# Nilpotent implies intersection of normal subgroup with upper central series is strictly ascending till the subgroup is reached

## Statement

Suppose $G$ is a nilpotent group and $H$ is a normal subgroup of $G$. Suppose $Z^{(i)}(G)$ denotes the $i^{th}$ member of the upper central series of $G$, i.e., $Z^{(0)}(G)$ is the trivial subgroup, $Z^{(1)}(G)$ is the center, and $Z^{(i+1)}(G)/Z^{(i)}(G)$ is the center of $G/Z^{(i)}(G)$.

Let $r$ be the smallest nonnegative integer such that $H \le Z^{(r)}(G)$. Then $H \cap Z^{(i)}(G)$ is a proper subgroup of $H \cap Z^{(i+1)}(G)$ for all $0 \le i < r$.

## Related facts

### Weaker facts

• Nilpotent implies center is normality-large: A special case, where we only use that $H \cap Z^{(0)}(G) \le H \cap Z^{(1)}(G)$ for $r \ge 1$.