# Multiplicity of indecomposable stable summands of $BG^{\wedge}_p$

I am reading the article Homotopy stable classification of $$BG^{\wedge}_p$$ by Martino-Priddy. Let $$P_u$$, $$P_v$$ be $$p$$-subgroups of a finite group $$G$$, such that $$P_u\leq x^{-1}P_v x$$ for some $$x\in G$$, if $$X$$ is a common indecomposable stable summand of $$BP_u$$ and $$BP_v$$, $$\iota: X\rightarrow BP_u$$ is an inclusion map of $$X$$ as a summand of $$BP_u$$, the proof of proposition 3.2 part b) seems to say implicitly that the composite $$X\xrightarrow{\iota} BP_u\xrightarrow{Bi\circ Bc_x} BP_v$$ is an inclusion map of X as a summand of $$BP_v$$. Is it true?.

P.D: All objects here are $$p$$-completed spectra, $$X$$ is also a summand of $$BG$$.

Here is a counterexample: Let $$P_u = C_2$$, $$P_v = G = C_4 \times C_2$$, and $$i: C_2 \rightarrow C_4 \times C_2$$ inclusion into the $$C_4$$ summand. Let $$X = BC_2$$, obviously a stable summand in both $$BC_2$$ (!) and $$B(C_4 \times C_2)$$. However, $$X=BC_2 \xrightarrow{Bi} B(C_4 \times C_2)$$ does not have a left inverse, since if it did, then $$BC_2$$ would be a summand in $$BC_4$$. But it is not.
• Dear Prof. Kuhn, thank you for your counterexample. I had this suspicion since, under the aforementioned conditions and according to the proof, any map of the form $X\xrightarrow{\iota_1} BP_u\xrightarrow{Bi_u} BG\xrightarrow{tr} BP_v\xrightarrow{\pi} X$ can be seen of the form $X\xrightarrow{\iota_2} BP_v\xrightarrow{Bi_v} BG\xrightarrow{tr} BP_v\xrightarrow{\pi} X$ modulo an ideal $I_{uv}$. Maybe this ideal makes it possible, but I do not know how, any suggestion? please. Nov 15 '19 at 3:04