Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.5
Prove for all $n > 1$ that $\mathbb{Z}/(n)$ is not a group under multiplication of residue classes.
Solution: Note that since $n > 1$, $\overline{1} \neq \overline{0}$. Now suppose $\mathbb{Z}/(n)$ contains a multiplicative identity element $\overline{e}$. Then in particular, $$\overline{e} \cdot \overline{1} = \overline{1}$$ so that $\overline{e} = \overline{1}$. Note, however, that $$\overline{0} \cdot \overline{k} = \overline{0}$$ for all k, so that $\overline{0}$ does not have a multiplicative inverse. Hence $\mathbb{Z}/(n)$ is not a group under multiplication.
