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Consider the extensions of type $0\rightarrow \mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/p^2\mathbb{Z}\rightarrow \mathbb{Z}/p\mathbb{Z}\rightarrow0$, consider two of them, explicitely the ones with $\varphi:\mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/p^2\mathbb{Z}$, where $\varphi(1)=ip$ with $1\leq i\leq p-1$, and another extension with $\psi:\mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/p^2\mathbb{Z}$ such that $\psi(1)=jp$ with $1\leq j\leq p-1$. Suppose that $i\neq j$ I want to prove that these two extensions aren't equivalent, could ou give me some hints?

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What makes you think they are not equivalent? –  Thomas Andrews Mar 14 '12 at 0:59
because I showed that the number of extensions of $\mathbb{Z}/p\mathbb{Z}$ by $\mathbb{Z}/p\mathbb{Z}$ is $p$. We have the split one and those $p-1$, so I think that using this it's clear that they are not equivalent, but I'd like something more direct, like a diagram chase that gives a contradiction. –  Alex M Mar 14 '12 at 1:04
It just seems that if $ik\equiv j\pmod p$ and $kk'\equiv 1\pmod p$ then the maps from $\mathbb Z/p^2\mathbb Z\rightarrow \mathbb Z/p^2\mathbb Z$ corresponding to multiplication by $k$ and $k'$ seem like an equivalence between $\phi$ and $\psi$. –  Thomas Andrews Mar 14 '12 at 1:59
you're right... maybe I did something wrong –  Alex M Mar 14 '12 at 2:29
The problem is that the epimorphisms have not been defined! If they are defined to map $1 \to 1$ in both cases, then the two resulting extensions are inequivalent. –  Derek Holt Mar 14 '12 at 8:42

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