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

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