remove the useless center

chapter-normal-subgroups-homomorphisms.tex

@@ -95,11 +95,8 @@
     k_2\inv h_2\inv = h' k_1 k_2\inv$. So this means $ab\inv = h_1 h' k_1
     k_2\inv$ which is in $HK$.
 
-    We have thus proven the following proposition. We state it here; but it is
-    not used much and thus does not go into a proposition box.
-    \begin{center}
-        Let $H$ be normal in $G$ and $K$ be a subgroup of $G$. Then, $HK$ is a subgroup of $G$.
-    \end{center}
+    We have thus proven the following proposition: Let $H$ be normal in $G$ and
+    $K$ be a subgroup of $G$. Then, $HK$ is a subgroup of $G$.
 \end{example}