Powers of Element form Subgroup
Jump to navigation
Jump to search
Theorem
Let $\struct {G, \circ}$ be a group.
Then:
- $\forall a \in G: H = \set {a^n: n \in \Z} \le G$
That is, the subset of $G$ comprising all elements possible as powers of $a \in G$ is a subgroup of $G$.
Proof
Clearly $a \in H$, so $H \ne \O$.
Let $x, y \in H$.
\(\ds \) | \(\) | \(\ds x, y \in H\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \exists m, n \in \Z: x = a^m, y = a^n\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x^{-1} y = \paren {a^m}^{-1} a^n\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x^{-1} y = a^{-m} a^n = a^{n-m}\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x^{-1} y \in H\) |
Thus by the One-Step Subgroup Test:
- $H \le G$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 37.4$ Some important general examples of subgroups