Group Extensions

In this post we discuss extensions of groups.

Group Extensions

Definition 1 For two groups $G, F$, an extension of $G$ by $F$, $\mathcal{E}$, means the following pair satisfying the condition $\ker p=\im i$:

\[\mathcal{E}: F\overset{i}{\hookrightarrow}E\overset{p}{\twoheadrightarrow}G\]

The intuition for this is the following pair

\[\mathcal{E}_0: F \rightarrow F\oplus G \rightarrow G\]

which we call the trivial extension. However, in the above situation, while the relation

\[G\cong E/\ker p=E/\im i\]

holds in general by the first isomorphism theorem, one must be careful because this does not imply that

\[E\cong (E/i(F))\oplus i(F)\]

holds. In any case, from this computation we see that for $E$ to be an extension of $G$ by $F$, it is equivalent that $E$ has a suitable normal subgroup $F'$ such that $F'$ is isomorphic to $F$ and $E/F'$ is isomorphic to $G$.

Now, for fixed $G$ and $F$, the collection of all extensions of $G$ by $F$ forms a category. Morphisms here are given as follows.

Definition 2 For two extensions $\mathcal{E}_1: F \rightarrow E_1 \rightarrow G$ and $\mathcal{E}_2:F \rightarrow E_2 \rightarrow G$, a morphism from $\mathcal{E}_1$ to $\mathcal{E}_2$ means $u:E_1 \rightarrow E_2$ making the following diagram

commute.

Then if $u:E_1 \rightarrow E_2$ is an isomorphism as a group homomorphism, the inverse of $u$, $u^{-1}: E_2 \rightarrow E_1$, also satisfies the condition of Definition 2, and therefore we can verify that $u$ is an isomorphism as a morphism between extensions.

Extending the previous definition, we define the following.

Definition 3 If an extension of $G$ by $F$, $\mathcal{E}:F \rightarrow E \rightarrow G$, is isomorphic to the extension

\[\mathcal{E}_0:F \rightarrow F\oplus G \rightarrow G\]

then we call it a trivial extension.

Then the following holds.

Proposition 4 For an extension $\mathcal{E}:F \rightarrow E \rightarrow G$, the following are all equivalent.

$\mathcal{E}$ is a trivial extension.
A retraction $r: E \rightarrow F$ exists.
A section $s: G \rightarrow E$ exists such that $s(G)$ can be chosen to lie in the centralizer of $i(F)$.

Of course, here retraction and section mean group homomorphisms, not merely functions. ([Set Theory] §Retraction and Section, ⁋Definition 2)

Proof of Proposition 4

First, assume the first condition and consider the following diagram:

Then from this, we may define the retraction $r:E \rightarrow F$ as $\pr_1\circ u$, and $s:G \rightarrow E$ as $u^{-1}\circ\iota_2$.

Conversely, assume that the second condition holds. Then $(r,p): E \rightarrow F\oplus G$ is an isomorphism between the given extension and $F \rightarrow F\oplus G \rightarrow G$. Similarly, assume the third condition. Then, since $s(G)$ is contained in the centralizer of $i(F)$, from $F\oplus G$, via the weak direct product of $F$ and $G$, we can construct a morphism to $E$.

If $i(F)$ is contained in the center of $E$, $Z(E)$, then in the third condition we may ignore the relationship between $s(G)$ and $i(F)$. ([Algebraic Structures] §Group Action, ⁋Definition 12)

Definition 5 An extension $\mathcal{E}:F \rightarrow E \rightarrow G$ is called a central extension if the image of $F$ in $E$ is contained in the center of $E$.

Semidirect Products of Groups

On the other hand, the reason extensions other than the trivial extension exist is that, as seen above, for a group $G$ and any of its normal subgroups $N$, the formula

\[G\cong (G/N)\oplus N\]

does not always hold. However, the converse does not always hold either.

Definition 6 Let two groups $N,H$ and a group homomorphism $\tau:H \rightarrow \Aut(N)$ be given. Then the semi-direct product of $N$ and $H$ with respect to $\tau$, $N\rtimes_\tau H$, is the group given by the set $N\times H$ with the following operation:

\[(x_1,y_1)(x_2,y_2)=(x_1\tau(y_1)(x_2), y_1y_2)\]

Then one can show that $N\rtimes_\tau H$ has a group structure under the above operation; in this case the identity element of $N\rtimes_\tau H$ is $(e_N, e_H)$ and the inverse of $(x,y)$ is $(\tau(y^{-1})(x^{-1}), y^{-1})$. Moreover, the following holds.

Proposition 7 Define two maps $i: N \rightarrow N\rtimes_\tau H$ and $p: N\rtimes_\tau H\rightarrow H$ by the formulas

\[i(x)=(x, e_H),\qquad p(x,y)=y\]

Then these maps are group homomorphisms, and

\[\mathcal{E}_\tau: N \overset{i}{\rightarrow} N\rtimes_\tau H\overset{p}{\rightarrow} H\]

obtained from them is an extension of $H$ by $N$. Moreover, if we define a map $s: H \rightarrow N\rtimes_\tau H$ by the formula

\[s(y)=(e_N, y)\]

then $s$ is a section of $p$. In particular, if $\tau$ is a trivial homomorphism, then since $s(H)$ is contained in the centralizer of $i(N)$, by Proposition 4, $\mathcal{E}_\tau$ is a trivial extension.

The proof of this is a straightforward computation.

Now suppose the $N,H$ examined above are subgroups of a specific group $G$. If $N$ is a normal subgroup of $G$, then for each $h\in H$, the inner automorphism $\rho_h$ defined by it is an automorphism of $N$, and thus $\rho: H \rightarrow \Aut(N)$ is defined. ([Algebraic Structures] §Group Action, ⁋Definition 10) Then from the above proposition we obtain the following.

Corollary 8 Let a group $G$, and in $G$ a normal subgroup $N$, and in $G$ a subgroup $H$ be given. If $N\cap H=\{e_G\}$ and $NH=G$ hold, then the group homomorphism defined by the formula

\[N\rtimes_\rho H \rightarrow G;\qquad (x,y)\mapsto xy\]

is an isomorphism.

Proof

It suffices to construct an inverse. To do this, for an arbitrary element $g$ of $G$, we must find $x\in N$ and $y\in H$ satisfying $g=xy$; such $x, y$ exist by the condition $NH=G$, and they are unique by the condition $N\cap H={e_G}. The rest is a straightforward computation.

In this case we say that $G$ is the (internal) semi-direct product of $N$ and $H$. The difference between the external semi-direct product and the internal semi-direct product is merely where we started from, and is not important.

K

Group Extensions

Group Extensions

Semidirect Products of Groups

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