Order of Conjugate of Subgroup

Theorem

Let $G$ be a group.

Let $H$ be a subgroup of $G$ such that $H$ is of finite order.


Then $\order {H^a} = \order H$.


Proof

From the definition of Conjugate of Group Subet we have $H^a = a H a^{-1}$.

From Set Equivalence of Regular Representations:

$\order {a H a^{-1} } = \order {a H} = \order H$

$\blacksquare$


Sources

  • 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $5$
  • 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions: Exercise $1$
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