I want to prove that there does not exist a surjective group algebra homomorphism from $FS_5$ (the group algebra (or group ring) of the symmetric group, $S_5$, over the field $F$) to $M_2(F)$, where $F$ is the field $\mathbb{Z}_7$ and $M_2(F)$ denotes the matrix algebra of $2\times 2$ matrices over the field $F$.
Using representation theory, I have been able to solve this problem by calculating the irreducible representations of $S_5$. However, I want to show it in the following way:
I know that for $a=(1,2)$, $b=(1,2,3,4,5)$, $$S_5=\langle a,b \mid a^2,b^5,(ab)^4,(bab^{-2}ab)^2 \rangle.$$ Let $\phi$ be an onto $F$-algebra homomorphism, i.e. $$\phi:FS_5\to M_2(F).$$ Note that $|FS_5|=7^{120}$ and $|M_2(F)|=7^4$. Since $S_5$ is generated by two elements, $\phi$ is defined if it is known on $a$ and $b$. Further, as $M_2(F)$ has no elements of order $5$ or no element whose order divides $5$ (except identity), we have that $$\phi(b)=Id_{2\times 2}.$$ Now, as $a^2=(ab)^4=e$, we must have $\phi(a)$ be such that its order divides $2$ and $4$ (since $\phi(a)=\phi(ab)$ as $\phi(b)=Id_{2\times 2}$). Take $$\phi(a)=\begin{bmatrix}0&1\\ 1&0\end{bmatrix}.$$ Clearly $\phi(a)$ has order $2$. As, by assumption, $\phi$ is onto, we must have $$|\text{Kernel}(\phi)|=7^{116}.$$
How to reach at a contradiction? Please help.