Homomorphism $f:S_{2009} \rightarrow S_{2009}$

Question

Is there a non-trivial homomorphism $f:S_{2009} \rightarrow S_{2009}$ with $(1,2)$ in the kernel?

Are there more than $2009$ different homomorphisms $f:S_{2009} \rightarrow S_{2009}$ with $(1,2,3)$ in the kernel?

Prime factorization: $2009 = 7^2*41$

Tips and/or answers are welcome, thanks in advance!

Answer 1

If the kernel contains $(1\;2)$, it also contains every conjugate of it, that is every transposition, hence also every product of transpositions, i.e. all of $S_{2009}$. Remark: Here, $2009$ can be replaced by any $n\ge 2$.

If the kernel contains $(1\;2\;3)$ it also contains every 3-cycle and products therof, including $(1\;2\;3)(1\;2\;4)=(1\;3)(2\;4)$ and again conjugates and products, hence all even permutations. So if the homomorphism is nontrivial the kernel must be $A_{2009}$, the image cyclic of order $2$. There are more than $2009\choose 2$ subgroups of order $2$ in $S_{2009}$, as there is already one for each two-cycle $(a\;b)$. Together with th etrivial homomorphism this gives us enough different homomorphisms. Remark: So here $2009$ can be replaced by any $n\ge 3$ (and in fact for $n=3$ there are exactly ${3\choose 2}=3$ subgroups of order $2$, but note that we still have the trivial homomorphism)

Answer 2

Every non-trivial homomorphism of $S_n$ ($n>4$) has the kernel $A_n$, so $ (1,2)$ is not in the kernel.

Answer 3

Hint: What are the normal subgroups of $S_{2009}$?