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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!

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    $\begingroup$ It’s true that $|H|=|gHg^{-1}|$ always. Isomorphisms require a bijective underlying function, and a bijection establishes equality of cardinality. $\endgroup$ – Arturo Magidin Apr 7 at 17:56
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There are three facts needed, here:

  • Every group isomorphism is a bijection.
  • Given two sets $A$ and $B,$ $A$ and $B$ are of equal cardinality if and only if there exists a bijection between them.
  • The order of a group is the cardinality of the set of its elements.

Consequently, we will always have $|H|=\left|gHg^{-1}\right|.$

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