1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.