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I need to find the the possible homomorphisms from $A_n$ (Alternating group) for $n \geq 5$ to $C^*$ (the group of non-zero complex numbers).

My take:

Fact: $A_n$ for $n \geq 5$ is simple

For any group homomorphism from a simple group to an arbitrary group, the kernel being normal is either trivial or the group itself.

Also we see that even cycles can not map to -1 but only to 1.

So the kernel is the group $A_n$ for $n \geq 5$ itself which in turn implies the homomorphism to be trivial by the first isomorphism theorem.

Is my argument valid?

Any suggestions, improvements or an easier approach would be of much help.

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Your argument is not correct. Of course there can be an injective group homomorphism from a finite group into an infinite one. Take the inclusion from$\{-1,1\}$ into $\mathbb{C}\setminus\{0\}$, for instance.

However, $\mathbb{C}\setminus\{0\}$ is abelian, and therefore, the kernel cannot be $\{e\}$. Sinece $A_n$ is simple, then the kernel has to be the whole $A_n$. So, the only such homomorphism is the trivial one.

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    $\begingroup$ @ShatabdiSinha Re the first comment in José's answer, the (oriented) symmetries of the dodecahedron define an injective group homomorphism $A_5 \hookrightarrow GL(3, \Bbb R)$. $\endgroup$ Nov 11 '18 at 18:51

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