Amount of homomorphisms $S_5$ to $C_6$ I think I understand the amount  of homomorphisms $C_6$ to  $S_5$:
Because $C_6$ is cyclic, we only have to look where we can send $\langle a\rangle$, because $f(\langle a\rangle)$ must have order $1, 2, 3$, or $6$ we have:
-for $1$: trivial homomorphism.
-for $2$: the ten $(xy)$ cycles in $S_5$, the fifteen $(xy)(ab)$  cycles in $S_5$
-for $3$: the twenty  $(xyz)$ cycles in $S_5$
-for $6$: the twenty $(xyz)(ab)$ cycles in  $S_5$
so total $= 66$
the other way around I find more difficult.
Can anyone help?
Bonus question: is the size of the conjugacy class of an element in the group always the amount of elements of the same shape? so in $S_5$, $(12)$ has conjugacy class size 10, so there are ten $(xy)$ in $S_5$?
 A: The kernel of a homomorphism $f: S_5 \rightarrow C_6$ must be a normal subgroup of $S_5$. There are only three such: $S_5, A_5$ and $\{1\}$. Since $|S_5| > |C_6|$ we cannot have $\ker f = \{1\}$. If $\ker f = S_5$ then $f$ just sends everything to the identity. If $\ker f = A_5$ then by first isomorphism theorem im$f \cong C_2$, which there is only one copy of inside $C_6$, so we have $f(a) = g^3$ if $a \notin A_5$, $f(a) = 1$ if $a \in A_5$.
A: Suppose $h:S_5\to C_6$ is a homomorphism.  $S_5$ has elements of order $5$ (namely cycles), which get mapped by $h$ to the identity, since that's the only element of $C_6$ whose order divides $5$.   But the elements of order $5$ in $S_5$ generate the whole alternating subgroup $A_5$, so all of $A_5$ is sent by $h$ to the identity.  So $h$ factors as the projection $S_5\to S_5/A_5$ followed by a homomorphism from the $2$-element cyclic group $S_5/A_5$ into $C_6$.  Since $C_6$ has only two elements whose order divides $2$, there are just $2$ such homomorphisms.
