Question: I started with the claim that: If $n$ is even, $n^k$ is even.

But I have only gotten this far:
$$n = 2a,$$
$$n^k = (2a)^k,$$
$$n^k = 2^k a^k.$$

I am trying to reach the format where I can show that
$$n^k = 2a,$$
so that $n^k$ is even.

Answer: 
We want $S\_3$ to act transitively on 6 points. Now $|S\_3| = 6$ and a group acts transitively on its own elements.

Let us make this explicit, label each of the elements of $S\_3$ with a number from 1 to 6:

* $\color{green}1 = (\color{blue}{})$
* $\color{green}2 = (\color{blue}{1\,2})$
* $\color{green}3 = (\color{blue}{1\,3})$
* $\color{green}4 = (\color{blue}{2\,3})$
* $\color{green}5 = (\color{blue}{1\,2\,3})$
* $\color{green}6 = (\color{blue}{1\,3\,2})$

We understand the action of $S\_3$ on the blue objects and we can use it to write out the action of the green objects. For example $(1\,2)$ acts like so:

```
gap> (1,2)*();
(1,2)      { which is green 2 }
gap> (1,2)*(1,2);
()         { which is green 1 }
gap> (1,2)*(1,3);
(1,2,3)    { which is green 5 }
gap> (1,2)*(2,3);
(1,3,2)    { which is green 6 }
gap> (1,2)*(1,2,3);
(1,3)      { which is green 3 }
gap> (1,2)*(1,3,2);
(2,3)      { which is green 4 }

```

Therefore $\sigma (1\,2) = (\color{green}{1\,2})(\color{green}{3\,5})(\color{green}{4\,6})$ and you can compute the $S\_6$ permutation representation of each element of $S\_3$ in a similar way. Giving you a subgroup with a transitive action on 6 points.
A subgroup of order $6$ of $S\_6$, say $\Sigma$, which acts transitively on the set $X:=\{1,\dots,6\}$, must have all the stabilizers trivial (Orbit-Stabilizer Theorem): $ \forall i\in X, \operatorname{Stab}(i)=\{Id\_X\}$, or equivalently: $ \forall i\in X, \sigma\in \Sigma\mid\sigma(i)=i\Rightarrow \sigma=Id\_X$. Therefore, we need to look at the elements of $S\_6$ which *do not fix any* element of $X$, namely:

1. $6$-cycles;
2. products of a $2$-cycle with a disjoint $4$-cycle;
3. products of $2$ disjoint $3$-cycles;
4. products of $3$ disjoint $2$-cycles.

Now, narrowing the survey to $\Sigma$'s isomorphic to $S\_3$, options 1 and 2 are ruled out because $S\_3$ is made of (further than the identity) $3$ elements of order $2$ and $2$ elements of order $3$. Therefore, any closed $\Sigma\subseteq S\_6$ made of the identity, three products of $3$ disjoint $2$-cycles, and two products of $2$ disjoint $3$-cycles, is a transitive subgroup isomorphic to $S\_3$.