Prove that there is no homomorphism from $S_5$ onto a group of order $24$.
My solution:
Let $G$ be a group such that $|G|=24.$
Denote $\phi: S_5\to G$.
The normal subgroups of $S_5$ are $S_5,A_5,\{e\}$.
Let $\phi$ be a groups homomorphism.
Denote $\operatorname{Ker(\phi)}=S_5 , A_5, \text{or } \{e\}$
Using the first isomorphism theorem, $S_5/{\rm Ker}(\phi) \cong{\rm Im}(\phi) \implies \frac{|S_5|}{|{\rm Ker}(\phi)|}=|{\rm Im}(\phi)|$.
In case $\operatorname{Ker(\phi)}=S_5 \implies |{\rm Im}(\phi)|=1\neq 24$.
In case $\operatorname{Ker(\phi)}=A_5 \implies |{\rm Im}(\phi)|=2\neq 24$.
In case $\operatorname{Ker(\phi)}=\{e\} \implies |{\rm Im}(\phi)|=5!\neq 24$.
Hence, there is no homomorphism $\phi$.
Is my solution correct?