 Solution

For an arbitrary subgroup $A$, the full inverse image of its image is $AK$, not generally $A$. In fact, $\theta (g)\in \theta (A)$ exactly when $g=ak$ for some $a\in A,k\in K$. Since $K$ is normal, $AK$ is a subgroup. By the subgroup correspondence for a surjective group homomorphism,Multiplicativity of finite index givesThe map $k(K\cap A)\mapsto kA$ bijects the left cosets of $K\cap A$ in $K$ with those of $A$ in $AK$, since $AK=KA$. ConsequentlyThis need not equal $i$. For example, take $\theta :\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ to be reduction modulo $2$ and $A=3\mathbb{Z}$. Then $[G:A]=3$, but $\theta (A)=H$ has index one. If $A$ contains $K$, however, the denominator is one andFor finite $i$, equality holds exactly when $K\le A$.