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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 $a=1$.

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 in part (b). Now B is also SQ universal (e.g. since it has G as a quotient.) I agree 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 is an easier question) are there infinitely many non-isomorphic (perhaps one should require non-epimorphic) quotients of B using these words as relators?

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 (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 $a=1$.

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 $a=1$.

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 in part (b). Now B is also SQ universal (e.g. since it has G as a quotient.) I agree 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 is an easier question) are there infinitely many non-isomorphic (perhaps one should require non-epimorphic) quotients of B using these words as relators?

My apologies about the lack of type-setting software!

The answer to part (a) of your question is that the normal closure of a^k1.t.a^k2.t.a^k3 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.a^m.t.a^n=1, so t.a^m=a^-n.t^-1. Then b=t.a.t^-1=(t.a^m)a(t.a^m)^-1, so also b=(a^-n.t^-1).a.(t.a^n) and hence t^-1.a.t=a^n.b.a^-n.

It follows that not only does one have b.a.b^-1=a^2, but (after conjugating all letters in the last written relation through by a^-n) we can hence deduce (t^-1.a.t)a(t^-1.a.t)^-1=a^2. Now conjugating (every letter of) this last relation through by t we obtain a.b.a^-1=b^2.

It is easy to see that b.a.b^-1=a^2 and a.b.a^-1=b^2 imply a=1.

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 $a=1$.