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.

isall of $G$, simply because the Higman-like group with 3 instead of 4 in the definition is trivial. $\endgroup$ – H A Helfgott Oct 23 '15 at 16:012more comments