# Normal subgroups of an extension of the Higman group

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).

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?
• Presumably you exclude $k_1=k_2=k_3=0$? – მამუკა ჯიბლაძე Oct 23 '15 at 8:13
• 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 – Mark Grant Oct 23 '15 at 9:21
• 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. – H A Helfgott Oct 23 '15 at 16:01
• 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 $G$. – მამუკა ჯიბლაძე Oct 23 '15 at 20:46