【Basic Abstract Algebra】Exercises for Section 2.2 — Subgroups

1. Let \(H\) be a subgroup of \(G\), if \(g\in G\), show tha

   \[gHg^{-1}=\{g^{-1}hg\mid h\in H\} \]

   is also a subgroup of \(G\).
   Proof: Since \(e~(\text{identity})\in gHg^{-1}\subseteq G\), \(gHg^{-1}\) is nonempty. For any \(g^{-1}h_1g,~g^{-1}h_2g\in gHg^{-1}\), note that

   \[(g^{-1}h_1g)(g^{-1}h_2g)^{-1}=g^{-1}h_1gg^{-1}h_2^{-1}g=g^{-1}h_1h_2^{-1}g, \]

   and \(h_1h_2^{-1}\in H\) by \(h_1,h_2\in H\le G\). It follows that \((g^{-1}h_1g)(g^{-1}h_2g)^{-1}\in gHg^{-1}\). Thus \(gHg^{-1}\le G\).

2. Let \(G\) be a group and \(g\in G\). Show that the center of \(G\): \(\mathcal Z(G)=\{x\in G\mid gx=xg,~g\in G\}\) is a subgroup of \(G\). And compute the center of \(GL_n(\mathbb R),SL_n(\mathbb R)\).
   Proof: Clearly, the identity \(e\in\mathcal Z(G)\), i.e. \(\mathcal Z(G)\) is not empty. For any \(x_1,x_2\in\mathcal Z(G)\), we have \(gx_1=x_1g,~gx_2=x_2g\). Then

   \[g(x_1x_2^{-1})=x_1gx_2^{-1}=x_1gx_2^{-1}g^{-1}g=x_1g(gx_2)^{-1}g=x_1g(x_2g)^{-1}g=x_1gg^{-1}x_2^{-1}g=(x_1x_2^{-1})g. \]

   So \(x_1x_2^{-1}\in \mathcal Z(G)\). Thus \(\mathcal Z(G)\le G\).

   (1) The center of \(GL_n(\mathbb R):~\mathcal Z(GL_n(\mathbb R))=\{cE\mid c\in\mathbb R,E ~\text{is the identity matrix}\}\).

   • Let \(P\in\mathcal Z(GL_n(\mathbb R))\), then for any \(A\in GL_n(\mathbb R)\), we have \(AP=PA\). Suppose that \(A=\left(\begin{matrix} -1&0&\cdots&0\\0&1&\cdots&0\\\vdots&\vdots&&\vdots\\0&0&\cdots&1\end{matrix}\right)\in GL_n(\mathbb R)\), then by \(AP=PA\), we obtain that the first row and the first column of \(P\) are all \(0\) except the main diagonal element. Similarly, let \(A=(e_1,-e_2,\cdots,e_n),\cdots,(e_1,e_2,\cdots,-e_n)\), we can obtain that \(P\) is a diagonal matrix.
   • Moreover, let \(A\) be a permutation elementary matrix. By simple calculation, we can obtain \(P=cE\), where \(E\) is the identity matrix and \(c\in\mathbb R\).

   (2) The center of \(SL_n(\mathbb R):~\mathcal Z(SL_n(\mathbb R))=E\).

   • \(|cE|=1\Rightarrow c=1\).