# When does a short exact sequence of groups imply it is isomorphic to direct product group [closed]

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

• Try to use the splitting lemma! It's very useful here, Regards! Jun 16, 2019 at 9:05
• @AlexisOsorio isn't that for Abelian groups? May 12, 2022 at 17:51

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.

• In all these statements the isomorphisms should be related to the maps in the original sequence! Jun 16, 2019 at 10:36
• Yes, you are right. Let me add that.
– Con
Jun 16, 2019 at 10:39
• Let me add a reference as to why this is important: math.stackexchange.com/questions/135444/… Jun 16, 2019 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? Feb 23 at 21:02
• @ReaderManifold Oh yes, this is indeed written quite poorly. Let me quickly change it.
– Con
Feb 24 at 9:15