I read it in the above context (from the book "Endomorphism Rings of Abelian Groups"). Hence $\mathbb{Z}(n)$ is an abelian group. However it is not being defined in the book itself.
I am guessing $\mathbb{Z}(n)$ is the same as $\mathbb{Z}/n\mathbb{Z}$ considered as an abelian group?
Is there such a result $\text{End}(\mathbb{Z}/n\mathbb{Z})\cong\mathbb{Z}_n$?
Thanks.
Yes, indeed, $\mathbb{Z}(n)$ seems to be $\mathbb{Z}/n\mathbb{Z}$ as an abelian group, since if you look at the next line, it says that there is a generator meaning that the group is cyclic. My guess is that the author uses this notation to make clear that this is an abelian group, and distinguish it from the ring $\mathbb{Z}/n\mathbb{Z}$.
The fact that End$(\mathbb{Z}(n)) \cong \mathbb{Z}/n\mathbb{Z}$ is very general (this can be seen as a very special case of the enriched Yoneda embedding for example).
A less extreme generalization is as follows: given any ring $R$, we can consider $R$ to be an $R$ module, and then End$(R) \cong R$ as a ring. This is just the case $R = \mathbb{Z}/n\mathbb{Z}$.