Given a group $H$, define $d(H)$ to be the minimum number of generators of $H$. Define $m(G)$ as: $$m(G) = \max d(A) : \text{$A \leqslant G$ and $A$ is abelian}$$ The Thompson subgroup $J(G)$ is defined as: $$J(G) = \langle A \mid \text{$d(A)=m(G) $ and $A$ is abelian } \rangle$$
The Quasidihedral Group of Order $16$ is defined as: $$QD_{16} = \langle \sigma,\tau \mid \sigma^8=\tau^2=1 , \sigma \tau = \tau \sigma^3 \rangle$$
I am to show that $J(QD_{16})$ is a subgroup of order $8$. However, I am getting this subgroup to be $J(QD_{16})$ and I am not able to spot my mistake.
First, we claim that $m(QD_{16})=2$. One example for such an abelian subgroup $A$ is $\langle \tau , \sigma^4 \rangle$. To prove that $m(QD_{16})<3$, we can see that an example for such a $A$ must satisfy $|A| \geqslant 2^{m(QD_{16})}$ due to the decomposition of elements of $A$ in terms of generators. Since $QD_{16}$ is not abelian, if we assume the contrary, then $m(QD_{16})=3$. We can clearly only have at most one of the generators of the form $\sigma^x$. Thus, this forces two generators of the form $\tau\sigma^a$ and $\tau\sigma^b$. Since these commute, we can see that $a \equiv b \pmod{4}$, which forces $a \equiv b+4 \pmod{8}$ for the generators to be distinct. The final generator must then be of the form $\sigma^x$, but this would force $2x \equiv 0 \pmod{8} \implies x=4$. But $\sigma^4$ is already generated by the other two generators, which is a contradiction. Thus, $m(QD_{16})=2$.
Now, we can see that both $\langle \tau,\sigma^4 \rangle$ and $ \langle \tau\sigma,\sigma^4 \rangle$ are abelian groups with two generators. Hence: $$\langle \langle \tau, \sigma^4 \rangle \langle \tau\sigma, \sigma^4 \rangle \rangle \leqslant J(QD_{16})$$
However, clearly $\tau \cdot \tau \sigma = \sigma$. As we have generated both generators of $QD_{16}$ as elements, this concludes: $$J(QD_{16})=QD_{16}$$ Where is the fallacy? Have I misunderstood any definition? Thank you in advance.