# Is $D_8$ a normal subgroup of $S_4$?

I am trying to show that $$D_8$$ is not a normal subgroup of $$S_4$$, I have shown that $$D_8$$ is Subgroup of $$S_4$$.Then, let $$r=\langle(1234)\rangle$$ and $$s =\langle(14)(23)\rangle$$.Then, we see that $$D_8$$ is a subset of $$S_4$$ and since it a group so $$D_8$$ is a subgroup of $$S_4$$.

Proving that $$D_8$$ is not a normal subgroup. In $$D_8$$ there are two elements of order $$4$$.So, we take $$\sigma (1234) \sigma^{-1}$$ which should be in $$H$$.Now conjugacy preserves the cycle structure so we choose $$\sigma$$ in such a way so that conjugate element does not belong to $$H$$.Choosing such a cycle is possible as there are $$6$$ 4 cycles in $$S_4$$ and there are only two of them in $$D_8$$.Since I have shown it for one element, that the conjugate is not in $$D_8$$.Do I need to show it for the other elements?

A subgroup $$H$$ of a group $$G$$ is normal if and only if for all $$g\in G$$ and $$h\in H$$, $$ghg^{-1}\in H$$.
Since you showed that there exists $$g\in S_4$$ such that $$g(1234)g^{-1}\notin D_8$$, it is sufficient to conclude that $$D_8$$ is not a normal subgroup of $$S_4$$.
If we view the elements $$1,2,3,4$$ as vertices of a square, and consider the dihedral group of that square, this gives us one copy of $$D_8$$. We can obtain other copies by reordering.
There are $$6$$ ways to order $$4$$ vertices (assuming we view cyclic shifts such as $$1,2,3,4$$ and $$2,3,4,1$$ as the same ordering), resulting in $$3$$ distinct copies of $$D_8$$ (because "opposite" pairs of orderings such as $$1,2,3,4$$ and $$1,4,3,2$$ result in the same group; just flip the square upside-down to go from one ordering to the other).
These three distinct copies of $$D_8$$ are conjugate to each other, e.g. because they are Sylow $$2$$-subgroups (they have the correct order), and all Sylow subgroups of a given order are conjugate. In particular they are not normal.