Why must the subgroup generated by $x$ contain $x^{-1}$ and $1$?

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 $1$.

A group generated by one element is called cyclic because the elements of the generated subgroup can be viewed as cycling through powers (or multiples) of the generator. In finite cases like $\mathbb{Z}/n\mathbb{Z}$, the sequence of multiples of the generator repeats periodically, forming a cycle.

Conditions: The group is generated by a single element.; In the finite case, the modulus $n$ determines the length of the cycle.

$\langle x \rangle$ 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 $G = \langle x \rangle$.

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 $+$.; $y \in H$.