Question

Let Hbe a subgroup of a group Gand assume that ${\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathit{H}\mathit{x}\mathbf{\subseteq }\mathit{H}$for every$\mathit{x}\mathbf{\in }\mathit{G}$ (notation as in Exercise 27). Prove that ${\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathit{H}\mathit{x}\mathbf{=}\mathit{H}$for each$\mathit{x}\mathbf{\in }\mathit{G}$ .

Short answer

It is proved that $x^{-1}Hx=H$.

Step-by-step solution

Consider the given assumption

Let H be a subgroup of a group G. Assume that ${\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathit{H}\mathit{x}\mathbf{\subseteq }\mathit{H}$for every $\mathit{x}\mathbf{\in }\mathit{G}$.

Prove that x-1Hx=H

Consider the given group:

$$\begin{gathered} I=x^{-1}Hx \\ =\left\{x^{-1}ax|a\in H\right\} \end{gathered}$$

Consider an element $h$from group$H$. This implies $x^{-1}hx\in H$. Therefore, $xH{x}^{-1}\in H$.

It is sufficient to prove that group $H$is a subset of group$I$, that is, the condition$H\subseteq I$to prove the required results.

Perform the following operations on the element $h$representing the identity element.

$$\begin{gathered} h=ehe \\ =x^{-1}xh{x}^{-1}x \\ =x^{-1}Hx \end{gathered}$$

Therefore, the condition $H⊑x^{-1}Hx$is proved and $x^{-1}Hx=H$.