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

I am asking for help on the following exercise:

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

as taken from D. Robinson, A Course in the Theory of Groups, see my other recent post for the notation.

There are five groups of order $$27$$. I found example extensions with the same groups that are not equivalent, but I was unable to give an isomorphism of extensions. Namely $$E = \mathbb Z_3 \oplus \mathbb Z_9$$ with $$\mu(1) = (0,3)$$ and $$\overline \mu(1) = (1,3)$$. Then an automorphism $$\beta : E \to E$$ giving equivalent extensions has to fulfill $$\beta(0,3) = (1,3)$$, but as $$\beta(0,3) = \beta(0,1) + \beta(0,1) + \beta(0,1) = (0, x) \ne (1,3)$$ this is not possible. The same argument works for the semidirect product $$\mathbb Z_9 \rtimes \mathbb Z_3$$, but both could not be made isomorphic extensions if we use the only non-trivial automorphism $$\alpha(1) = 2$$ on $$\mathbb Z_3$$. In $$\mathbb Z_{27}$$ we have only one way to embed $$\mathbb Z_3$$, so this will not work either. And $$G = \mathbb Z_3 \oplus \mathbb Z_3 \oplus \mathbb Z_3$$ will not work either, as we can alway write $$\mu(Z_3) \oplus U = G$$ as this is an $$\mathbb Z_3$$-vector space, hence make every isomorphism $$\overline \mu(\mathbb Z_3) = \beta(\mu(\mathbb Z_3)$$ into an automorphism of $$G$$. So the only promising candidate seems to be $$UT(3,3)$$, but here as every element has order three I am somewhat unable to go along the same lines above to show that certain maps could not be extended to isomorphisms.

So I am asking for help on this exercise! Any hints?

EDIT: It is not a duplicate question, the other had a very specific question concerning understanding what is asked for. This one asks for help on the exercise themselve. Should not be difficult to tell that they are both different!