  </em>
</p>
<p>
  <em>Proof.</em>

  For all \( \gamma \in M^* \), \( x' \in \tau(M) \), we use the
  notation \( x = \tau^{-1}(x') \in M \) and get

  \[
    (\tau \gamma \tau^{-1})(x') = (\tau \gamma)(x) = \tau(x) = x'.
  \]

  In the second equality above, we have used the fact that \( \gamma
  \in M^* \) which implies that \( \gamma \) is an \( M \)-automorphism
  of \( L \) which allows us to conclude that \( \gamma(x) = x \) for
  \( x \in M \).  Since every \( \tau \gamma \tau^{-1} \in \tau M^*
  \t