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Suppose that $1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1$ is a short exact sequence of groups.

Then, what is a (necessary and )sufficient condition for $G\cong N\times Q$.

In other words, let $N$ be a normal subgroup of $G$, then what is a (necessary and )sufficient condition for $G\cong G/N\times N$.

This is true for vector spaces (cf. complementary subspace), but false in general for groups. For example, if $2\mathbb{Z}\subset \mathbb{Z}$ had a complementary, it would be $\cong \mathbb{Z}/2\mathbb{Z}$ and this had non zero torsion.

user682141
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1 Answers1

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For abelian groups (or even in any abelian category) you can use the splitting lemma:

Proposition 1. Let $0 \longrightarrow A' \stackrel{f}\longrightarrow A \stackrel{g}\longrightarrow A'' \longrightarrow 0$ be a short exact sequence of abelian groups. Then the following are equivalent:

(i) There exists a homomorphism $h \colon A \rightarrow A'$ such that $h \circ f = \text{id}_{A'}.$

(ii) There exists a homomorphism $j \colon A'' \rightarrow A$ such that $g \circ j = \text{id}_{A''}.$

(iii) We have an isomorphism $h \colon A \rightarrow A' \oplus A''$, such that $h \circ f$ is the natural inclusion of $A'$ into the direct sum $A' \oplus A''$ and $g \circ h^{-1}$ is the natural projection from $A' \oplus A''$ onto $A''$.

For non-abelian groups the splitting lemma does not hold in general. Consider for example the short exact sequence $1 \longrightarrow A_n \stackrel{\iota}\longrightarrow S_n \stackrel{\text{sgn}}\longrightarrow C_2 \longrightarrow 1$. We can send the generator of $C_2$ to any $2$-cycle to get (ii), but (i) and (iii) do not hold.

Let me now rephrase the splitting lemma for general groups:

Proposition 2. Let $1 \longrightarrow G' \stackrel{f}\longrightarrow G \stackrel{g}\longrightarrow G'' \longrightarrow 1$ be a short exact sequence of groups. Then the following are equivalent:

(i) There exists a homomorphism $h \colon G \rightarrow G'$ such that $h \circ f = \text{id}_{G'}.$

(ii) There exists an isomorphism $\alpha \colon G \rightarrow G' \times G''$ such that $\alpha \circ f = i_1$ and $p_2 \circ \alpha = g$. Here $i_1 \colon G' \rightarrow G' \times G''$ is the inclusion into the first factor and $p_2 \colon G' \times G'' \rightarrow G''$ is the projection onto the second factor.

You can also get another version:

Proposition 3. Let $1 \longrightarrow G' \stackrel{f}\longrightarrow G \stackrel{g}\longrightarrow G'' \longrightarrow 1$ be a short exact sequence of groups. Then the following are equivalent:

(i) There exists a homomorphism $j \colon G'' \rightarrow G$ such that $g \circ j = \text{id}_{G''}.$

(ii) There exists a homomorphism $\varphi \colon G'' \rightarrow \text{Aut}(G')$, such that $\beta \colon G' \rtimes G'' \rightarrow G$, $(a,b) \mapsto f(a)j(b)$ is an isomorhpism.

Con
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  • In all these statements the isomorphisms should be related to the maps in the original sequence! – Christoph Jun 16 '19 at 10:36
  • Yes, you are right. Let me add that. – Con Jun 16 '19 at 10:39
  • Let me add a reference as to why this is important: https://math.stackexchange.com/questions/135444/a-nonsplit-short-exact-sequence-of-abelian-groups-with-b-cong-a-oplus-c – Christoph Jun 16 '19 at 10:45
  • @Con Is the Proposition 2 true for any isomorphism $\alpha$ of groups from $G$ to $G' \times G''$? Specifically, I wanted to know more about how do you say the second part implies the first, when the $h$ itself was defined in the first part. Can you please clarify this? – Reader Manifold Feb 23 '24 at 21:02
  • @ReaderManifold Oh yes, this is indeed written quite poorly. Let me quickly change it. – Con Feb 24 '24 at 09:15
  • Before I just constructed $\alpha$ in terms of $h$, which shows how (i) implies (ii). – Con Feb 24 '24 at 09:23