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Question: Can there be a split extension with $G=N\times H$? Also can a split extension be exact? If so, when?(From bottom for increased context)

Definition of exact sequence (in the context of group theory):

A sequence of group homomorphisms is exact if $\text{Im}(\psi_i)=\ker(\psi_{i+1})$ for all group homomorphisms.

The following sequence is exact, where we have $G=NH$, $N\cap H=\{1\}$, and $N,H$ are subgroups of $G$, where $N$ is normal. $$1\hookrightarrow N\hookrightarrow N\rtimes H\twoheadrightarrow H \twoheadrightarrow 1$$

Where the first two maps are monomorphisms, and the latter two are epimorphisms.

We can see this is a short exact sequence, since:

\begin{align} \text{Im}(\psi_1)=&\,\,1=\ker(\psi_2)\\ \text{Im}(\psi_2)=&\,\,\text{N}=\ker(\psi_3)\\ \text{Im}(\psi_3)=&\,\,\text{H}=\ker(\psi_4) \end{align}


Definition of a split extension:

If we have a short exact sequence: $$1\hookrightarrow K \hookrightarrow G \twoheadrightarrow H \twoheadrightarrow 1$$ And we also have some group homomorphism $s_1:H\hookrightarrow G$, such that the composition of $s$ and $\psi_3$, gives us an identity map across $H\hookrightarrow G \twoheadrightarrow H$. Now regardless of whether this $G$ is a direct or semi-direct product, it seems impossible that this is itself a short exact sequence. Why?

\begin{align} \text{Im}(s_1)=H\\\text{but}: \ker(s_2)=K \end{align}

Regardless of $G=K\rtimes H$ or $G=K\times H$. This is just an observation.


Can there split extensions for groups that are the direct product of two subgroups?

I can see that for a semi-direct product we have a split extension, since:

$$1\hookrightarrow N\hookrightarrow N\rtimes H\twoheadrightarrow H \twoheadrightarrow 1$$

We don't need this split extension to be exact, but we need it to induce the identity. E.g. $$H\hookrightarrow N\rtimes H \twoheadrightarrow H $$

Simply take $\psi_1$ as the natural embedding, e.g. the identity map. $\psi_2: N\rtimes H =G \to G/N$, this quotient map simply takes $\psi_2: hN \mapsto h$

Hence we can see $\psi_2(\psi_1(h))=\psi_2(h)=h$ and this is thus an identity map.


For a direct product however I am confused. The direct product however I view as a two tuple, and as a result, I have no idea how to find a split extension for such a case. Can there be a split extension with $G=N\times H$? With either of the following short exact sequences? $$1\hookrightarrow N \hookrightarrow N\times H \twoheadrightarrow H \twoheadrightarrow 1$$ $$1\hookrightarrow H \hookrightarrow N\times H \twoheadrightarrow N \twoheadrightarrow 1$$

Also can a split extension be exact? If so, when?

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  • $\begingroup$ You can't verify $N\to N\times H\to H$ is exact? Also I am confused by your explanation of $K\to G\to H$. Clearly the image of $K\hookrightarrow G$ is $K$, no? $\endgroup$
    – anon
    May 2, 2015 at 2:12
  • $\begingroup$ @anon But that isn't a split extension? For the second point, indeed, but I believe that the split extension is on $H\to G\to H$ $\endgroup$
    – anon
    May 2, 2015 at 2:14
  • $\begingroup$ Sure, $N\times H$ is split, the map from either of $N$ or $H$ back into $N\times H$ (the section $s$) is the obvious embedding, and it should also be straightforward to verify e.g. $N\to N\times H\to N$. $\endgroup$
    – anon
    May 2, 2015 at 2:16
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    $\begingroup$ The extension that we are testing for split-ness is $1\to K\to G\to H\to 1$, not $H\to G\to H$ (which is not an extension anyway). $\endgroup$
    – Billy
    May 2, 2015 at 2:16

1 Answer 1

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Now regardless of whether this G is a direct or semi-direct product, it seems impossible that this is itself a short exact sequence. Why?

$\mathrm{Im}(\psi_1) = H$

This is not true: $\mathrm{Im}(\psi_1) = K$. (Also, I think you are renumbering your maps on-the-fly, which may be causing part of the confusion.)

Can there be a split extension with $G=N×H$?

Every $N\times H$ is a split extension of $N$ by $H$ (or vice-versa). All semidirect products are also split extensions (but the order matters).

In fact, a short exact sequence of groups splits iff the group in the middle is a semidirect product of the other two. Can you prove this?

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  • $\begingroup$ Is it wrong to think that $1\to N\to N\times H \to H \to 1$ to have a split extension, we need $H\to N\times H \to H$ to be the identity map? $\endgroup$
    – anon
    May 2, 2015 at 2:18
  • $\begingroup$ No, that is correct. The map $N\times H\to H$ is the obvious one, $(n,h)\mapsto h$, and as a 'splitting' map, we may take $H\to N\times H$ to be $h\mapsto (1_N,h)$. $\endgroup$
    – Billy
    May 2, 2015 at 2:19
  • $\begingroup$ I don't know what you mean by either "equivalent" or "isomorphic" here, because elements can't be those things. You are right that they have to be equal. We have a composite map $H\to N\times H \to H$ which sends $h\mapsto (1,h) \mapsto h$, and $h$ is definitely very much equal to $h$. $\endgroup$
    – Billy
    May 2, 2015 at 2:22

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