Question

N1: Prove that the specified subset is a subgroup of the given group…

a.) The set of complex numbers of absolute value 1, (i.e. the unit circle in the complex plane (under multiplication)

b.) For fixes n ε Z+ the set of rational numbers, whose denominators are relatively prime to n (under addition).

Answer 1

if i remember correctly since complex numbers are an abelian all you have to show it that the set is closed under inverses.
the inverse of
\[e^{i\theta}\] is
\[e^{-i\theta}\] which also has absolute value 1

Answer 2

second one is also commutative, but you need i suppose to show it is closed. i.e. if
\[\frac{a}{b},\frac{c}{d}\] then so is
\[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] and this amounts to showing that if
\[b, d\] are relatively prime to n, then so is
\[bd\] which you can do by the fundamental theorem of arithmetic. I hope i didn't leave anything out, but i may have

Answer 3

Oh ok, it all sounds good to me, thank you so much.

Answer 4

N4: Show that {x ε Dn | X\[^{2}\] = e} is not a subgroup of Dn (n ≥ 3).