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Let $R$ be some (possibly non-commutative) ring, and let $X$ be an $R$-module with dual $X^\ast=\mathrm{Hom}_R(X,R)$. I've shown both $X,X^\ast$ exist in an exact sequence of the form

$$0\to B\to X \to A\to 0 \text{ and } 0\to B\to X^\ast\to A\to 0.$$

For ease, I'll denote this by $X,X^\ast\in\mathrm{Ext}^1(A,B)$ even though I actually mean their respective classes. In addition, I have shown $\mathrm{Ext}^1(A,B)\cong\mathbb{Z}/3$ so, up to isomorphism, we have three classes. Now, if $X$ is indecomposable then does it follow that $X\cong X^\ast$? Here is my argument:

If $X,X^\ast$ are in the same class, then clearly they are isomorphic by the Five Lemma, so suppose not. As neither is indecomposable then one belongs to the class $c_1$ (say) and the other to $c_2$. In particular, $c_1=-c_2$ so I have

$$0\to B\to X \stackrel{f}{\to} A\to 0 \text{ and } 0\to B\to X^\ast\stackrel{-f}{\to} A\to 0.$$

Let $g:X\to X^\ast$ be the map sending an element to its dual, then surely this gives us a commutative diagram from which we can once more use the Five Lemma? Or am I missing something?

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    What is the natural map $X \to X^*$? – 2017-02-21
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    The map sending an element $e$ to it's dual $e^\ast$. In my case, $R$ is a group ring $S[G]$ and so $X$ in fact has an $S$-basis. So I send each basis element to it's dual. – 2017-02-21
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    This is far from being natural. – 2017-02-21
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    Okay sorry, I fixed it in my question. – 2017-02-21
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    @MooS This is something I have recently been wondering myself, however surely it is true for any modules $X, X^\prime$ such that neither splits as a direct sum of $A\oplus B$? Using your notation above, Sam, if $X, X^\prime$ belong to the same class $c_1$, then they are isomorphic by the Five Lemma. Alternatively, one belongs to $c_1$ (say $X$), the other to $c_2=-c_1$. This surely implies $X^\prime$ is the middle term in an exact sequence congruent to $-c_1$. As the middle term in $-c_1$ is $X$, it follows from the Five Lemma that $X\cong X^\prime$. – 2017-10-17

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