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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.

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2 Answers 2

up vote 5 down vote accepted

In general, it isn't true. Let $G=\mathbb Z/\left<2\right>\times \mathbb Z/\left<4\right>$. Then the subgroups generated by $(1,0)$ and $(0,2)$ are isomorphic, but the isomorphism cannot be extended to all of $G$ because there is an element $g\in G$ such that $g+g=(0,2)$, but there is no such element such that $g+g=(1,0)$

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That was helpful, thanks a lot. I should have tried harder finding a counterexample. –  Andy Brandi Dec 18 '12 at 17:16

I think this is not true in general.

$G=\mathbb Z \,;\, M=2 \mathbb Z \,;\, N= 3\mathbb Z$

and $\phi(2x)=3x$. This cannot be extended, and this can be proven easely by contradiction (what can $\phi(1)$ be?).

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Ah, right. We would have $2\phi(1) = \phi(1) + \phi(1) = \phi(2) = 3$ which is a contradiction. Thanks! –  Andy Brandi Dec 18 '12 at 17:16

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