Why is a group generated by one element called cyclic?

To show that the integers $\mathbb{Z}$ under addition are cyclic, demonstrate that $\mathbb{Z} = \langle 1 \rangle$. The subgroup generated by $1$ contains $1$, its inverse $-1$, the identity $0$, and all integer multiples of $1$, which exhausts all integers.

Conditions: The group is $\mathbb{Z}$ with operation $+$.; The candidate generator is $1$.

$\langle x \rangle$ is the subgroup generated by $x$, which is the smallest subgroup of $G$ containing $x$. It is built by including $x$, its inverse $x^{-1}$, the identity $1$, and all integral powers of $x$ and $x^{-1}$ required for closure.

Conditions: $G$ is a group with operation $\times$.; $x \in G$.

A group is cyclic if it can be generated by a single element. In the video, this is written as $G = \langle x \rangle$, meaning every element of the group belongs to the subgroup generated by $x$.

Conditions: $G$ is a group.; There exists an element $x \in G$ such that $G = \langle x \rangle$.

Any subgroup containing $x$ must also contain its inverse $x^{-1}$ and the identity element $1$ because these are required by the definition of a subgroup. Additionally, closure under the group operation forces the inclusion of all integral powers of $x$ and $x^{-1}$.

Conditions: $G$ is a group.; $H \le G$ is a subgroup.; $x \in H$.

In additive notation, for a group $H$ with operation $+$ and an element $y \in H$, the subgroup generated by $y$ is the smallest subgroup of $H$ containing $y$. It consists of all integer multiples of $y$, written as $\langle y \rangle = \{\dots, -3y, -2y, -y, 0, y, 2y, 3y, \dots\}$.

Conditions: $H$ is a group with operation $+$.; $y \in H$.