Normal Subgroups and Center of a Nilpotent Group (6.1.2a)

Dummit and Foote Abstract Algebra, section 6.1, exercise 2a:

Let $G$ be finite and nilpotent, and $1 < H \trianglelefteq G$. Prove $H \cap Z(G) ≠ 1$.

Proof: Let $|G| = p_1^{\alpha_1}...p_r^{\alpha_r}$.

By theorem 3(4), we have $G ≅ P_1 \times ... \times P_r$. Associate $H$ under this isomorphism, and for some nontrivial element in $H$ choose a coordinate (the $i^{th}$) in which a nontrivial element resides; call the associated Sylow $p_i$-subgroup $P_i$. Due to the way by which direct products work, for any $x∈A$ we are able to identify some element in $H$ with $x$ in the $i^{th}$ coordinate (*), where $A$ is the group generated by the $i^{th}$ coordinates of the elements of $H$. Since $H \trianglelefteq G$, we must have $A \trianglelefteq P_i$, since an arbitrary element of $P_i$ associated within $G$ may be used to conjugate the elements of $H$; if one of these conjugations results in an element in the $i^{th}$ coordinate not contained in $A$, then since $H$ is normal, this has contradicted the construction of $A$. Therefore, by theorem 1, choose a nontrivial element of $A \cap Z(P_i)$ and associate it with an element in $H$ (*), and raise this to the power of $|G|/p_i^{\alpha_i}$. This returns an element of $H$ that has the identity in every coordinate besides the $i^{th}$, and in the $i^{th}$ an element within $Z(P_i)$ that is nontrivial by Lagrange. Therefore, this nontrivial element is within $Z(G)≅Z(P_1) \times ... \times Z(P_r)$ by 5.1.1. $\square$