(2): In the case $bg(R)=0$, $R$ is Gorenstein and has minimal multiplicity and thus $e(R)\leq 2$. The converse also holds. Now suppose $bg(R)=1$. Then by Theorem \[thA\], $\m$ is isomorphic to $\m^\dag$. Therefore, $R$ has minimal multiplicity if and only if $R$ is almost Gorenstein.

\(1) $\implies$ (3): Clear.

Now assume the residue field $R/\m$ is infinite.

\(6) $\implies$ (4): Obviously, $S/\mathfrak{n}$ is also infinite. If $e(S)\leq \edim S$, then $e(S)\leq 2$ and $R$ also has an inequality $e(R)\leq 2$. This says that $R$ is Gorenstein and has minimal multiplicity. So we may assume $e(S)=\edim S+1$.

Take a minimal reduction $(t)$ of $\mathfrak{n}$ and a preimage $\delta\in \mathfrak{n}^2$ of a generator of the socle of $S/(t)$. Then $\mathfrak{n}^3=t\mathfrak{n}^2$ (see [@Sally2 Proof of (3.4)]), $\mathfrak{n}^2=t\mathfrak{n}+S\delta$ and $(t):_S \mathfrak{n}=(t)+S\delta$. Therefore $$R\cong \End_S(\mathfrak{n})\cong (t):_S \mathfrak{n}/t=S+S(\delta/t).$$ Identify $R$ with $S+S(\delta/t)$. Since $R$ is local and $\delta^2\in\mathfrak{n}^4=t^2\mathfrak{n}^2$, $(\delta/t)$ cannot be a unit of $R$. This shows $\m=\mathfrak{n}+S(\delta/t)$. By this equality, we also have an isomorphism $R/\m\cong S/\mathfrak{n}$ induced by $S\subseteq R$. Observe the following equalities $$t\m=t\mathfrak{n}+S\delta=\mathfrak{n}^2$$ and $$\m^2=(\mathfrak{n}+S(\delta/t))^2=\mathfrak{n}^2+\mathfrak{n}(\delta/t)+S(\delta/t)^2.$$ Then $\delta^2\in\mathfrak{n}^4=t^2\mathfrak{n}^2$ implies $S(\delta/t)^2\subseteq \mathfrak{n}^2$, and $\mathfrak{n}\delta\subseteq \mathfrak{n}^3=t\mathfrak{n}^2$ implies $\mathfrak{n}(\delta/t)\subseteq \mathfrak{n}^2$. So $\m^2=\mathfrak{n}^2=t\m$. This means that $R$ has minimal multiplicity.

It remains to show that $\m\cong \m^\dag$. 