Powers of Element form Subgroup

From ProofWiki
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