*Algebra*

## Section 6.8. The Operation on Cosets

Exercise 6.8.1 Yes. \(I * A = I A I^t = A\) and \((PQ) * A = (PQ)A(PQ)^t = PQAQ^tP^t = P(QAQ^t)P^t = P * (Q * A)\).

Exercise 6.8.2 If \(g\) is in the stabilizer of some subset \(S \subseteq G\), then it's not hard to see that \(aga^{-1}\) is in the stabilizer of \(aS\). Conversely, every element \(g'\) that stabilizes \(aS\) must have this form, since \(a^{-1}g'a\) would stabilize \(S\), implying that \(g' = aga^{-1}\) where \(g\) stabilizes \(S\). By Proposition 6.8.1, the stabilizer of the coset \(aH\) is therefore the conjugate subgroup \(aHa^{-1}\).

Exercise 6.8.3 Let \(s\) be some vertex in \(S\). The stabilizer of \(s\) consists of the identity operation and the reflection across the diagonal containing \(s\). Call this reflection \(r\), so \(H = \{1, r\}\). The orbit is all of \(S\). There are \(8 \div 2 = 4\) left cosets \(aH\), namely \(\{1, r\}, \{\rho, \rho r\}, \{\rho^2, \rho^2 r\}, \{\rho^3, \rho^3 r\}\), which are respectively mapped by \(\epsilon\) to the vertices \(1s = rs = s, \rho s = \rho r s, \rho^2 s = \rho^2 r s, \rho^3 s = \rho^3 r s\), thus hitting all four vertices exactly once. That is, \(\epsilon\) is bijective as expected.

Exercise 6.8.4 The subgroup \(H\) consists of all \((n-1)!\) permutations on the indices \(\{2, 3, \ldots, n\}\). Let \(a \in S_n\). Then the coset \(aH\) consists of permutations \(\sigma\) such that \(\sigma(1) = a(1)\), since \(ah(1) = a(1)\) for all \(h \in H\), since \(h(1) = 1\). Since the coset \(aH\) contains \((n-1)!\) elements and there are \((n-1)!\) permutations with \(\sigma(1) = a(1)\), the coset \(aH\) must contain all of them. The map (6.8.4) sends the coset \(aH\) to \(a(1)\), by definition.