Dummit and Foote: conjugate elements and conjugate subgroups have the same order.

Question

Source: Abstract Algebra, 3rd edition, Dummit and Foote.

The statement is from Corollary 14 in Section 4.4 "If $K$ is any subgroup of the group $G$ and $g\in G$, then $K\cong gKg^{-1}$. Conjugate elements and conjugate subgroups have the same order."

My Question: Does "the order of conjugate elements" refer to the number of conjugates of a given element $g\in G$ (i.e., the cardinality of the orbit of $g$)? If yes, then what is the conjugate subgroups here then? This is not the first time I got confused by Dummit and Foote-any clarification would be greatly appreciated.

Answer 1

Does "the order of conjugate elements" refer to the number of conjugates of a given element $g \in G$ (i.e., the cardinality of the orbit of $g$)?

No. The "order of an element" refers to the number of its powers (finite order is the least $n > 0$ making the $n$th power trivial). The order of an element is not usually equal to the number of elements conjugate to it: the order of $g$ counts the number of different $g^n$ as $n$ runs through $\mathbf Z$, while the order of the conjugacy class of $g$ refers to the number of different $tgt^{-1}$ as $t$ runs over $G$. It's not the same thing. Nobody ever calls the number of conjugates of an element in a group the "order" of that element.

If yes, then what is the conjugate subgroups here then?

The "order of a subgroup" is the number of elements in it. If $K = gHg^{-1}$ then $H$ and $K$ have the same cardinality (the same order). If $y = gxg^{-1}$ then $x$ and $y$ have the same number of different powers (the same order as elements of the group).