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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 $G\cong G/N\times N$.

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    $\begingroup$ Try to use the splitting lemma! It's very useful here, Regards! $\endgroup$ Jun 16 '19 at 9:05
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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) We have that $\alpha \colon G \rightarrow G' \times G''$, $a \mapsto (h(a),g(a))$ is an isomorphism.

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.

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  • $\begingroup$ In all these statements the isomorphisms should be related to the maps in the original sequence! $\endgroup$
    – Christoph
    Jun 16 '19 at 10:36
  • $\begingroup$ Yes, you are right. Let me add that. $\endgroup$
    – Con
    Jun 16 '19 at 10:39
  • $\begingroup$ Let me add a reference as to why this is important: math.stackexchange.com/questions/135444/… $\endgroup$
    – Christoph
    Jun 16 '19 at 10:45

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