Symmetric Group $S_3$ with mapping $\psi: S_3 \to S_3$ with $x \mapsto x^3$. Prove $\psi$ is not onto.

Question

Let $S_3$ be the symmetric group on 3 letters. Consider the mapping

$\psi : S_3 \rightarrow S_3 $ that sends $x \mapsto x^3$

Prove that $\psi$ is not onto.

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I am having trouble trying to figure out what to do to finish the proof as I just have the opening argument of

Let $x,y \in S_3$ We wish to show that $\psi (x) = y $ such that x is a pre-image of y

Do I just say that $x^{1/3}$ would have to be the pre-image but since this is a permutation group the cube power will not give us y?

Answer 1

The map $x\mapsto x^3$ cannot be onto, since the $3$-cycles are not cubes in $S_3$. Indeed, assume that $x^3=(123)$. Then $x$ is not a transposition, and also no $3$-cycle, because $(123)^3=(132)^3=id$. So there is no element $x$ in $S_3$ with $x^3=(123)$.

Answer 2

This group is small enough where, if you had no other options or ideas, you could just brute force check it. Take elements, cube them, and see what you get. As soon as you double-up, it can't be onto anymore.

Answer 3

The map $f:S_3 \to S_3$ is onto if and only if it is injective (that is one-to-one), as the cardinality of the domain and co-domain are the same.

But $f(x)=x^3$ is not one-to-one because for $a=(1 2 3)$ and $b = (2 3 1)$ we have $f(a)=f(b)=e$.

Answer 4

HINT.-Try to verify that $$((1)(2)(3))^3=(1)(2)(3)=(123)^3$$ This is enough because $S_3$ being finite a function of $S_3$ into itself that is not injective can not be surjective.