2.9 Centralizers and Center
Fix a magma \(M\) with binary operation \((x,y)\mapsto xy\).
The centralizer of a subset \(S\subseteq M\) is the set \(C(S)\) of all elements \(x\in M\), which commute with all elements of \(S\). So, explicitly,
If \(s\in S\), then, by abuse of notation, we set \(C(s) = C(\{ s\} )\} \). So, expicitly, \(C(s) = \{ x\in M: xs = sx\} \).
Suppose \(S\subseteq T\subseteq M\). Then \(C(T) \subseteq C(S)\).
Obvious (by reading the definition).
Suppose \(S, T \subseteq M\). Then the following are equivalent:
\(S\subseteq C(T)\).
\(T\subseteq C(S)\).
For all \(s\in S\) and \(t\in T\), \(st = ts\).
Exercise.
Suppose \(S\) is a nonempty subset of \(M\). Then \(C(S) = \cap _{s\in S} C(s)\).
Obvious (once you understand the definition and the abuse of notation).
Suppose \(S\subseteq M\)
If \(M\) is associative, then \(C(S)\) is a submagma of \(M\).
If \(M\) is a monoid, then \(C(S)\) is a submonoid of \(M\).
If \(M\) is a group, then \(C(S)\) is a subgroup of \(M\).
1 Suppose \(x,y\in C(S)\) and \(s\in S\). Then \((xy)s = x(ys) =x (sy) = (xs)y = (sx)y = s(xy)\).
2 Since we already know that \(C(S)\) is a submagma of \(M\), we just need to show that \(e \in C(S)\), where \(e\) denotes the identity element of \(M\). For that, pick \(s\in S\) and note that \(es = se\) by definition.
3 Since we already know that \(C(S)\) is a submonoid of \(M\), we just need to show that \(x\in C(S)\Rightarrow x^{-1}\in C(S)\). For that, pick \(s\in S\). Then \(x^{-1}s = x^{-1}se = x^{-1}sxx^{-1} = x^{-1}xsx^{-1} = e sx^{-1} = sx^{-1}\). So we’re done.
The center of a group \(G\) is the set of elements, which commute with all elements of \(G\). In other words,
If \(G\) is a group, then \(Z(G) = C(G)\). In other words, the center of \(G\) is just the centralizer of \(G\) itself viewed as a subset of \(G\).
Obvious from the definitions.
The center \(Z(G)\) of a group \(G\) is a subgroup of \(G\).
Obvious.
Suppose \(G\) is a group and \(S\subseteq G\). Then \(C(S) = C(\langle S\rangle )\).
Since \(S\subseteq \langle S\rangle \), it’s clear from Lemma 2.62 that \(C(\langle S\rangle ) \leq C(S)\). So we just need to prove that \(C(S) \leq C(\langle S\rangle )\).
To see this, suppose \(h\in C(S)\). Then \(S\subseteq C(h)\). But, since \(C(h)\leq G\) and \(\langle S\rangle \) is the smallest subgroup of \(G\) containing \(S\), this implies that \(\langle S\rangle \leq C(h)\). But this, in turn, implies that \(h\in C(\langle S\rangle )\). So, as \(h\) was arbitrary, \(C(S) \leq C(\langle S\rangle )\).
Suppose \(S\subseteq G\) is a set of commuting elements. In other words, suppose that, for all \(s,t\in S\), \(st = ts\). Then the subgroup \(\langle S\rangle \) is abelian.
By assumtion, \(S\subseteq C(S)\). And \(C(S) = C(\langle S \rangle )\) by Proposition 2.69. So \(S\subseteq C(\langle S\rangle )\). But this implies that \(\langle S\rangle \leq C(\langle S\rangle )\), which directly implies that \(\langle S\rangle \) is abelian.
Let \(G = \mathbf{GL}_2(\mathbb {R})\), the group of invertible \(2\times 2\) matrices with integer coefficients, and let \(J\) be the matrix
Let’s compute the centralizeer \(C(J)\) of \(J\). Suppose \(X\) is the matrix
Then
So we see that
Show that the center of \(\mathbf{GL}_2(\mathbb {R})\) consists of the set of diagonal matrices of the form
with \(a\neq 0\).