Surjectivity problem involving permutations I am preparing for a contest and I have just started learning permutations. I found this interesting problem: Let $S_4$ be the set of all permutations of length $4$. Consider the function $f:S_4 \to S_4$, $f(\sigma)=\sigma^3$. Is the function $f$ surjective?  
 A: There are only 24 permutations to think about. Why not cube them all and see what happens? That's not the fastest way to solve this particular problem, but it's a good way to learn about permutations.
Be sure to learn about cycle notation.
A: Hint : consider $\sigma : (1,2,3,4) \rightarrow (1,4,2,3) $
What is $\sigma^3$? And what is $Id^3$ (identity cubed)?
A: Every permutation is a product of disjoint cycles, so there are only a few categories that an element $\sigma$ of $S_4$ can fall into:
(1) $\sigma$ = the identity permutation, in which case $\sigma^3 = \sigma.$
(2) $\sigma$ is a transposition $(a\;b)$, in which case we also have $\sigma^3=\sigma.$
(3) $\sigma$ is a 3-cycle $(a \; b \; c),$ in which case $\sigma^3$ is the identity.
(4) $\sigma$ is a 4-cycle $(a \; b \; c \;d),$ in which case $\sigma^3$ is $(a\;d\; c\; b),$ another 4-cycle.
(5) $\sigma$ is a product $(a\;b)(c\;d)$ of two disjoint 2-cycles, in which case $\sigma^3$ is $\sigma$ again.
Note that in none of these cases is $\sigma^3$ a 3-cycle, so the function $f$ is not surjective.
