Hw11 Problem 5

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### Problem 5

Show that the set of all $g\in G$ such that $i_g:G\leftrightarrow G$ is the identity inner automorphism is a normal subgroup of a group $G$.

**Solution**

Let $H = \{g \in G | i_{g}=i_{e}\}$ and $a,b\in H$. Then for $x \in G$ we have

(1)\begin{equation} (ab)x(ab)^{-1} = a(bxb^{-1})a^{-1} =axa^{-1} =x \end{equation}

so $i_{ab}=i_{e}$ and $ab \in H$. Clearly, $e \in H$, and $axa^{-1}=x$, then

(2)\begin{equation} x=a^{-1}xa=a^{-1}x(a^{-1})^{-1} \end{equation}

so $a^{-1} \in H$. Therefore, $H$ is a subgroup of $G$.

Now that it is a subgroup, we need to show that $H$ is a normal subgroup of $G$. Let $a\in H$ and $x\in G$. We must show that $xax^{-1}\in H$ or $i_{xax^{-1}} = i_{e}$. For any $y \in G$, we have

(3)\begin{equation} i_{xax^{-1}}(y)=(xax^{-1})y(xax^{-1})^{-1}=x[a(x^{-1}yx)a^{-1}]x^{-1}=x(x^{-1}yx)x^{-1}=y=i_e(y) \end{equation}

so, $i_{xax^{-1}}=i_e$.