$hKh^{-1}=K$ if and only if $h^{-1}Kh=K$?

Question

Suppose $G$ is a group and $K$ is a subgroup of $G$. If $h\in G$, is it true that

$$hKh^{-1}=K \Leftrightarrow h^{-1}Kh=K?$$

I am reading a proof on page 94 of Dummit and Foote's Abstract Algebra, 3rd edition.

I am guessing we have $h^{-1}hk\in K$ (the highlighted part) because the above statement holds. Could someone please verify?

Answer 1

Your statement is true.
Since $N_G(K)$ is a subgroup, $$hKh^{-1}=K\iff h\in N_G(K) \iff h^{-1}\in N_G(K) \iff h^{-1}Kh=K$$ As stated in comment, you just have to notice that for $h\in H$, $h^{-1}\in H\le N_G(K)$. Hence $h^{-1}kh\in K$.

Answer 2

If $$hKh^{-1}=K$$ then $$h^{-1}(hKh^{-1})h=h^{-1}Kh,$$ $$(h^{-1}h)K(h^{-1}h)=h^{-1}Kh,$$ $$eKe=h^{-1}Kh,$$ $$K=h^{-1}Kh.$$

If $$h^{-1}Kh=K$$ then $$h(h^{-1}Kh)h^{-1}=hKh^{-1},$$ $$(hh^{-1})K(hh^{-1})=hKh^{-1},$$ $$eKe=hKh^{-1},$$ $$K=hKh^{-1}.$$

Answer 3

For every $x\in G$, the maps $\lambda_x\colon G\to G$ and $\rho_x\colon G\to G$ defined by $\lambda_x(g)=xg$ and $\rho_x(g)=gx$ are bijections.

Thus, for $A,B\subseteq G$, the following are equivalent:

1. $A=B$;
2. $\lambda_x(A)=\lambda_x(B)$ (in other notation $xA=xB$);
3. $\rho_x(A)=\rho_x(B)$ (in other notation $Ax=Bx$).

Now it's just computing and observing that $x^{-1}xA=A$ and the other similar identities.