I was writing lecture notes on nilpotent groups, and wanted to show the following well-known basic result.
Theorem: Let $G$ be a nilpotent group and let $A$ be an abelian normal subgroup of $G$, maximal amongst such subgroups. Then $A=C_G(A)$.
The standard proof of this is easy: certainly $C=C_G(A)$ contains $A$, so if this is proper then consider $C/A$. As this is a non-trivial normal subgroup of $G/A$ it intersects the centre non-trivially, so let $x\neq 1$ be in this intersection. The subgroup $\langle A,x\rangle$ is abelian, since $A$ commutes with $x$, and normal since $\langle A,x\rangle/A$ is central, hence normal in $G/A$. Since $A$ is maximal, this yields a contradiction.
Now, the problem is that when I was writing this result, I had not yet proved that normal subgroups intersect the centre non-trivially, so I tried to come up with a proof that did not use that. In the end I gave up and rearranged the results so the other one came first. But is there a proof that doesn't essentially prove the normal subgroups result along the way? (Note that I am not assuming that $G$ is finite, or even finitely generated.)
Best Answer
We can reason like this.
Let $A$ be a maximal normal abelian subgroup of nilpotent group $G$ and $C=C_G(A)$. Note at once that $C$ is a normal subgroup of $G$. Let $$ 1=Z_0\leq Z_1\leq\ldots\leq Z_s=G $$ be the upper central series of $G$.
Prove by induction that $C\cap Z_i\leq A$ for all $i$. It is clear that $C\cap Z_0\leq A$. Let $C\cap Z_i\leq A$ and let $h\in C\cap Z_{i+1}$.
For any $g\in G$ we have $[g,h]\in Z_i\cap C\leq A$. Hence, the group $\langle A,h\rangle$ is a normal Abelian subgroup of $G$. Then $h\in A$.
Finally we obtain that $C\cap G=C\cap Z_s\leq A$, hence $C=A$.