# The order of conjugate subgroups.

Let $$G$$ be a group, and let $$H$$ be a subgroup of $$G$$.

Then, for any fixed $$g\in G$$, i know that $$H\cong gHg^{-1}$$ by inner-automorphism.

I have some question about the 'order':

(1) Is it true that $$|H|=|gHg^{-1}|$$ if $$|H|<\infty$$ or $$|G:H|<\infty$$?

(2) If (1) is true, give some counterexample for the case that $$|H|=\infty$$ or $$|G:H|=\infty$$.

Thank you!

• It’s true that $|H|=|gHg^{-1}|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality. – Arturo Magidin Apr 7 at 17:56

• Given two sets $$A$$ and $$B,$$ $$A$$ and $$B$$ are of equal cardinality if and only if there exists a bijection between them.
Consequently, we will always have $$|H|=\left|gHg^{-1}\right|.$$