We know that when a group $G$ has order $2^k m$, where $m$ is an odd integer, $G$ should have a normal subgroup with order $m$ from here. When $k=1$, this implies the index of the normal subgroup is $2$. However, when $k=2$, $m$ is a prime, can we also find a normal subgroup which has index $2$?
That is to say, when $|G|=4p$, where $p$ is a prime congruent to $1$ or $3$ modulo $4$, does a normal subgroup of index $2$ always exists? I don't know how to construct the contourexample, any idea or solutions are welcomed, thanks!