4
$\begingroup$

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 $G\cong A_5$?

$\endgroup$
8
$\begingroup$

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 $\phi(G)\subseteq A_6$.

$\endgroup$
  • $\begingroup$ Wonderful solution! I just asked myself why does G have six Sylow 5-subgroups and not just one, and I realised G wouldn't be simple in that case. :) $\endgroup$ – Leaning Oct 23 '18 at 19:47
  • 3
    $\begingroup$ How do we know $\phi(G)$ will be contained in $A_6$? $\endgroup$ – Al Jebr Oct 23 '18 at 19:55
  • 3
    $\begingroup$ @AlJebr for otherwise $\phi(G)\xrightarrow{sign}\{\pm 1\}$ would define a normal subgroup of $\phi(G)$. $\endgroup$ – Leaning Oct 23 '18 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.