# If $G$ is a simple group of order $60$, how can we there exists some $G\to A_6$ not trivial without using the fact that $G\cong A_5$?

If $$G$$ is simple, any homomorphism out of $$G$$ must be trivial or injective since the kernel of a homomorphism is a normal subgroup.

If $$G$$ is simple of order $$60$$, how can I show that there exists an injective map from $$G\to A_6$$ without using the fact that $$G\cong A_5$$?

A simple group $$G$$ of order $$60$$ will have six Sylow $$5$$-subgroups. Then $$G$$ acts transitively on them (by conjugation), so there's a homomorphism $$\phi:G\to S_6$$ with transitive image. From the simplicity of $$G$$ it readily follows that $$\phi$$ is injective, and $$\phi(G)\subseteq A_6$$.
• How do we know $\phi(G)$ will be contained in $A_6$? – Al Jebr Oct 23 '18 at 19:55
• @AlJebr for otherwise $\phi(G)\xrightarrow{sign}\{\pm 1\}$ would define a normal subgroup of $\phi(G)$. – Leaning Oct 23 '18 at 20:05