Question

Let $\mathit{H}$be a subgroup of a group $\mathit{G}$and, for$\mathit{x}\mathbf{\in }\mathit{G}$ , let ${\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathit{H}\mathit{x}$denote the set role="math" localid="1651629379568" $\left\{x^{-1}ax|a\in H\right\}$. Prove that a subgroup of G.

Short answer

It is proved that $x^{-1}Hx$ is a subgroup of G.

Step-by-step solution

Subgroup Theorem

A non-empty subset $H$of a group$G$is a subgroup of the latter if the elements $a$and $b$belong to $H$, then the element $ab$belongs to $H$, and if the element belongs $a$to $H$, then the element $a^{-1}$belongs to$H$ .

Show that x-1Hx is a subgroup of G

Consider the following group:

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

Consider two elements:

$a=x^{-1}{h}_{1}x$

$b=x^{-1}{h}_{2}x$

Here, the elements$h_1$and$h_2$belong to group $H$.

Find the product of the elements $a$and $b$.

localid="1651628363882" $$\begin{gathered} ab=x^{-1}{h}_{1}x{x}^{-1}{h}_{2}x \\ =x^{-1}{h}_1{h}_{2}x \end{gathered}$$

Since $H$is a group, the product $h_1{h}_2$is present in group $H$. This implies that the product is $ab=x^{-1}{h}_1{h}_{2}x$present in set $I$.

Find the inverse of the element $a=x^{-1}{h}_{1}x$.

$$\begin{gathered} a^{-1}={\left(x^{-1}{h}_{1}x\right)}^{-1} \\ ={\left(x^{-1}\right)}^{-1}{\left(h_1\right)}^{-1}{\left(x\right)}^{-1} \\ =x{\left(h_1\right)}^{-1}{\left(x\right)}^{-1} \end{gathered}$$

Since is a group, the ${h_1}^{-1}$inverse is present in group H. This implies that the $a^{-1}$inverse is present in set $l$.

Hence, the set $\mathbf{\left\{}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{ax}\mathbf{|}\mathbf{a}\mathbf{\in }\mathbf{H}\mathbf{\right\}}$is a subgroup of G