Extensions of $\mathbb{Z_3}$ by $\mathbb{Z_9}$ I'm working through MacLane's Homological Algebra, and I was trying to compute some easy examples of Ext groups, but just confused myself:
I'm trying to compute all three (I know there's three by the first theorem in the third chapter) extensions of $Z_9$ by $Z_3$.
If I resolve $Z_3$ and set it up, it looks like this:
$$\begin{matrix}
0 & \to & 3Z & \xrightarrow{i} & Z & \xrightarrow{\pi} & Z_3 & \to & 0 \\
\ & \   & \downarrow^{h^i} & \ & \downarrow^{\psi} \ & \ & \downarrow^{\simeq} \\
0 & \to & Z_9 & \xrightarrow{i'} & P & \xrightarrow{\pi'} & Z_3 & \to & 0
\end{matrix}$$
Now as he proves, the class of the extension we get by filling in the bottom given $h^i$ is a function of the homotopy class of $h^i$ (considered as a morphism of complexes lifting the identity on $Z_3$). To fill in the bottom (can't figure out how to make $\psi, i', \pi'$ dashed). We compute the pushout of the left square.
Ok, so starting with $h^0 = 0$, it's easy to see that I just get the split extension as expected.
Then taking $h^1(3n) = n \text{ mod } 9$, I get $Z_{27}$, no problem. Next I choose $h^2(3n) = 3n \text{ mod } 9$. Then to compute the pushout $P \simeq \frac{Z_9 \bigoplus Z}{N}$ for $N = \{ (-3n \text{ mod } 9, 3n) : 3n \in 3Z \}$.
So after starting at this for a day or two, I believe this is metacycic? Anyways, this group of order 27 is something like (here's where I get v. confused):  $P = \langle a,b \mid 3a = 3b, 9a = 9b = 0, b^{-1} a b = a\rangle$.
But looking at grouppros groups of order 27, the semidirect product I think this should be (https://groupprops.subwiki.org/wiki/M27) doesn't have these relations (or atleast it's not obvious how to change basis), and moreover isn't abelian. What's going on? What is the third isomorphism class of extensions here?
Thanks!

As I was told in the comments, there are only two abelian extensions of $Z_9$ by $Z_3$. I was going based on MacLane's Proposition 1.1 in Chapter 3, which says that for Abelian A, $\text{Ext}_Z(Z_m,A) \simeq A/mA$. Still not sure why I interpreted that wrong, but I'll update this when I figure that out.
Also, not sure why P as defined above isn't a group, but that's probably not too hard once I think about it in a less braindead state (thought it does seem to fit the description of metacyclics by presentations here https://mathworld.wolfram.com/MetacyclicGroup.html).
 A: Note first of all that the extensions classified by Ext are extensions within the category of abelian groups.  So, $P$ should definitely be abelian, and your description of it as a pushout is correct (but note that this is a pushout in the category of abelian groups, not in the category of all groups).
I'm not sure exactly what you mean when you talk about the homotopy class of $h^i$ (you talk about a morphism between complexes, but which two complexes are you talking about?).  In any case, though, your $h^2$ actually gives the same extension as $h^0$ (so, the trivial extension).  Indeed, two homomorphisms $3\mathbb{Z}\to\mathbb{Z}_9$ give the same extension iff they differ by a homomorphism that can be extended to $\mathbb{Z}$.  In this case, $h^2-h^0=h^2$ can be extended to $\mathbb{Z}$ by having it map each $n$ to $n$ mod $9$.  You can see explicitly that the inclusion $i':\mathbb{Z}_9\to P$ splits for your extension by mapping $(a,b)\in \mathbb{Z}_9\oplus\mathbb{Z}$ to $a+b$ mod $9$ (this vanishes on $N$, and so is well-defined on $P$).
To get the third extension that really is different, you would instead want to take $h(3n)=2n$ mod $9$.  Since this $h$ does not extend to $\mathbb{Z}$ it gives a nontrivial extension, and it also gives a different extension than your $h^1$ does since $h-h^1=h^1$ does not extend to $\mathbb{Z}$.  Note though that the group $P$ you will get from this $h$ is actually isomorphic to $\mathbb{Z}_{27}$.  So up to isomorphism there are only two possibilities for the group $P$, even though there are three different extensions.  This is because for two extensions to be equivalent, you need not just an isomorphism of the $P$'s but an isomorphism that is compatible with the short exact sequences.  See Can two elements of an Ext group come from the same middle object of an SES? for more discussion.
