Alternative way of proving the subgroup of rotations is normal in $\mathbb D_4$

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I've just solved a basic group theory exercise which is: decide if $\{1,r,r^2,r^3\}$ is a normal subgroup of $\mathbb D_4$ (I mean the dihedral group of $8$ elements, not the one of $4$). I've used the following result: if $H$ is a subgroup of index $2$, then $H \lhd G$. Since $|\mathbb D_4:R|=\dfrac{|\mathbb D_4|}{|R|}=\dfrac{8}{4}=2$, then $ R \lhd \mathbb D_4$.

I was looking for an alternative solution to mine without using the concept of index. For example, a direct proof calculating $(r^js^k)r^p(r^js^k)^{-1}$ for arbitraries $r^p \in R$, $r^js^k \in \mathbb D_4$, or an intuitive idea of why the rotations are invariant under conjugation. Thanks in advance.

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The intuition for the subgroup of rotations being normal is actually quite nice in this example, and easy to imagine visually. These are the symmetries of a square, and the element $s$, of order $2$, is a "flip". (Okay, I'm assuming that is what your $s$ denotes.) If you flip a square over, then rotate it, then flip it back again ($s$ followed by some $r^i$ followed by $s^{-1}=s$ again), the net result is the same as if you had just performed a rotation. This shows that the rotation subgroup is invariant under conjugation by $s$. Since it is also, clearly, invariant under conjugation by $r$, and $s$ and $r$ together generate the entire group, it must be normal. This generalises to other dihedral groups too!