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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 $BG$.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Victor TC
    Nov 15 '19 at 3:04

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