In a somewhat different direction from Alireza: the conjecture is true for a large family of groups, including all abelian groups and many supersolvable groups.
Let me start with the abelian case. Pick an element $g$ of highest possible prime-power order in an abelian group $G$. Then $\langle g \rangle$ has a complement: that is, there is a subgroup $K$ such that $K \langle g \rangle = G$ and $K \cap \langle g \rangle = 0$. In particular, $K \cong G / \langle g \rangle $ by the Isomorphism Theorems, and for any subgroup $X$ with $\langle g \rangle \subseteq X$ there is a corresponding subgroup $X \cap K$ which does not contain $\langle g \rangle$.
In fact, the same argument applies to any $G$ and prime-power order element $g$ if 1) $g$ generates a normal subgroup, and 2) we can find a complement $K$ to $\langle g \rangle$ in G. In this situation, $[\langle g \rangle, G] \cong [1,K]$.
When can we find a complement to $\langle g \rangle \triangleleft G$? Hall showed that if $G$ is supersolvable with all Sylow subgroups elementary abelian then all subgroups are complemented, so this condition would suffice. Gaschütz showed that a normal $p$-subgroup $H$ is complemented in $G$ if and only if $H$ is complemented in a Sylow $p$-subgroup containing it. It would thus suffice for $G$ to be supersolvable with an elementary abelian Sylow $p$ subgroup for $p$ the largest prime dividing $| G |$. (Since a supersolvable group has a normal subgroup of largest-dividing-prime order.)