Subgroup of index $n$ induces subgroup of size $n$. In working with groups I have come across a lot of special cases of this theorem which seems like it should be true, but I can't seem to prove it (even in the Abelian case): 
Given a group $G$ with $H \triangleleft G$ and $[G:H] = n$, then $\exists K \leq G$ such that $G/H \cong K$, and therefore $|K| = n$. 
Does anyone know if this is true? 

This is not true.  As a counterexample, consider the group $G=SL_2(\mathbb{F}_3)$ and the center $H=\{\pm I\} \triangleleft G$.  Then $G$ has order $24$, but has no subgroup of order $|G|/|H|=12$ (see http://groupprops.subwiki.org/wiki/Special_linear_group:SL(2,3)#Table_classifying_subgroups_up_to_automorphism; probably the quickest way to verify there is no subgroup of order $12$ is observe that such a subgroup would be a normal subgroup of index $2$, so it would need to be a union of conjugacy classes which includes all elements of odd order, but there is no collection of such conjugacy classes whose sizes add up to $12$).
