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I can see that $S_3$ is not the direct sum of $A_3$ and $\{1,-1\}$, so that the sequence doesn't split. However, we were told that there still exist a map $\beta:\{1,-1\} \to S_3$ such that $g \beta=I_{\{1,-1\}}$, where $g:S_3 \to \{1,-1\}$ is the last map in the SES. And that there is no map $\alpha : S_3 \to A_3$ satisfying $\alpha f=I_{A_3}$.

Why if the sequence splits from the right does it not imply that the SES is split exact?

Also, how should one construct the map $\beta$ that then becomes a homomorphism and why doesn't no $\alpha$ exist?

Mark
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If such an α existed, $S_3$ would be isomorphic to the direct product $A_3\times\{1,-1\}$ hence would be commutative since $A_3$ is a cyclic group of order $3$.

$\beta$ may be constructed with $$\beta(1)=\operatorname{id}=(),\quad\beta(-1)=(12)\quad\text{(or any transposition)}.$$

Bernard
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  • Thank you for your answer. I don't follow how we get that $S_3$ would be isomorphic to the direct product from $\alpha$. – Mark Oct 16 '16 at 18:53
  • Also, I was reading about this just now on Wikipedia here: https://en.wikipedia.org/wiki/Splitting_lemma#Counterexample , can you also please explain their reasoning. They're saying the only possibility for $\alpha$ is that it be the trivial map. It's not so clear how this is so. – Mark Oct 16 '16 at 18:55
  • @user264885: Here are some details: for any permutation σ, the order of α(σ) is a divisor of the order of σ since α is a group homomorphism. Hence for any transposition τ, α(τ) has order $1$ or $2$. Besides, this order is also a divisor of the order of $A_3$, which is $3$. As a conclusion, α(τ) has order $1$, i.e. is the identity. As any permutation σ is a product of transpositions, and α is again a homorphism, α(σ) is the identity. – Bernard Oct 16 '16 at 19:47