# Isomorphic extensions of $\mathbb Z_3$ by $\mathbb Z_3 \times \mathbb Z_3$ which are not equivalent

The following exercise from D. Robinson, A Course in the Theory of Groups confuses me:

Find two isomorphic extensions of $$\mathbb Z_3$$ by $$\mathbb Z_3 \times \mathbb Z_3$$ which are not equivalent.

For two group extensions $$N, E, G$$ and $$\overline N, \overline E, \overline G$$ with injections $$\mu : N \to E, \overline\mu : \overline N \to \overline E$$ and projections $$\varepsilon : E \to G, \overline\varepsilon : \overline E \to \overline G$$ a homomorphism of extensions is a triple of homomorphism $$(\alpha, \beta, \gamma)$$ from $$N$$ to $$\overline N$$, $$E\to\overline E$$ and $$G \to \overline G$$ such that the resulting diagram commutes, see this image from the book.

An equivalence between extensions $$N, E, G$$ and $$N, \overline E, G$$ is a triple of morphisms $$(1_N, \beta, 1_G)$$, an isomorphism between extension $$N, E, G$$ and $$\overline N , \overline E, G$$ is a triple of morphisms $$(\alpha, \beta, 1)$$ where $$\alpha$$ (and hence $$\beta$$) are isomorphisms. Note that in an isomorphism in general $$N \ne \overline N$$ , where they are equal for the notion of equivalence.

Coming back to the exercise, first as $$\mathbb Z_3$$ is a field, the only isomorphism from $$\mathbb Z_3$$ to $$\mathbb Z_3$$ is the identity map. So, as I see it, there is no way to give an isomorphism $$\alpha \ne 1_{\mathbb Z_3}$$, hence not possible to give an isomorphism which is not also an equivalence?

Or do I misunderstand something? Could you please clarify what this exercise asks for?

There is only one ring automorphism of $$\Bbb Z_3$$, but the topic at hand is group extensions, and so group automorphisms, and there is a non-identity group automorphism of $$\Bbb Z_3$$ (switching the images of $$1$$ and $$2$$).