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Let $G$ be a group and let $H$ be a subgroup. Prove that the following are equivalent. (1) $H$ is normal in $G$.

$(1) \Rightarrow (2)$: Assume that $H$ is a normal subgroup of $G$. We need to show that for every $g \in G, g H g^{-1}=H$. Let $g\in G$ and $h \in H$ be arbitrary. Since $H$ is normal in $G$, we have $ghg^{-1}\in H$. Hence, $gHg^{-1} \subseteq H$. To show the other inclusion, let $x\in H$. Then $g^{-1}xg \in H$ since $H$ is normal, and so $x=gg^{-1}xgg^{-1}=g(g^{-1}xg)g^{-1}\in gHg^{-1}$. Therefore, we have $H\subseteq gHg^{-1}$, which implies that $gHg^{-1}=H$.

(2) For every $g \in G, g H g^{-1}=H$.

$(2) \Rightarrow (3)$: Assume that for every $g\in G, g H g^{-1}=H$. We need to show that for every $a \in G, a H=H a$. Let $a \in G$ be arbitrary. We want to show that $aH\subseteq Ha$. Let $h\in H$ be arbitrary. Then, we have $ah=g(g^{-1}ag)h(g^{-1}ag)^{-1}\in gHg^{-1}=H$, where we use that $g^{-1}ag$ is just some element of $G$ that we can conjugate with. Therefore, $ah\in Ha$ for all $h \in H$, which implies that $aH \subseteq Ha$. By a similar argument, we can show that $Ha \subseteq aH$, and so we have $aH=Ha$.

\begin{prob}

(3) For every $a \in G, a H=H a$.

\end{prob}

\begin{proof}

$(3) \Rightarrow (1)$: Assume that for every $a \in G, a H=H a$. We need to show that $H$ is normal in $G$. Let $g\in G$ and $h\in H$ be arbitrary. We want to show that $ghg^{-1}\in H$. To do this, note that $g^{-1}hg \in H$ since $H$ is closed under conjugation by elements of $G$. Therefore, $ghg^{-1}=g(g^{-1}hg)g^{-1} \in gHg^{-1}$, and so $H$ is normal in $G$.