Let \(G\) be a group. For \(X\subseteq G\), define \[ C(X)=\{y\in G:xy=yx\text{ for every }x\in X\}. \] Prove:
If \(X\subseteq Y\), then \(C(Y)\subseteq C(X)\).
\(X\subseteq C(C(X))\).
\(C(X)=C(C(C(X)))\).
Solution
1. If \(X\subseteq Y\), then\[ C(Y)\subseteq C(X). \]
Let \(g\in C(Y)\). By definition, \(g\) commutes with every element of \(Y\). Since \(X\subseteq Y\), it therefore commutes with every element of \(X\). Hence \[ g\in C(X). \] Thus \(C(Y)\subseteq C(X)\).
2. For every subset \(X\subseteq G\),\[ X\subseteq C(C(X)). \]
Let \(x\in X\). If \(y\in C(X)\), then by definition \[ xy=yx. \] Thus \(x\) commutes with every element \(y\) of \(C(X)\), which means \[ x\in C(C(X)). \] Therefore \(X\subseteq C(C(X))\).
3. One has\[ C(C(C(X)))\subseteq C(X). \]
Step 2 gives \[ X\subseteq C(C(X)). \] Applying the order-reversing property from step 1 to this inclusion yields \[ C(C(C(X)))\subseteq C(X). \]
4. One has\[ C(X)\subseteq C(C(C(X))). \]
Apply step 2 to the subset \(C(X)\) in place of \(X\). This gives \[ C(X) \subseteq C(C(C(X))). \]
5. Therefore\[ C(X)=C(C(C(X))). \]
Combine the two inclusions from steps 3 and 4.
6. Q.E.D.
Steps 1, 2, and 5 prove the three requested assertions.