Prove / disprove that [tex] \left< U_n, \cdot \right> [/tex] is a group. The elements of [tex] U_n [/tex] is the solutions to [tex] x^n = 1 [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

Example:

[tex] \left< U_4, \cdot \right> [/tex] is the solutions to [tex] x^4 = 1 [/tex], [tex] U_4 = \left\{ 1, -1, i, -i \right\} [/tex]. And here [tex] \cdot [/tex] is multiplication. So I'm wondering if this is enough to prove that [tex] \left< U_n, \cdot \right>[/tex] is a group ...

1. There exists [tex] e \in U_n [/tex] such that ae=ea=a for all a. This can be shown to be e=1 since [tex] 1\cdot a = a \cdot 1 = a [/tex]. We know that for all n then 1^n = 1 ... hence [tex] 1 \in U_n [/tex]

2. Closure, if [tex] a,b, \in U_n [/tex] then [tex] a \cdot b \in U_n [/tex]. This must be true since if [tex] a^n = b^n = 1 [/tex] then [tex] \left( a \cdot b \right)^n = a^n \cdot b^n = 1 [/tex] hence it is closed under multiplication

3. Existence of inverse for all elements, this must be true since if [tex] a^n = 1 [/tex] then we know from the fact that the elements of [tex] U_n [/tex] is the ones of the form x^n = 1. So all x satisfying this must have |x| = 1. Hence if [tex] a^n = 1 [/tex] for [tex] a = e^{iv} [/tex] for some v then [tex] a' = e^{-iv} [/tex] and this is the conjugate of a. Ie [tex] a' = \bar{a} [/tex]. It can be prove that if a is a root of a polynom with real coefficients then [tex] \bar{a} [/tex] must also be a solution. Hence a' exists.

4. Associative, this is true due to that normal multiplication is assocative

The one that I'm not sure about is 2, the one about closeure ... but I don't know.

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# The group U_n

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