For $N \unlhd S_4$, $\sigma = (12)$, is $H=N \cup N\sigma$ a subgroup?

Question

Tl;dr - that first & last 2 lines.

I'm trying to answer the above question wisely (i.e. without checking naively that H is a closed subset), for $N=\{(1), (12)(34), (13)(24), (14)(23)\}$.

For the first 6 members, I managed to prove that their products are within H: I know already that N is a normal subgroup (since the product of two permutations of the structure (ab)(cd) is a permutation of the same struture, all its conjugates are of the same structure as well, and N contains all the elements of this structure).
Now, $H = N \cup \{(12), (34), (13)(24)(12)=(1423), (14)(23)(12)=(3241)\}$. Every element in H has an inverse in H (the last two elements are one each other's inverses, and all of the rest are elements of order 2).
I know that N is closed, and $\forall n \in N,\ n(12) \in N(12) \subset H,\ (12)n = (n(12))^{-1} \in H$.
For (13): $(12)(34) \in N \rightarrow \forall n \in N,\ n(12)(34) = n' \in N$, so $n' = n(34)(12)=n(12)(34)$. We also know $n'(12) \in H \rightarrow n(34) = n'(12) \in H$.
We can conclude that for $(34)n$ as we did for $(12)n$.

For the last two terms, let's call one of them x, I managed to prove only that $nx \in H$. Since $x=n'(12)$ for some $n'\in N$, then $\forall n \in N,\ nx = (nn')(12) \in N(12) \subset H$.
Also: $(12)x=(12)n'(12) \in N$ because is conjugate, $x(12)=n' \in N$, $x(34) = n'((12)(34)) \in N$. I can't find how to show that for $xn$ as well, nor for $(34)x$. Is there any "abstract" way to do that, beside multiplying and checking?

Thanks a lot

Answer 1

Hint: Since $N\sigma = \sigma N$, any expression of the form $n\sigma \in N\sigma$ can be written as $\sigma n'$ for $n' \in N$. Use this fact to show that $N\cup N\sigma$ is a subgroup.

Answer 2

Work in $G/N$. Obviously we have that $(N\sigma)^2 = N$, as $\sigma^2 = e$. Therefore $\langle N\sigma\rangle = \{N, N\sigma\} < G/N$. Now using the canonical homomorphism $\gamma: G \to G/N$ we have that $\gamma^{-1} [\langle N\sigma\rangle]$ is a subgroup of $G$. But obviously $\gamma^{-1} [\langle N\sigma\rangle] = N \cup N\sigma$, as two cosets are nonintersecting.