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? 
 A: 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 $12$).
A: 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:  $$ C_2 \rightarrow C_4 \stackrel{a^2}\rightarrow  C_2 $$.
