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If $A,B,C$ are groups and I have $A / B = C$:

Is it then obvious that $A = C \oplus B$? I'm asking because I think this is used in the explanation of reduced homology on page 110 in Hatcher.

Thanks for your help!

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This is false. Let $A$ be the cyclic group of order $4$, and $B$ and $C$ be the cyclic groups of order $2$. – Zhen Lin Aug 3 '11 at 14:18
It's false in general, but it's true if $C$ is free (exercise). – Qiaochu Yuan Aug 3 '11 at 14:19
To connect Qiaochu's point to the original context in Hatcher, $\mathbb{Z}$ is free, so if $H_0 / \tilde{H}_0 \cong \mathbb{Z}$ then $H_0 \cong \tilde{H}_0 \oplus \mathbb{Z}$. – Zhen Lin Aug 3 '11 at 14:24
Cool, thanks. Maybe one of you could post it as an answer then I'll accept it. – Rudy the Reindeer Aug 3 '11 at 14:32
Technically, you should ask whether $A$ is isomorphic to $C \times B$, rather than equal. As other mentioned, this is false in general. Note that even when this is true, there need not be a "natural" isomorphism (even when the groups are vector spaces) – Mark Aug 3 '11 at 15:11
up vote 3 down vote accepted

The claim is known as the splitting lemma for exact sequences. A proof can be found on Wikipedia, but I give one below which does not chase elements.

Proposition. Let $0 \to B \mathrel{\overset{f}{\to}} A \mathrel{\overset{g}{\to}} C \to 0$ be an exact sequence in an abelian category. The following are equivalent:

  1. $g$ is a split epimorphism, i.e. there is $s : C \to A$ such that $g \circ s = \mathrm{id}_C$.
  2. $f$ is a split monomorphism, i.e. there is $r : A \to B$ such that $r \circ f = \mathrm{id}_B$.
  3. $A \cong B \oplus C$, with $f : B \to A$ a coproduct insertion and $g: A \to C$ a product projection.


  • A homomorphism $f : B \to A$ is monic if it admits left cancellation, i.e. if $f \circ b = f \circ b'$, then $b = b'$. A monomorphism is a monic homomorphism.

  • A homomorphism $g : A \to C$ is epic if it admits right cancellation, i.e. if $c \circ g = c' \circ g$, then $c = c'$. A epimorphism is an epic homomorphism.

  • The category of abelian groups is an abelian category. Note that in this category, a homomorphism is monic if and only if it is injective, and it is epic if and only if it is surjective.

Proof. (3) certainly implies (1) and (2): let $s: C \to A$ be the other coproduct insertion and $r : A \to B$ be the other product projection.

We now show (1) implies (2). Let $h = \mathrm{id}_A - s \circ g : A \to A$. Then, $g \circ h = g - g \circ s \circ g = 0$, so $h$ must factor through $\ker g$: let $r : A \to B$ be the map such that $h = f \circ r$. On the other hand, $h \circ f = f - s \circ g \circ f = f$, so $f \circ r \circ f = f$. But $f$ is monic, so $r \circ f = \mathrm{id}_B$. The dual proof shows (2) implies (1).

Finally, (1) and (2) together imply (3). Since we have maps $f : B \to A$ and $s : C \to A$, we have a map $f + s : B \oplus C \to A$, and similarly we have a map $(r, g) : A \to B \times C$. But we may take $B \oplus C = B \times C$, and it is evident that $(r, g) \circ (f + s)$ is the identity map. Let $k = (f + s) \circ (r, g) = f \circ r + s \circ g$. Then $k \circ f = f$ and $g \circ k = g$, so $k$ is both monic and epic, hence an isomorphism. These altogether imply $(f + s)$ and $(r, g)$ are isomorphisms.

Corollary. Let $0 \to B \mathrel{\overset{f}{\to}} A \mathrel{\overset{g}{\to}} C \to 0$ be an exact sequence of abelian groups. If $C$ is free, then $A \cong B \oplus C$.

Proof. Since $g : A \to C$ is surjective, we may use the universal property of $C$ to construct the required $s : C \to A$.

Exercise. Find where the proof of the proposition fails when $A$ is non-abelian. (Notice that there are maps $f, g, r, s$ satisfying the hypotheses of (1) and (2) for the sequence $0 \to A_3 \to S_3 \to C_2 \to 0$.)

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To explain Zhen Lin's example:

let $\mathbb Z_4=\{1,g,g^2,g^3\}$ and $\mathbb Z_2=\{1,g^2\}$ then $\mathbb Z_4/\mathbb Z_2$ has two cosets $\mathbb Z_2.1=\{1,g^2\}$ and $\mathbb Z_2.g=\{g,g^3\}$ and it is clear that $(\mathbb Z_2.g)(\mathbb Z_2.g)=\mathbb Z_2.1$ which means that $\mathbb Z_4/\mathbb Z_2\cong \mathbb Z_2$. But it is clear that $\mathbb Z_4 \not \cong \mathbb Z_2 \oplus \mathbb Z_2$ as $\mathbb Z_4$ has an element of order 4 while $\mathbb Z_2 \oplus \mathbb Z_2$ has not.

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