Lemma- Suppose $P$ is a $p$-group contained in $G$ and $u\in N_U(G)$ where $U=U(\Bbb{Z}G)$. Then there exist $y\in G$ such that $u^{-1}gu=y^{-1}gy\ \forall\ g\in P$.
We use this lemma to prove Normalizer problem i.e. $N_U(G)=G\cal{z}$ where $\cal{z}$ $=Z(U(\Bbb{Z}G))$, where $G$ is a nilpotent group.
I was doing the proof from sehgal's book (Units in Integral group rings).
I know that we can write $G=\prod P_i$ , direct product of sylow p subgroups. Then if we let $u\in N_U(G)$ then by lemma there exist $x_i\in G$ such that $u^{-1}gu=x_i^{-1}gx_i$ for all $g\in P_i$. But the next line says that we can pick $x_i\in P_i$ which is not very clear to me that how is it done. And then proof is concluded by saying that it follows that $u^{-1}gu=x^{-1}gx$ for all $g\in G$ with $x=\prod x_i$. If some can explain last two lines, it will be very helpful. Thanks
As you have pointed out we have that $G = \Pi P_i $ is a direct product of its Sylow $p$-subgroups. Now we can take the $x_i \in P_i$ as you mentioned above, by noticing that we can write $x_i$ in the direct product decomposition of $G$: $$x_i = (x_{i,1},...,x_{i,n})$$ Now since a $g \in P_i$ has as decomposition $$ g = (e,...,e,g,e,...,e) $$ (where the $g$ is in the $i$-th component). We can see that all entries of $x_i$, that are not in the $i$-th component, are annihilated when we conjugate. So we may replace $x_i$ by $(e,...,e,x_{i,i},...,e)$. For every $g \in G$ we can then notice that $$ u^{-1} g u = (u^{-1} g_1 u , ..., u^{-1} g_n u) = (x_1^{-1} g_1 x_1,...,x_n^{-1} g_n x_n) = \otimes$$ Now recall that elements of different $p$-sylow subgroups commute in a nilpotent group, so we can introduce all other $x_j$ in every component (to form $x$). Thus we may continue with $$ \otimes = ((x_1 ... x_n)^{-1} g_1 (x_1 ... x_n), ..., (x_1 ... x_n)^{-1} g_n (x_1 ... x_n)) = (x^{-1} g_1 x, ..., x^{-1} g_n x) = x^{-1} g x $$ Which is exactly what we wanted.