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).
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.
 A: 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?
A: 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$.
A: 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$.
A: 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. 
