Surjective Homomorphism Symmetric group For $G=S_4$ i'm having a bit of trouble following the solution.
For the blue underline I was wondering if there is a strategy for spotting this relatively quickly.
For the green underline I understand why $x^{-1}gx$ is even but I don't understand why it must be of order 2 and why that must mean it's in V again, when there are other elements of order 2 in $S_4$ i.e $(12)$.
Regarding the yellow part I understand that all non-identity rotations of $D_6%$ are of order 3 and non-identity are of order 2 and  $y^{-1}xy=x^{-1}$ is the equation relating elements in $D_6$, but i'm confused as to whether the choices of $x$ and $y$ given allow for this.
For the blue part, do I have to check both $(123)$ and $(132)$ or is just one sufficient?


 A: I'm not fully sure what you're asking for the first blue underline and the yellow underline, but will explain the others.
The key word here for the green underline is that the permutation is even and of order 2. It should be easy to see that conjugation preserves both order of an element and whether or not it's even - think about the homomorphism $\varepsilon : S_n \rightarrow C_2$ given by $\varepsilon(\sigma) = \mathrm{sign}(\sigma)$. $(12)$ - whilst of order 2 - is not even, so cannot be conjugate to an element in $V$.
In fact, two elements in $S_n$ are conjugate if and only if they have the same cycle type, and in $S_4$, the conjugacy class of elements of cycle type $2^2$ (my notation for two 2-cycles) has size 3. So you've accounted for all such elements in that conjugacy class, and $V$ is a disjoint union of conjugacy classes ($V = Cl_G(1) \sqcup Cl_G((12)(34))$) and hence normal (the last sentence should make it easy to see why $V$ is closed under conjugation by elements of $G$).
For the second blue underline, you need to check both. We know that conjugation preserves the order (and cycle type) of an element, so for any $x \in G$, $x^{-1}(123)x$ is going to be a 3-cycle. Since the only elements of order 3 in $N$ are necessarily 3-cycles, we need to find an $x\in G$ that conjugates one of these 3-cycles (in this case $(123)$) out of $N$; i.e., an $x$ such that $x^{-1}(123)x \neq (123), (132)$.
