Let $M,N\subset G$ be subgroups of an abelian group $G$ with $M\neq N$ and $M \cong N$ by an isomorphism $\phi: M\rightarrow N$.
I'm trying to show that there exists an automorphism $\psi$ of $G$ which extends $\phi$, i.e. $ \psi(m) = \phi(m) \quad\forall m\in M. $
I managed to proof the special case where $M+N=G$, as then each $g\in G$ can be uniquely written as $g = m + n$ where $m\in M$ and $n\in N$, and $ \psi(m+n) := \phi(m) + \phi^{-1}(n) $ has the requested properties (we even have $\psi(N)=M$).
In the general case, however, I cannot find a canonical choice for $\psi$ and I am unsure if the assertion is still true here actually, although it feels "right" and I could not come up with any counterexamples. Using an approach along the lines of the one above by writing $g = g' + m + n$ where $g'\in S:= G-(M+N)\cup \{0\}$ (or some similar subset of $G$), I don't get an unique representation as $S$ need not be a subgroup of $G$.
If anyone could give me a hint on how to proceed in the general case, or provide counterexamples or references, I would be grateful. Thanks.