Is the decomposition of the homotopy type of a complex into a bouquet unique? 
*

*Is it true that if $A_1 ​​\vee A_2 \vee .. \vee A_n = B_1 \vee B_2 \vee .. \vee B_m$, where $A_i, B_j$ are homotopy types of complexes not decomposable into a bouquet, then the multisets $A_i$ and $B_j$ coincide? That is, is it true that a commutative monoid of homotopy types decomposable into a bouquet of a finite number of indecomposable ones is freely generated by indecomposable ones.


*Is it true that any finite complex decomposes into a finite number of indecomposable ones (and thus all finite complexes are included in the monoid above)? Countable complexes are not necessarily included in it, for example, any countable bouquet. But, for example, all spaces $K(G, n)$ are included, it seems.
 A: There is a substantial literature on the subject.   You should probably start with
C. W. Wilkerson, Genus and cancellation, Topology {\bf 14}
(1975), 29-36.
In particular, he shows
Definition:
Let $X$ be a 1-connected $p$-local space of finite type.  Then $X$ is said
to be $prime$ if for every self-map $ f:X \rightarrow X $, either
$i$) $ f $ induces an isomorphism in mod-$p$ homology, or
$ii$) for every $n$, there exists an $m$ such that the $m$-fold iterate of
$f$ induces the zero map on $H_{i}(X; {\bf Z}_p)$ for $ 0 < i \leq n$.
Then, we have
Theorem 3:  (Wilkerson)
$i$) Any finite dimensional 1-connected $p$-local co-$H$-space is
equivalent to a wedge of prime spaces.
$ii$) If a 1-connected $p$-local space of finite type is equivalent to a
wedge of primes, then the prime wedge summands are unique up to
order.
$iii$)  A prime space which is a retract of a wedge of 1-connected $p$-local spaces of finite type is a retract of one of the wedge summands.
A: In Hilton&Roitberg paper "On principal $S^3$-bundles over spheres" it's proven that if you have a prime order $p \neq 2,3$ class $\alpha$ in $\pi_k(S^n)$ that is a suspension, then for a prime $q \neq \pm 1 \, mod \, p$ mapping cones $C(\alpha)$ and $C(q \cdot \alpha)$ satisfy $C(\alpha) \vee S^n \cong C(q \cdot \alpha) \vee S^n$.
Now it's easy to see that those cones are indecomposable, and give counterexample to 1.
For 2, I think Grushko theorem gives that all but finite number of summands are simply-connected, and then you have only finitely many generators in homology so overall any such sum has to be finite. I'll think about that, but my reasoning seems  correct to me now.
