# central extension of $\mathbb Z_3$ by $\mathbb Z_3$

Suppose we are interested in all possible central extensions of $\mathbb Z_3$ by $\mathbb Z_3$, i.e. in all possible groups $E$ satisfying a short exact sequence $1 \to \mathbb Z_3 \to E \to \mathbb Z_3 \to 1$. This is actually trivial: since $|E| = 9$, and we know there are only two groups of that order, and both are such extensions (i.e. $\mathbb Z_3 \times \mathbb Z_3$ and $\mathbb Z_9$).

My confusion is the following: the central extensions are in 1-to-1 correspondence with the second group cohomology group $H^2(\mathbb Z_3; \mathbb Z_3)$. According to the groupprops wiki, we have $H^2(\mathbb Z_3; \mathbb Z_3) = \mathbb Z_3$. But that implies there must be at least three distinct groups of order nine... Hence clearly at least one of the things I have said is wrong. Which is it? Most likely I have misinterpreted the result on the groupprops wiki

• No, there are two non-equivalent extensions (this is different from non-isomorphic groups!) for $\mathbb Z/9$, can you find them?
– Pedro
Commented Aug 22, 2017 at 17:32

There can be non-isomorphic extensions with the same middle term, in your case there is one cohomology class corresponding to the trivial extension and two classes corresponding to extensions with middle term $\mathbb Z/9$.
Concretely, consider the extensions $\xi_1$ and $\xi_2$ where the projection $p_1$ in $\xi_1$ is 'multiplication by 1' and that the projection $p_2$ in $\xi_2$ is 'multiplication by 2', so the kernel of the both is $\{3,6\}\simeq \mathbb Z/3$. Let $j$ denote the inclusion of this subgroup into $\mathbb Z/9$. Note that the only nontrivial maps $\mathbb Z/9\to\mathbb Z/3$ are these two.
To see these extensions are not isomorphic, note we need an isomorphism $\theta$ of $\mathbb Z/9$ that satisfies $\theta j = j$ and also $p_2 = p_1\theta$. If $1$ is the generator of $\mathbb Z/9$ then $\theta(1)=u$ must also be a generator, i.e. an element of $\{1,2,4,5,7,8\}$.
Now if we want $2 = p_2(1) = p_1(u) = u \mod 3$ then certainly $u\in \{2,5,8\}$, and if we want $3 = 3u \mod 9$ then none of $2,5$ or $8$ work, since in each case $3u=6$, so $\theta$ does not exist.
What the cohomology group measures are the isomorphism classes of extensions. There are two isomorphism classes of extensions $$0\to\newcommand{\Z}{\Bbb Z}\Z_3\to\Z_9\to\Z_3\to0.$$ In such an extension the injection takes $1$ to $a\in\{3,6\}$ and the surjection takes $1$ to $b\in\{1,2\}$. The extensions with $(a,b)=(3,1)$ and $(6,2)$ are isomorphic, also the extensions with $(a,b)=(3,2)$ and $(6,1)$ are isomorphic. But the extensions with $(a,b)=(3,1)$ and $(3,2)$ are not isomorphic.