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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.