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I am learning about exact sequences that split, in the context of modules. In this context, as I understand it, sequences that split on the left are the same as sequences that split on the right. But in non Abelian groups, is there an easy example of an exact sequence $1 \rightarrow G \rightarrow H \rightarrow K \rightarrow 1$ that splits on the right and not on the left, and one that splits on the left but not on the right?

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  • $\begingroup$ I haven't learned "splits on the right" and "splits on the left". For non-abelian groups, I've only learned about splitting as what you would call "splitting on the right", which implies that $H$ is a semidirect product of $G$ and $K$. $\endgroup$
    – Arthur
    Oct 11, 2018 at 13:56
  • $\begingroup$ To say $H$ is a semidirect product of $G$ and $K$ means that the sequence splits on the right (not on the left). In general, the "splitting" (on either side) is just a one-sided inverse of that arrow (although the side on which it is an inverse is different on the right than on the left). $\endgroup$
    – Lee Mosher
    Oct 11, 2018 at 13:58
  • $\begingroup$ See Theorems 3.2 and 3.3 in kconrad.math.uconn.edu/blurbs/grouptheory/splittinggp.pdf: splitting on the left is equivalent to the group in the middle looking like the direct product of the other two groups (with the two maps in and out of the middle looking like a standard embedding into one factor of a direct product and projection to the other factor), while splitting on the right is equivalent to the group in the middle looking like a semidirect product of the other two groups (where the group on the right acts as automorphisms of the group on the left). $\endgroup$
    – KCd
    Feb 2, 2020 at 3:34
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    $\begingroup$ Since a direct product is a special case of a semidirect product, splitting on the left always implies splitting on the right. The converse is false. For example, if the first and third groups are abelian then their direct product is abelian, so if the group in the middle is a nontrivial semidirect product of the first and third groups (so it is nonabelian) then the exact sequence splits on the right but not on the left. $\endgroup$
    – KCd
    Feb 2, 2020 at 3:39

3 Answers 3

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For examples that split on the right but not on the left, one can take any right split short exact sequence where $H$ is finitely generated but $G$ is not. For example, start with the free group $F\langle a, b \rangle$ with free basis $a,b$, and let $h : F\langle a, b \rangle \to \mathbb Z \oplus \mathbb Z = \langle a\rangle \oplus \langle b\rangle$ be the abelianization homomorphism, let $\pi_a : \langle a\rangle \oplus \langle b\rangle \to \langle a\rangle = \mathbb Z$ be the projection homomorphism, and consider the composition $h_a = \pi \circ h : F \langle a,b \rangle \to \mathbb Z$, which is surjective. Let $K = \text{Kernel}(h_a)$. We get a short exact sequence $$1 \to K \to F \langle a,b \rangle \to \mathbb Z \to 1 $$ However, $K$ is not finitely generated, so there is no retraction homomorphism $F \langle a,b \rangle \mapsto K$. However, every short exact sequence with $K=Z$ is split.

There are also examples that split on the right but not on the left where $G$ and $H$ are both finitely generated. The key idea here is that if a subgroup injection $G \hookrightarrow H$ is split, i.e. if there is a retraction $H \mapsto G$, then that injection is undistorted with respect to word metrics on $G$ and on $H$. This means that if I pick a generating set $S$ for $G$ and if I extend $S$ to a generating set $T$ for $H$, and if I pick $g \in G$, then the length of the shortest word in the generators $S$ that represents $g$ is comparable (up to a constant multiplicative bound) to the length of the shortest word in the generators $T$ that represents $g$. Now let's take this solvable group expressed as a semi direct product $$1 \to\mathbb Z \oplus \mathbb Z \to H \to \mathbb Z \to 1 $$ where the action of $\mathbb Z$ on $\mathbb Z \oplus \mathbb Z$ is generated by the matrix $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$. One can show without too much trouble that the normal subgroup injection $\mathbb Z \oplus \mathbb Z \hookrightarrow H$ is distorted, hence this example does not split on the left. But, it does split on the right, because again the group $K$ is just infinite cyclic.

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  • $\begingroup$ I think splitting on the right but not the left is possible even for finite groups, for instance with the dihedral group $1\to\Bbb Z_n\to D_n\to \Bbb Z_2\to 1$. There is no natural map $D_n\to \Bbb Z_n$ but there is one $\Bbb Z_2\to D_n$. $\endgroup$
    – Arthur
    Oct 11, 2018 at 13:59
  • $\begingroup$ But what is this natural map? $Symmetry \rightarrow 0$ and $Rotation \rightarrow 1$. This wouldn't work, no? $\endgroup$
    – roi_saumon
    Oct 11, 2018 at 14:51
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At above comment, I think the map you described would work.

To clarify what Arthur said, I’d like to give you detailed answer of

why there is no such a natural map.

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If you consider the left inverse map from $D_n \rightarrow \mathbb{Z}_n$,

it has to map $r^i \mapsto i \ , \ r^{i}s \mapsto 0$ for each $i \in \left\{1,2,...,n\right\}$.

then $r \mapsto 1$ and $s \mapsto 0$.

But their product is mapped to $0$ in $\mathbb{Z}_n$, on other hand, the product of images of each element in $D_n$ is not $0$.

Hence no such map exist.

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In the full category of groups, $\bf Grp,$ left and right split aren't equivalent.

If it's left split then it's right split, and then a direct product.

There's a little example on Wikipedia of one that is right but not left split, with $S_3$ and $A_3.$ But, any semi-direct product that is not a direct product should work.

An important example is the semi-direct product, $G=H\ltimes N.$ It's equivalent to having a short exact sequence $$1\to N\to G\to H\to 1,$$ which is the identity on $H.$

In general it's right split, but not always left split.

This is also called a group extension ($G$ is an extension of $H$ by $N.$)

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