Suppose that $k$ is a field of prime characteristic $p$. Let $P$ be a finite $p$-group, and let $U$ be a finitely generated $kP$-module. Then:
(i) We have $soc(U) = U^P$ ($P$ fixed points in $U$). (ii) If $U$ has no nonzero projective direct summand, then $U_1^P = (\sum _{x \in P}x)U={0}$. (iii) The $kP$-module $U$ is projective if and only if $U^P = U_1^P$.
When proving only if of (iii) the author states if $U$ is not projective, then $U_1^P$ is a proper subspace of $U^P$ by (ii).
I am totally lost by his statement. I understand it is a subspace. But how this is connected to (ii)? Any hints would be sufficient to me. Thank you!
I am finding the notation a bit peculiar, but here goes. Let $U$ be non-projective, and consider its largest possible projective summand $V$. Then $U\cong V\oplus W$ where $W$ is non-zero but has no non-zero projective summand. But $(V\oplus W)_1^P= V_1^P\oplus W_1^P=V_1^P\oplus\{0\}$ is a proper subspace of $V\oplus W$.