Question

For x belongs to G , define H _x = { g^(-1) x g | g belongs to G } . Under what conditions on x will H_x <= G ? Further, if H _x <= G , will H_x ∆= G ? Give reason for your answer.

Answer 1

Answer on Question #78752 - Math - Abstract Algebra

June 30, 2018

Question. For $x$ belonging to $G$, define $H_{x} = \{g^{-1}xg \mid g \in G\}$. Under what conditions on $x$ will $H_{x} \leq G$? Further, if $H_{x} \leq G$, will $H_{x} \triangleleft G$? Give reason for your answer.

Solution. Assume that $H_{x}$ is a subgroup of $G$ (denoted by $H_{x} \leq G$). Then $e \in H_{x}$. So there is $g \in G$ such that $e = g^{-1}xg$, but then $e = geg^{-1} = gg^{-1}xgg^{-1} = x$.

On the other hand, assume that $x = e$.

- We have $e = e^{-1}xe \in H_{x}$.

- For every $y \in H_{x}$, $y = g^{-1}xg$ for some $g \in G$, so $y = g^{-1}eg = e$.

Hence $H_{x} = \{e\}$. This is a trivial subgroup of $G$, so $H_{x} \leq G$.

We see that $H_{x} \leq G$ if and only if $x = e$.

The subgroup $\{e\}$ is also normal in $G$ because for every $g \in G$, $g^{-1}eg = e \in \{e\}$.

Answer. $H_{x} \leq G$ if and only if $x = e$, and in that case, $H_{x} \triangleleft G$.