# In a Simplicial group the Degenerate elements don't intersect the kernel of the face maps.

Let $G_{\bullet}$ be a simplical group. I denote by $D_n \subset G_n$ the subgroup generated by all the degeneracy maps $s_i:G_{n-1} \to G_n$. I also denote $$M_n = \bigcap_{i>0} \mathrm{Ker} d_i \subset G_n$$ and $$Z_n = \bigcap_{i\geq 0} \mathrm{Ker} d_i \subset G_n$$ When $G_{\bullet}$ is a simplicial abelian group it is well known and there many references to the fact that

$$M_n \cap D_n = \{e\}$$

I know how to prove it for the non-abelian case, but I believe there should be a reference somewhere. In fact I'm actually interested in the weaker claim that :

$$Z_n \cap D_n = \{e\}$$

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Tomer: The claim you make about $M_n\cap D_n$ being trivial in the non abelian case is not true. In fact that condition is equivalent to the Moore complex of $G$ being a crossed complex in the sense of Brown and Higgins. This seems to be first proved in the thesis of Ashley (N. Ashley, Simplicial T-Complexes: a non abelian version of a theorem of Dold-Kan, Dissertations Math., 165, (1989), 11 – 58). It is also a lemma in one of the two papers by Brown and Loday that if $G_2=D_2$, so there are no new non-degenerate elements in $G_2$, just ones there because of $G_1$, then $\partial(M_2\cap D_2)=[Ker d_0,Ker d_1]$. This condition implies that $M_1\to M_0$ is a crossed module. The kernel of that crossed module will not be trivial in general but is $Z_2\cap D_2$.
Minor correction: having vanishing Whitehead products does not imply a space is a product of Eilenberg-MacLane spaces. Consider the space $X = P_3 Q S^2$. Since it is a 3-type, the only Whitehead product to worry about is $\pi_2 \times \pi_2 \to \pi_3$, which vanishes since it agrees with the one for $QS^2$, which is an infinite loop space. On the other hand $X$ is not equivalent to $K(\pi_2X,2) \times K(\pi_3X,3)$ because the operation $\pi_2 \to \pi_3$ given by precomposing with the generator of $\pi_3(S^2)$ is non-zero on $X$ (but vanishes for the product of Eilenberg-MacLane spaces). – Omar Antolín-Camarena Sep 24 '14 at 15:16
Thanks for the clarification. Can you add what is $Q$ to help `the reader'? Has anyone calculated the algebraic models (crossed square, or whatever) for this 3-type? – Tim Porter Sep 25 '14 at 6:12
$Q=\Omega^\infty\Sigma^\infty$ so that $QS^2 = \mathrm{colim}_{n\to\infty} \Omega^n S^{n+2}$. I'd like to know about algebraic models for its 3-type too. (I'm sadly ignorant about that sort of thing, all I know is that this one is not modeled by a strict 3-groupoid.) – Omar Antolín-Camarena Sep 25 '14 at 12:51