|x−y|= 1, we have that the determinate of the derivative of fis ±1. So
f(x) = Rx +uwhere R∈GL(n). To show Ris orthogonal, we note that 4(Rx, Ry) = |Rx +Ry|2−
|Rx −Ry|2=|R(x+y)−R(0)|2− |Rx −Ry|2=|x+y|2− |x−y|2= 4(x, y), where ( ,) is the
normal inner product on Rn. So computation on an orthonormal basis yields that the matrix for R
is orthogonal.
The Galilei Group
5. Given ((R, u), v, s),((R0, u0), v0, s0)∈G(n+1), then F((R,u),v ,s)◦F((R0,u0),v0,s0)=F(RR0,Ru0+u,Rv0+v,s+s0).
Solution: Computation gives F((R,u),v,s)◦F((R0,u0),v0,s0)(x, t) = F((R,u),v,s)(R0x+u0+v0t, t +s0) =
(R(R0x+u0+v0t)+u+vt, t+s0+s) = (RR0x+Ru0+u+Rv0t+vt, t+s0+s) = F(RR0,Ru0+u,v0+v,s+s0)(x, t).
6. Given ((R, u), v, s)∈G(n+ 1), then F−1
((R,u),v,s)=F((R∗,−R∗u),−R∗v,−s).
Solution: The computation gives F((R,u),v,s)◦F((R∗,−R∗u),−R∗v,−s)(x, t) =
(RR∗x−RR∗u+u−RR∗vt +v t, t−s+s) = (x, t). Similarly F((R∗,−R∗u),−R∗v,−s)◦F((R,u),v,s)(x, t) =
(x, t).
7. With multiplication and inverses as defined, G(n+ 1) is a group.
Solution: Since the set of all invertible functions from Rnto Rnis a group under composition,
and since G(n+ 1) is nonempty, closed under composition and contains all inverses, it is subgroup,
and therefore a group.
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