I'm stuck with the following I hope someone could help me.
Let $N$ a normal subgroup of $G$. Show that if $[G:N]=4$, exists a normal subgroup $M$ of $G$ s.t. $[G:M]=2$. My idea: Since $G/N$ has orden 4, either is isomorphic to $\mathbb{Z}/4$ or $\mathbb{Z}/2 \times\mathbb{Z}/2$ and in any case we have a subgroup of order $2$ and for instance the same applies to $G/N$ but from here I'm not sure how to get the normal subgroup $M$ of $G$ which have exactly two cosets. any idea?
I think I got it. $G \to_\pi G/H\to _f\{\text{klein 4 group or cyclic group}\}$. Either if we have a cyclic group or the Klein 4 group (both commutatives) we can find a subgroup of order $2$, say $H$ (which obviously is normal), then $f^{pre}[H]=K\lhd G/H$, and define $M =\pi^{pre}[K]\lhd G$ (because the canonical map is a surjecion). To conclude we claim that $[G:M]= 2$. Since $4=[G:N]=[G:M][M:N]=[G:M]2$, we're done.
I think is correct, thanks
Use the homomorphism theorems. By the correspondence theorem, if $N \trianglelefteq G$ and $\pi:G \rightarrow G/N$ is the canonical projection, then $\pi$ respects indices, factor groups, containment, and normality. Thus, if $H'$ is the image of $H$, then $H'$ is normal in $G':=G/N$ iff $H$ is normal in $G$, and the index $|G':H'|$ equals $|G:H|$. In your case $G/N \cong C_4$, which has a normal subgroup $H'$ of index 2. By the correspondence theorem, the (complete) preimage of $H'$ is a normal subgroup in $G$ and has index 2 in $G$.