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?
On
In case like you mention you would be interested in the notion of (short) exact sequence. In the case you mention one has the exact sequence: $$0 {\rightarrow} H \stackrel{i}\rightarrow G \stackrel{\pi}\rightarrow G/H \rightarrow 0 $$ In some cases the sequence is split (see split extension). This means that there exists a homomorphism $\sigma: G/H \rightarrow G$ such that $\pi \circ \sigma = 1_{G/K} $. If this is the case then your "theorem" is true, with $K = \sigma(G/K)$. Moreover $G$ will be expressible as a semi-direct product $G = H \rtimes K$. But there are cases when the sequence does not split as in the examples given in the comments. The simplest abelian case I know is: $$ C_2 \rightarrow C_4 \stackrel{a^2}\rightarrow C_2 $$.
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 $12$).