September 29th, 2026

Let $A, B, C$ be arbitrary groups, together with exact sequences $f: A \to B$ and $g: B \to C$. We write this as a sequence of objects and maps as such.

We say the sequence is exact if $\im(f) = \ker(g)$. In general, a longer sequence is exact if all parts of the sequence are exact wherever possible.

Consider the sequence of three objects again. We split (heh, see what I did there) the condition $\im(f) = \ker(g)$ into the conditions $\im(f) \subseteq \ker(g)$ and $\ker(g) \subseteq \im(f)$. The first says that, for all objects “entering” at $B$ (i.e., for all objects that are sent from $A$ to $B$ by $f$), when these objects “leave” $B$ (i.e., the objects are sent to $C$ by $g$), they are killed. Conversely, the only objects killed when leaving $B$ are precisely the objects that enter $B$. In sum, all objects that enter $B$ are killed when leaving, and these are the only objects killed when exiting $B$.

By the definition of exactness, we can construct short sequences that encode whether a map is injective or surjective. For example, consider the following sequence:

This sequence is exact at $A$ if and only if $f$ is injective. The unique map $\text{¡} : 1 \to A$ from the zero (initial) object sends the trivial element of $1$ to the identity element of $A$. Because no other elements are mapped, its image is simply the trivial subgroup: $\im(\text{¡}) = \{1_A\}$. By definition, exactness at $A$ requires everything entering to match everything killed when leaving, meaning $\im(\text{¡}) = \ker(f)$. Substituting our image, we get $\ker(f) = \{1_A\}$, which is precisely the condition for $f$ to be an injective map.

Moreover, consider the following sequence:

Similarly, this sequence is exact at $C$ if and only if $g$ is surjective. The unique map $! : C \to 1$ into the zero (terminal) object sends every single element in $C$ to the trivial element in $1$. Because everything in $C$ is “killed” by this map, its kernel is the entire object: $\ker(!) = C$. Exactness at $C$ requires that $\im(g) = \ker(!)$. Substituting our kernel, this reduces to $\im(g) = C$, which is precisely the condition for $g$ to be a surjective map.

Consider the sequence as follows.

We say this sequence is short exact if the sequence is exact. In other words, the sequence is short exact if $f$ is injective, $\im(f) = \ker(g)$, and $g$ is surjective.

Suppose we have a short exact sequence given as follows.

First, note that $f$ is injective, so $\ker(f) = 1$. Then $f$ is an isomorphism onto its image, and so $A \cong f(A)$. But note the image of a subgroup of the domain is a subgroup in the codomain, so we have that $f(A)$ is a subgroup of $B$. Since $A \cong f(A)$, it has all the structural properties of $f(A)$, and so we can also treat $A$ as a subgroup of $B$! Moreover, note that the sequence is exact, so $[\im(f) = f(A)] = \ker(g)$, and so $f(A)$ is a normal subgroup of $B$. Then, $A$ is realized as a normal subgroup of $B$. Finally, recall that $g$ is surjective. Then $\im(g) = C$. This culminates into an encoding of the First Isomorphism Theorem. Indeed, since $B/{\ker(g)} \cong \im(g)$, we have that $B/A \cong C$. Noting this, we can actually rewrite the exact sequence as follows.

From this, we will realize the First Isomorphism Theorem for an arbitrary homomorphism $\phi: G \to G'$. Indeed, note that

is a short exact sequence. Recalling the fact that every normal subgroup $N$ of a group $G$ is the kernel of some homomorphism, namely of the canonical projection $\pi: G \to G/N$, we can make any exact sequence of the form $1 \to A \to B \to B/A \to 1$ as usual.

Consider, once again, the following exact sequence.

A section of the surjection $p: G \to H$ is a homomorphism $s: H \to G$ such that $p \circ s = \id_H$.

Consider the exact sequence as follows.

Define $p: \bR^3 \to \bR^2$ by $p(x, y, z) = (x, y)$. We will observe what a section of $p$ truly is.

First, a section $s: \bR^2 \to \bR^3$ must be a right-inverse. For $p \circ s = \id_{\bR^2}$ to be true, we need that $s(x, y) = (x, y, f(x, y))$ for any function $f(x, y)$. But we also need $s$ to be a homomorphism. For example, consider the right-inverse

Then note that $s$ is not a homomorphism of vector spaces (i.e., a linear map), since $s$ does not preserve the origin. Instead, consider the following right-inverses.

Note they are all proper sections, since these represent homomorphisms.

We note one more thing: geometrically, all of these sections work by physically embedding $\bR^2$ into $\bR^3$. The entire $xy$-plane is mapped into $\bR^3$ with a potential tilt, angle, or general transformation that respects linearity. If, instead, we did not demand that a section be a homomorphism, a map like $(x, y) \mapsto (x, y, x^2)$ would be problematic. Indeed, the image of $\bR^2$ under this map would no longer be a plane, but an ugly, mangled, grotesque paraboloid. We illustrate this property in more generality as follows.

Suppose $p: G \to H$ is a projection of groups as usual, and $s: H \to G$ a section. Note that $s$ admits a left-inverse by construction, namely $p$, and so $s$ is injective. Recall that an injective homomorphism $s: H \to G$ tells us that $(H \cong s(H))$ is a subgroup of $G$. Thus the map $s$ embeds a nice copy of $H$, namely $s(H)$, inside $G$. The point is that, in enforcing the section $s: H \to G$ be a homomorphism, we are able to establish $H \cong s(H)$, and so we can embed a perfect copy of $H$ in $G$ as desired.

A short exact sequence $1 \to N \to G \xrightarrow{p} H \to 1$ is said to be split if it admits a section.

Suppose $1 \to N \to G \xrightarrow{p} H \to 1$ is a short exact sequence of groups. Then the following statements are equivalent:

1. The exact sequence splits.

2. $p$ has a section $s: H \to G$.

3. There is a subgroup $H' \le G$ such that $p|_{H'}: H' \to H$ is an isomorphism of groups.

4. There is some group action $\phi: H \to \Aut(N)$ such that $G \cong N \rtimes_{\phi} H$.

The equivalences $(1) \iff (2) \iff (3)$ follow readily from our earlier discussion, as any section $s: H \to G$ embeds an isomorphic copy $H' = s(H) \le G$ with inverse $p|_{H'}$. We now elaborate on how these conditions produce $(4)$.

Once again, let

be split, so that $p: G \to H$ admits a section $s: H \to G$, and let $H' = s(H) \le G$. First, we demonstrate that each $g \in G$ has a unique representation as $g = nh$, with $n \in N$ and $h \in H'$.

To show existence, take an arbitrary element $g \in G$, project it down to $p(g) \in H$, and pull it back into $G$ via the section by setting $h = s(p(g)) \in H'$. Since $p \circ s = \id_H$, we immediately see that $p(h) = p(s(p(g))) = p(g)$, meaning $g$ and $h$ share the exact same projection. Because $p$ is a homomorphism, we have

and so $gh^{-1} \in \ker(p)$. By exactness, recall that $\ker(p) = N$, which means $gh^{-1} = n$ for some $n \in N$. Rearranging yields $g = nh$, as desired.

To see that this expression is unique, suppose $g = n_1 h_1 = n_2 h_2$ for some $n_1, n_2 \in N$ and $h_1, h_2 \in H'$. Rearranging gives $n_2^{-1} n_1 = h_2 h_1^{-1}$. Notice that the left-hand side lives in $N$, while the right-hand side lives in $H'$, so this element must lie in the intersection $N \cap H'$. However, any element in $N = \ker(p)$ is killed by $p$, whereas $p|_{H'}$ is an isomorphism (and thus injective). The only element in $H'$ killed by $p$ is the identity, so $N \cap H' = \{\id_G\}$. This forces $n_2^{-1} n_1 = \id_G$ and $h_2 h_1^{-1} = \id_G$, proving $n_1 = n_2$ and $h_1 = h_2$.

Finally, we examine how multiplication behaves under this decomposition. Given two elements $g_1 = n_1 h_1$ and $g_2 = n_2 h_2$, we want to express their product $(n_1 h_1)(n_2 h_2)$ in our canonical form $(N\text{-part})(H'\text{-part})$. Because $G$ is not necessarily abelian, we cannot simply swap $h_1$ and $n_2$. Instead, we introduce $h_1^{-1} h_1 = \id_G$ to move $h_1$ past $n_2$:

Because $N \unlhd G$ is a normal subgroup, conjugation by $h_1 \in H'$ preserves $N$, guaranteeing that $h_1 n_2 h_1^{-1} \in N$. We record this twisting interaction between the two subgroups via the homomorphism $\phi: H' \to \Aut(N)$ defined by $\phi_h(n) = hnh^{-1}$. Substituting this in, the multiplication law becomes

Because $s$ is a homomorphism, $H'$ is a subgroup, meaning $h_1 h_2$ stays inside $H'$. Identifying $H'$ with $H$, this multiplication rule is precisely the semidirect-product law, giving $G \cong N \rtimes_{\phi} H$.