of) this last relation through by $t$ we obtain $a\cdot b\cdot a^{-1}=b^2$.

It is easy to see that $b\cdot a\cdot b^{-1}=a^2$ and $a\cdot b\cdot a^{-1}=b^2$ imply $a=1$.

P.S.The above combinatorial argument shows that in the one-relator group $B$ (generators $a,t$ with $bab^{-1}=a^{2}$ and $b=tat^{-1}$) the normal closure of the word in part (a) contains $a$. A similar type of argument (using quotients) gives the same result for the word $a^{k_{1}}ta^{k_{2}}t^{-1}a^{k_{3}}ta^{k_{4}}$ mentioned after part (b). Now $B$ is also SQ universal (e.g. since it has $G$ as a quotient.) I agree with Harald the problem starts to become more interesting with the words $a^{k_{1}}ta^{k_{2}}ta^{k_{3}}ta^{k_{4}}$ of part (b). To begin with (and I hope this may be an easier question) are there infinitely many non-isomorphic quotients of $B$ using these words as relators?


The answer to part (b) is that the normal closure of the word given in part (b) is always equal to $G$.

To see this, consider the quotient of $G$ by adding in this case a relation $a^{k}ta^{l}ta^{m}t=1$.

We will show that $a=1$, and then as $t^{4}=1$ it will also follow that $t=1$.

Let $b=tat^{-1},c=tbt^{-1},d=tat^{-1}$ with $bab^{-1}=a^{2},cbc^{-1}=b^{2},dcd^{-1}=c^{2},ada^{-1}=d^{2}$.

Write $a^{k}ta^{l}t^{-1}t^{2}a^{m}t^{-2}t^{3}=1$ so that $t=a^{k}b^{l}c^{m}=b^{k}c^{l}d^{m}=c^{k}d^{l}a^{m}=d^{k}a^{l}b^{m}$.

Case (1): Assume one of $k,l,m$ $\geqq0$ ; wlog assume $m\geqq0$.

Since $b^{m}ab^{-m}=a^{2^{m}}$ we have $b=tat^{-1}=$$d^{k}a^{l}b^{m}ab^{-m}a^{-l}d^{-k}=d^{k}a^{2^{m}}d^{-k}$. But $a^{2^{m}}da^{-2^{m}}=d^{2^{2^{m}}}$ so $b=d^{k}d^{-k2^{2^{m}}}a^{2^{m}}=d^{u}a^{2^{m}}$ where $u=k(1-2^{2^{m}})$.

Then $a^{2}=bab^{-1}=d^{u}a^{2^{m}}aa^{-2^{m}}d^{-u}=d^{u}ad^{-u}$. Using $ada^{-1}=d^{2}$ we have $a^{2}=d^{u}d^{-2u}a=d^{-u}a$ so that $a=d^{-u}$ and in particular $da=ad$. Then $ada^{-1}=d^{2}$ gives $d=1$.

Since $a$ and $d$ are conjugate $a=1$.

Case (2): Suppose $k,l,m$ all negative and not $0$.

Then $b=tat^{-1}=c^{k}d^{l}a^{m}aa^{-m}d^{-l}c^{-k}=c^{k}d^{l}ad^{-l}c^{-k}=c^{k}d^{l}d^{-2l}ac^{-k}=c^{k}d^{-l}ac^{-k}$ using $ad^{-l}a^{-1}=d^{-2l}$. Thus $c^{-k}bc^{k}=d^{-l}a$.

Since $-k$ is positive $c^{-k}bc^{k}=b^{2^{-k}}=b^{w}$ with $w=2^{-k}$. So $b^{w}=d^{-l}a$.

However, $b^{w}ab^{-w}=a^{2^{w}}$ and we see $d^{-l}ad^{l}=a^{2^{w}}$. Using $ad^{l}a^{-1}=d^{2l}$ we have $d^{-l}d^{2l}a=a^{2^{w}}$. Thus $d^{l}=a^{v}$ where $v=2^{w}-1$.

Consequently, from $a^{v}da^{-v}=d^{2^{v}}$ we can now deduce $d=d^{2^{v}}$ or $d^{2^{v}-1}=1$.

In any case $d$ has finite order, $n$. Since $d$ and $a$ are conjugate they have exactly the same order $n$.

It is now a standard argument originating with G. Higman that since $ada^{-1}=d^{2}$ we also have $d^{2^{n}-1}=1$ and so $n$ divides $2^{n}-1$. Then a simple number theory argument shows $n=1$.

Thus $a=1$.

  • $\begingroup$ Thanks! I guess you did not use the condition $t^4=e$ at all to obtain $a=e$? $\endgroup$ Oct 26, 2015 at 10:43
  • $\begingroup$ Ah, no, I see you do it tacitly at the very beginning. This makes me curious. If we remove the condition $t^4=e$ (but keep the other relations), must the group still be trivial? $\endgroup$ Oct 26, 2015 at 11:10
  • $\begingroup$ @HAHelfgott: I think the proof of $a=1$ here is still good when we don't have $t^4=1$. In line 5 of this answer, we have $t^{-3} = a^kb^lc^m = \dots$ instead of $t= \dots$, and by the other relations, we also have $b = t^{-3} a t^3$, which can be used instead of $b=tat^{-1}$. The rest of the proof does not depend on $t^4 = 1$, as far as I can see. But we have only the relation $t^3=1$, if we omit $t^4=1$, so we get a cyclic group of order $3$, not the trivial group. $\endgroup$ Oct 26, 2015 at 11:38

Let me try to answer the first query mentioned in part (c).

Let $G=<a,t|bab^{-1}=a^{2},b=tat^{-1},t^{4}=1>$ and $H=<a,b,c,d|bab^{-1}=a^{2},cbc^{-1}=c^{2},dcd^{-1}=c^{2},ada^{-1}=d^{2}>$ the Higman group, where $b=tat^{-1},c=t^{2}at^{-2},d=t^{3}at^{-3}$.

Consider the word $a^{-1}t^{2}at^{2}$, one of your words, and let $N$ be the normal closure in $G$ of $a^{-1}t^{2}at^{2}$.

Then $N$ is equal to the normal closure in $H$ of the two words $a^{-1}c$ and $b^{-1}d$ , so we can proceed to work in $H$.

Now $H$ is a free product with amalgamation (fpa) of $A=<a,b,c|bab^{-1}=a^{2},cbc^{-1}=b^{2}>$ and $B=<c,d,a|dcd^{-1}=c^{2},ada^{-1}=d^{2}>$ , amalgamating $a,c$ from $A$ with $a,c$ from $B$.

In turn $A$ is a fpa of $<a,b|bab^{-1}=a^{2}>$ and $<b,c|cbc^{-1}=b^{2}>$ where the infinite cyclic $b's$ are amalgamated, and $B$ is a fpa of $<c,d|dcd^{-1}=d^{2}>$ and $<a,d|ada^{-1}=d^{2}>$ with $d's$ amalgamated.

Firstly $N$ is not the trivial group because $a^{-1}c$ is in "standard form" in the fpa $B$.

Secondly $H/N\cong$$<a,b,c|bab^{-1}=a^{2},cbc^{-1}=b^{2}>(=A)$ as can be seen by adding$a=c,b=d$ to the relations of $H$. This shows $N$ is not equal to $H$ since $A$ is a non-trivial fpa.

  • $\begingroup$ In the last step, don't you get $H/N = \langle a,b|b a b^{-1} = a^2, a b a^{-1} = b^2\rangle$? And isn't that trivial? $\endgroup$ Oct 30, 2015 at 11:59
  • $\begingroup$ Another example to work on (given that I seem to have ungraciously broken yours): what about $atatat^2$? I suspect its closure does contain $H$. Is the same true for every word of the form $a^{k_1} t a^{k_2} t a^{k_3} t^2$? $\endgroup$ Oct 30, 2015 at 12:20
  • $\begingroup$ Of course. How careless of me. I had previously been looking hard at $(at)^{4}=abcd$, so that in $H/N$ just one relator was added, but could not decide whether the group was trivial or not. All I could say was that $H/N$ is generated by 2 elements. Perhaps someone with GAP experience might be more successful if indeed this is the trivial group. $\endgroup$
    – andrew
    Oct 30, 2015 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.