(Note: this started as a different question that soon changed form, thanks to the answers.)

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) $2\mathbb{Z}/4\mathbb{Z} \ltimes H_4$. 

(a) Can you prove that the normal closure in $G$ of any word of the form $a^{k_1} t a^{k_2} t a^{k_3} t a^{k_4} t$ ($k_1,\dotsc,k_4$ integers, not all $0$) necessarily contains $H_4$? 

(b) Can you prove that the normal closure in $G$ of any set consisting of two distinct words of the form $a^{k_1} t a^{k_2} t a^{k_3} t a^{k_4} t a^{k_5} t$ must contain $H_4$?

Note: I am saying "can you prove this" on purpose; a few lines of GAP code (sent to me by Kate Juschenko) suggest that what I am asking you to prove is in fact true.

(c) Can you prove that the normal closure in $G$ of $(a t)^5$ does *not* contain $H_4$?

(GAP fails to prove that this is false, on KJ's computer and mine, at least. Any further information on the normal closure is welcome.)