I am trying to find examples of pairs of short exact sequences (s.e.s) of abelian groups:
$0\rightarrow M\rightarrow M'\rightarrow M'' \rightarrow 0$ and
$0\rightarrow N\rightarrow N'\rightarrow N'' \rightarrow 0$
in the following cases
i) $M \cong N , M'\cong N' , M'' \ncong N''$;
ii) $M \cong N , M'\ncong N' , M'' \cong N''$;
iii) $M \ncong N , M'\cong N' , M'' \cong N''$;
I have an example for the first one
i)
For any two $p$ and $q$ prime numbers, $p \neq q$:
$0 \rightarrow \Bbb Z \xrightarrow{\times p} \Bbb Z \rightarrow \Bbb Z/p\Bbb Z \rightarrow 0 \tag{**} $
$0 \rightarrow \Bbb Z \xrightarrow{\times q} \Bbb Z \rightarrow \Bbb Z/q\Bbb Z \rightarrow 0 \tag{**} $
They are clearly s.e.s and $\Bbb Z/p\Bbb Z \ncong \Bbb Z/q\Bbb Z $
But I haven't been able to find any for the other two cases. Any help is appreciated. I would be nice to see other examples using other abelian groups as wel, the only one I could find in case i) was that one
Edit: Following the suggestion of lulu:
ii)
The groups of of order 4 that are not isomorphic are the cyclic group of order 4:$\Bbb Z/4\Bbb Z$ and the Klein 4-group $\Bbb Z/2 \Bbb Z \oplus \Bbb Z/2\Bbb Z$
Then, Would an example for ii) be the following ?
$0 \rightarrow \Bbb Z/2 \Bbb Z \rightarrow \Bbb Z/4\Bbb Z \rightarrow \Bbb Z/2 \Bbb Z \rightarrow 0\tag 1$
$0 \rightarrow \Bbb Z/2 \Bbb Z \rightarrow \Bbb Z/2 \Bbb Z \oplus \Bbb Z/2\Bbb Z \rightarrow \Bbb Z/2 \Bbb Z \rightarrow 0 \tag 2 $
I am not sure they are s.e.s though the quotient map $\Bbb Z \rightarrow \Bbb Z/4\Bbb Z $ is not injective
Edit 2: I am still trying to get a simpler example for iii) than that in the answer below
maybe this one works?
$0 \rightarrow \Bbb Z/2 \Bbb Z \xrightarrow{i_2} \Bbb Z \oplus \Bbb Z/4\Bbb Z \xrightarrow{p_2} \Bbb Z/2 \Bbb Z \rightarrow 0 \tag 3 $
$0 \rightarrow 2\Bbb Z \xrightarrow{i_1} \Bbb Z \oplus \Bbb Z/4\Bbb Z \xrightarrow{p_1} \Bbb Z/2 \Bbb Z \rightarrow 0 \tag 4 $
where $ i_j$ and $p_j$ ; j=1,2 are inclusion and projection maps in the corresponding components
To elaborate on the comments:
For $ii$: consider the two non-isomorphic groups of order $4$. Thus we let $M_1'=\mathbb Z/4\mathbb Z$ and let $M_2'=\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$. Both of these admit surjections to $\mathbb Z/2\mathbb Z$ with kernels also given by $\mathbb Z/2\mathbb Z$.
For another famous example of $ii$ (which uses non-abelian groups) consider $$0\to \mathbb Z/3\mathbb Z\to \mathbb Z/6\mathbb Z \to \mathbb Z/2\mathbb Z\to 0$$ $$0\to \mathbb Z/3\mathbb Z\to S_3 \to \mathbb Z/2\mathbb Z\to 0$$
For $iii$ let $M'=N'$ both be the countably generated free abelian group $G=\mathbb Z^{\mathbb N}$. For the first sequence we just consider the identity map from $G$ to itself, of course that has trivial kernel. For the second we consider the map from $G$ to itself given by $(a_1, a_2, \cdots)\mapsto (a_2, a_3, \cdots)$. That map is surjective, clearly, but the kernel is a copy of $\mathbb Z$. Thus we have $$0\to <e>\to \mathbb Z^{\mathbb N} \to \mathbb Z^{\mathbb N}\to 0$$ and
$$0\to \mathbb Z\to \mathbb Z^{\mathbb N} \to \mathbb Z^{\mathbb N}\to 0$$