Prove that every finite group of order \(n\) is isomorphic to a subgroup of \[ O(n), \] the group of real \(n\times n\) orthogonal matrices.
Solution
Let \[ G=\{g_1,\ldots,g_n\} \] be a finite group of order \(n\), and let \(V\) be the real vector space with basis \[ \{e_g:g\in G\}. \]
1. For each \(a\in G\), define a linear map\[ \rho(a):V\to V \]by\[ \rho(a)e_g=e_{ag} \]for every \(g\in G\).
Left multiplication \[ g\longmapsto ag \] is a permutation of the set \(G\). Hence the displayed prescription sends the basis of \(V\) to a basis and therefore extends uniquely to a linear automorphism of \(V\).
2. The assignment\[ \rho:G\to GL(V), \qquad a\longmapsto\rho(a), \]is a group homomorphism.
For \(a,b,g\in G\), \[ \rho(a)\rho(b)e_g = \rho(a)e_{bg} = e_{abg} = \rho(ab)e_g. \] Thus \[ \rho(a)\rho(b)=\rho(ab) \] because the two maps agree on a basis.
3. Each \(\rho(a)\) is orthogonal with respect to the inner product forwhich the basis \((e_g)_{g\in G}\) is orthonormal.
By step 1, \(\rho(a)\) permutes the orthonormal basis vectors. Hence for all \(g,h\in G\), \[ \inner{\rho(a)e_g}{\rho(a)e_h} = \inner{e_{ag}}{e_{ah}} = \delta_{ag,ah} = \delta_{g,h}. \] By bilinearity, \(\rho(a)\) preserves the inner product on all of \(V\). Therefore, after identifying \(V\) with \(\RR^n\) by this orthonormal basis, its matrix lies in \(O(n)\).
4. The homomorphism \(\rho\) is injective.
Let \(e\in G\) denote the identity. If \[ \rho(a)=\rho(b), \] then evaluating both maps on \(e_e\) gives \[ e_a = \rho(a)e_e = \rho(b)e_e = e_b. \] Distinct group elements label distinct basis vectors, so \(a=b\).
5. Therefore\[ \boxed{G\cong\rho(G)\leq O(n)}. \]
Steps 2 and 4 show that \(\rho\) is an injective group homomorphism, while step 3 shows that every matrix in its image is orthogonal.
6. Q.E.D.
Step 5 gives the required isomorphic embedding into \(O(n)\).