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Since $SL(2,5)$ has a subgroup of index $5$, I can use the left coset action to define a homomorphism between $SL(2,5)$ and $S_5$. How can I find the kernel and the image of this homomorphism?

Progress

The kernel would be a normal subgroup of $SL(2,5)$. Other than the trivial ones there's the cyclic group of order $2$, but I'm not sure these are all of them.

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    $\begingroup$ Well, what are the possible kernels (i.e. what are the normal subgroups)? $\endgroup$ Jan 8 '15 at 13:46
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    $\begingroup$ @TobiasKildetoft Other than the trivial ones there's the cyclic group of order 2, but I'm not sure these are all of them. $\endgroup$
    – highlander
    Jan 8 '15 at 13:57
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    $\begingroup$ $S_5$ has a trivial center. $\endgroup$ Jan 8 '15 at 14:08
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    $\begingroup$ related: Elementary isomorphism between $PSL(2,5)$ and $A_5$ etc $\endgroup$
    – Grigory M
    Jan 8 '15 at 14:09
  • $\begingroup$ This is appreciated, but I still can't quite see it. $\endgroup$
    – highlander
    Jan 11 '15 at 22:25

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