If $A$ is an abelian group, we have

$Aut\left(K\left(A,n\right)\right)=Aut(A) \ltimes K\left(A,n\right),$

where the left hand side is the space of self-homotopy equivalences. Is there an easy way to see this abstractly rather than by computation?

Also, it seems there is an analogy with the following: if $\mathbb{R}^{0|1}$ is the odd-line (regarded as a supermanifold),

$Aut\left(\mathbb{R}^{0|1}\right)=\mathbb{R}^\times \ltimes \mathbb{R}^{0|1}$

The vague analogy is that $\mathbb{R}^{0|1}=\mathbb{R}[1]$, and under Dold-Kan, $K\left(A,n\right)=A\left[-n\right].$

It seems to me there is unifying categorical principle at play. Can anyone shed some light on it? Thanks!

Edit: I have accepted the answer below as it does answer my first question. However I am still interested in seeing if there is any general principal at play.

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    $\begingroup$ The $\text{Aut}(A)$ part is hopefully easy to see. The other part is the translation action of $K(A, n)$ on itself. So this is at least a natural guess. In general there's always a natural action of $G \rtimes \text{Aut}(G)$ on $G$ (as a set). $\endgroup$ – Qiaochu Yuan Nov 7 '15 at 8:19
  • $\begingroup$ I see how both parts act. My question is how do we see this is everything? $\endgroup$ – David Carchedi Nov 7 '15 at 8:23
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    $\begingroup$ Could you explain your definitions please? I can't come up with candidates which make your first assertion correct in the case where $A=\mathbb{Z}$ and $n=1$. $\endgroup$ – HJRW Nov 8 '15 at 7:18
  • $\begingroup$ For Eilenberg-MacLane spectra the second term does not appear. $\endgroup$ – Fernando Muro Nov 8 '15 at 8:01

As per Qiaochu Yuan's comment we need to only understand the space of based maps between $K(A,n)$ with a chosen base point.

The loop-deloop pair of functors establish an equivalence between the categories of $A^\infty$-groups and connected spaces with a base point: $$\Omega: \mathrm{Top}_{*,\ \pi_0=0} \simeq \mathrm{Grp}: \mathbb{B}$$

More generally, $\Omega^n$ and $\mathbb{B}^n$ give an equivalence between $(n-1)$-connected basepointed spaces and $E_n$-groups. This means that $$\mathrm{Top}_*(K(A, n), K(A, n)) = \mathrm{Grp}_{E_n} (A, A) $$

But $A$ is discrete, so all higher coherence data vanishes and a map of $E_n$-groups is the same as a map of set-theoretic groups, thus the answer.

I don't think your second statement is formally analogous since it mixes up the additive and multiplicative groups and I can't think of any inductive construction that links them, but who knows.

EDIT: Regarding the semidirect product. I don't think it is actually true.

A group $G$ is a semidirect product of groups $H \rtimes F$ iff there is a split exact sequence $$0 \to H \to G \stackrel{\leftarrow}{\to} F \to 0$$

If X is a connected space with a chosen point $*\in X$ for the mapping space $\mathrm{Top}(X, X)$ we have in general a fiber sequence of spaces $$ \mathrm{Top}_* (X,X) \to \mathrm{Top}(X,X) \to X$$

The right arrow maps $f: X\to X$ to $f(*)$. There is the same sequence if $\mathrm{Top}(\cdot,\cdot)$ will denote invertible or homotopy invertible maps. Lets assume in the following that we consider invertible maps and all relevant spaces are topological spaces (we do not lose any generality since any homotopy fiber sequence is equivalent to a topological fibration and any homotopy group is equivalent to a topological group). Note that the left arrow in the sequence is always an inclusion of subgroup and the right arrow is the coset space of the action of based maps on all maps. If $X$ is a group $G$ (in particular, $G = K(A,n)$), then the right map has a section which maps an element $g: G$ to the shift operator $t_g : h: G \mapsto gh: G$ and this is a homomorphism of topological groups. Any $f: Aut(G)$ also can be uniquely represented as $f= t_g \circ \hat f$, where $g := f(*)$ and $\hat f : Aut_*(G)$.

On the level of spaces we thus certainly have that $Aut(G) = Aut_*(G) \times G$. However, while it looks like a semidirect product in fact it isn't, since neither $Aut_*(G)$ nor $G$ are normal subgroups in $Aut(G)$. While we have a factor map $Aut(G) \xrightarrow{/Aut_*(G)} G$ it is not a homomorphism of groups. If we represent some element of $f: Aut(G)$ as $t_g \cdot \hat f$, then the product of $f$ and $f^\prime$ looks like $$t_g \cdot \hat f \cdot t_{g^\prime} \cdot \hat f^\prime = t_g t_{g^\prime} \left( t_{g^\prime}^{-1} \hat f t_{g^\prime} \right) \hat f^\prime$$ which looks like a semidirect product but isn't, since conjugation by $t_{g^\prime}$ maps $Aut_*(G)$ to $Aut_{g^{\prime -1}}(G)$.

EDIT 2: As Achim Krause noted below, we actually have a homomorphism $Aut (K (A,n)) \to Aut (A) $ given by the action on n-th homology, thus the stated semidirect product decomposition $Aut (A) \ltimes K (A, n) $ is valid.

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  • $\begingroup$ Thanks for your answer Anton. I don't understand your argument however. Do you mind being a bit pedantic for me and spelling it out? So, from what you wrote, the space of self-maps of $K(A,n)$ in pointed spaces is equivalent to the discrete set $End(A).$ How do I go from here to concluding that we have the semi-direct product decomposition I'm after of the entire automorphism space? $\endgroup$ – David Carchedi Nov 8 '15 at 21:32
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    $\begingroup$ We do have a group homomorphism in the other direction, from $Aut(K(A, n)) \rightarrow Aut(A) $, which splits. By the long exact sequence of homotopy groups, the fiber is a $K(A, n)$. $\endgroup$ – Achim Krause Nov 9 '15 at 6:16
  • $\begingroup$ David may have confused you by writing the normal subgroup of the semidirect product on the right side in his question. $\endgroup$ – Achim Krause Nov 9 '15 at 6:17
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    $\begingroup$ Associating to a map its action on $\pi_n$, you get a map of sets $Aut(K(A,n)) \rightarrow Aut(A)$, and since two homotopic maps of induce the same map on $\pi_n$, it is constant on connected components of $Aut(K(A,n))$, therefore continuous when the right side is endowed with the discrete topology. $\endgroup$ – Achim Krause Nov 9 '15 at 12:25
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    $\begingroup$ (Say action on $H_n(\bullet; \mathbb{Z})$ to get around basepoints, although for $n\geq 2$ that's not an issue) $\endgroup$ – Achim Krause Nov 9 '15 at 12:32

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