# Group Extensions

Let $G$ and $H$ be groups. A group $E$ is called an extension of $G$ by $H$ iff there's a short exact sequence of the type $1\rightarrow G \rightarrow E \rightarrow H \rightarrow 1$. In that way, $G$ is isomorphic to a normal subgroup of $E$, and the quotient of $E$ by that subgroup is isomoprhic to $H$.

Next you define that two extensions $E_{1}$ and $E_{2}$ are equivalent when there's a group morphism sucht that certain diagram commutes (the commutative diagram can be found here, http://planetmath.org/encyclopedia/GroupExtension.html). It turns out that with this definition,if two extensions are equivalent the given morphism is an isomorphism.

My question is:

Suppose you have two extensions that are isomorphic, that is you have $E_{1}$ and $E_{2}$, which are extensions of $G$ by $H$, and $E_{1}$ is isomorphic to $E_{2}$ (but you don't know if the diagrams commute). Is $E_{1}$ equivalent to $E_{2}$? If not, why is the "correct" definition of extension equivalence the one above?

-
It is possible to have two extensions of $G$ by $H$ which are isomorphic as groups, but not as extensions of $G$ by $H$. Is that the question you are asking? If so, I can probably post an explicit example. The reason we have this definition is that we are interested not only in what the resulting group is "abstractly", but in how we are constructing it out of $G$ and $H$. The definition of equivalence takes into account how we are constructing the group, not just what we end up with. – Arturo Magidin Apr 14 '12 at 22:47
Yes, that was what i was asking for. You pretty much made it clear already, but if you want to post an example, I won't complain. Thank you. – Bill Apr 14 '12 at 22:53
Incidentally, the way you read the short exact sequence is the other way around: As you have put it, you should have said that "$E$ is an extension of $H$ by $G$". – Manolito Pérez Nov 25 '15 at 9:52

Here's a nice example of two extensions of $G$ by $H$ that are not equivalent, but with isomorphic middle groups. In fact, one of the extensions splits and the other does not.

Let $A=\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, let $B=\mathbb{Z}/p\mathbb{Z}$; let $b$ be a generator of $B$. Let $B$ act on $A$ by letting $(a_1,a_2)^{b^r} = (a_1,a_2+ra_1)$.

Let $E=(A\rtimes B)\times (\mathbb{Z}/p\mathbb{Z})$. Note that $A$ is a normal subgroup of $E$, and that $E$ can be realized as a semidirect product of $A$ by $B\times(\mathbb{Z}/p\mathbb{Z})$. Thus, we have a split exact sequence $$1 \longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}} \longrightarrow E \longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}}\longrightarrow 1.$$

On the other hand, let $D$ be the subgroup of $E$ generated by the second direct factor of $A$ (which is central in $A\rtimes B$, hence in $E$) and the direct factor $\mathbb{Z}/p\mathbb{Z}$. This subgroup is isomorphic to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, and is central in $E$. Thus, we have an exact sequence $$1\longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}} \longrightarrow E \longrightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}}\longrightarrow 1$$ induced by this subgroup. The extension cannot be split, because the subgroup $D$ is central, so if the extension were split it would be a direct product (which it is not). So we have two extensions of $(\mathbb{Z}/p\mathbb{Z})\times(\mathbb{Z}/p\mathbb{Z})$ by itself, which yield isomorphic groups, but one of the extensions splits and the other does not, so the two extensions are not equivalent.

As to why we use this definition: for extensions, we are actually interested in how we are building the extension, and not only on the resulting groups. The definition of equivalence is such that it captures the idea that the two groups are built up from $G$ and $H$ in "essentially the same way", which of course implies isomorphisms. But as the example above shows, we can end up with the same kind of group by considering essentially different ways of having $H$ act on $G$. It's certainly interesting to know that the resulting groups are isomorphic even though the extensions are not equivalent, but we don't want the extensions to be equivalent because we do care about how we arrived at the group $E$, not just that we arrived.

-

Here is another counterexample, with an infinite commutative extension group.

With $p$ a prime, $\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\NN}{\mathbb{N}}$consider the additive group $E = \ZZ_p\oplus(\ZZ_{p^2})^{(\NN)}$ and its (perforce normal) subgroups $G=0\oplus(p\ZZ_{p^2})^{(\NN)}$ and $G'=\ZZ_p\oplus(p\ZZ_{p^2})^{(\NN)}$. $\newcommand{\isomorph}{\cong}$Then $G\isomorph\ZZ_p^{(\NN)}\isomorph G'$ and $E/G\isomorph\ZZ_p^{(\NN)}\isomorph E/G'$, but there is no automorphism of $E$ mapping $G$ onto $G'$ because $G=pE\neq G'$ and $\varphi(pE)=pE$ for every automorphism $\varphi$ of $E$.

-