From the discussion in the comments two possible solutions arised, here is the first:
Since the group is nilpotent, upper central series terminates. Then for some $n$, $Z_n(G) = G$ and there must be a smallest $i$ such that $N \cap Z_i(G) = \langle e \rangle$ but $N \cap Z_{i+1}(G) \neq \langle e \rangle$. The goal is to show that $N \cap Z_{i+1}(G) \subseteq Z_1(G) = Z(G)$.
I will denote $Z_i(G),Z_{i+1}(G)$ simply by $Z_i, Z_{i+1}$.
We notice that the subgroup $[G,N]$ generated by the elements $\{ gng^{-1}n^{-1}\mid g \in G, n \in N\}$ is contained in the normal subgroup $N$. This is because for every generator $gng^{-1}n^{-1}$ of $[G,N]$, $gng^{-1}n^{-1} = (gng^{-1})n^{-1} \in N$. So in conclusion $[G,N] \subseteq N$.
Further we notice that since $Z(G/Z_i) = Z_{i+1}/Z_i$, then for any $c \in Z_{i+1}, g \in G$ we have that $Z_igc = Z_icg$, which implies $Z_igcg^{-1}c^{-1} = Z_i$ and $gcg^{-1}c^{-1} \in Z_i$ for all $g \in G, c \in Z_{i+1}$. These are exactly the generators of the group $[G, Z_{i+1}]$, therefore $[G, Z_{i+1}] \subseteq Z_i$
Now we look again at the nontrivial group $N \cap Z_{i+1}$. We established earlier that $[G,N] \subseteq N$ and $[G, Z_{i+1}] \subseteq Z_i$. This gives $[G, N \cap Z_{i+1}] \subseteq [G,N] \cap [G,Z_{i+1}] \subseteq N \cap Z_i = \langle e\rangle$. So actually the subgroup $[G, N \cap Z_{i+1}]$ is trivial, which happens when $N \cap Z_{i+1} \subseteq Z_1 = Z(G)$. This gives that $N \cap Z(G) \neq \langle e \rangle$.
And a perhaps a bit simpler one, relying on an alternative characterisation of nilpotent groups:
By an exercise in my book (Hungerford's Algebra, chapter 2, section 7, exercise 4), a group is nilpotent if and only if the $\gamma_m(G) = \langle e \rangle$ for some $m$, where $\gamma_1(G)=G,\gamma_2(G)=[G,G]$ and $\gamma_i(G)=[\gamma_{i−1}(G),G]$.
Given this, we define a sequence $N_1(G) = N, N_2(G) = [N,G]$ and $N_i(G) = [N_{i-1}(G),G]$, where $N$ is any proper normal subgroup of $G$.
Obviously for any $i$, $N_i(G) \subset N$ by normality of $N$.
It is also clear that $N_1(G) = N \subset G = \gamma_1(G)$. Assume inductively that $N_i(G) \subset \gamma_i(G)$. By definition, $N_{i+1}(G) = [N_{i}(G),G]$ and $\gamma_{i+1}(G)=[\gamma_{i}(G),G]$. It is clear then that $N_{i+1}(G) \subset \gamma_{i+1}(G)$.
But since $G$ is nilpotent, for some $m$, $N_m(G) \subset \gamma_m(G) = \langle e \rangle$ So in particular we get that $[N_{m-1}(G), G] = \langle e \rangle$. This can only be if $N_{m-1}(G) \subset Z(G)$, and since $N_{m-1}(G) \subset N$, we have that $N \cap Z(G) \neq \langle e \rangle$.
The space of marked groups can be used to show some properties of surface groups. For example, see the following question: Is the center of the fundamental group of the double torus trivial?
I found also some interesting results in the appendix of Model Theory by Hodges.
Consider the following problem: If $A,B,C$ are three groups such that $A \times C \simeq B \times C$; when $A$ and $B$ are isomorphic? For example:
Theorem: Let $G$ and $H$ be finitely generated finite-by-nilpotent groups. Then $G \times \mathbb{Z} \simeq H \times \mathbb{Z}$ iff $G$ and $H$ are elementary equivalent.
See Cancellation and Elementary Equivalence of Finitely Generated Finite-By-Nilpotent Groups or Cancellation of abelian groups of finite rank modulo elementary equivalence by Francis Oger.
Shelah solved the following problems using model theory:
Theorem: Any uncountable group $G$ of cardinality $\lambda$ has at least $\lambda$ subgroups not conjugate in pairs.
See Uncountable groups have many nonconjugate subgroups.
Theorem: There is an uncountable group whose proper subgroups are all countable.
See On a problem of Kurosh, Johnson groups, and applications.
In On some conjectures connected with complete sentences, Makowski relates the problem "Is there an infinite finitely presented group with a finite number of conjugacy classes?" to the existence of a specific kind of theory (theorem 2.6).
A group $G$ is said linear of degree $n$ if it is embedable into $GL(n,F)$ for some field $F$. In Barwise's book, Handbook of mathematical logic, there is a proof of
Theorem: Let $G$ be a group. If every finitely-generated subgroup of $G$ is linear of degree $n$, then $G$ is linear of degree $n$.
Best Answer
In group theory, this notation is normally used to define the normal closure in $G$ of the subset $X$. This is the smallest normal subgroup of $G$ containing $X$. With that definition you can easily prove that $X^G = <g^{-1}xg | g \in G, x\in X>$. It also equals the intersection of all normal subgroups of $G$ containing $X$.
Note that in other parts of mathematics the notion of closure is very common and defined in a similar way. For example in topology one has normal closures of sets, and in field theory, you will encounter algebraic closures.