- #1
hedlund
- 34
- 0
Prove / disprove that \left< U_n, \cdot \right> is a group. The elements of U_n is the solutions to x^n = 1.
Example:
\left< U_4, \cdot \right> is the solutions to x^4 = 1, U_4 = \left\{ 1, -1, i, -i \right\}. And here \cdot is multiplication. So I'm wondering if this is enough to prove that \left< U_n, \cdot \right> is a group ...
1. There exists e \in U_n such that ae=ea=a for all a. This can be shown to be e=1 since 1\cdot a = a \cdot 1 = a. We know that for all n then 1^n = 1 ... hence 1 \in U_n
2. Closure, if a,b, \in U_n then a \cdot b \in U_n. This must be true since if a^n = b^n = 1 then \left( a \cdot b \right)^n = a^n \cdot b^n = 1 hence it is closed under multiplication
3. Existence of inverse for all elements, this must be true since if a^n = 1 then we know from the fact that the elements of U_n is the ones of the form x^n = 1. So all x satisfying this must have |x| = 1. Hence if a^n = 1 for a = e^{iv} for some v then a' = e^{-iv} and this is the conjugate of a. Ie a' = \bar{a}. It can be prove that if a is a root of a polynom with real coefficients then \bar{a} 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.
Example:
\left< U_4, \cdot \right> is the solutions to x^4 = 1, U_4 = \left\{ 1, -1, i, -i \right\}. And here \cdot is multiplication. So I'm wondering if this is enough to prove that \left< U_n, \cdot \right> is a group ...
1. There exists e \in U_n such that ae=ea=a for all a. This can be shown to be e=1 since 1\cdot a = a \cdot 1 = a. We know that for all n then 1^n = 1 ... hence 1 \in U_n
2. Closure, if a,b, \in U_n then a \cdot b \in U_n. This must be true since if a^n = b^n = 1 then \left( a \cdot b \right)^n = a^n \cdot b^n = 1 hence it is closed under multiplication
3. Existence of inverse for all elements, this must be true since if a^n = 1 then we know from the fact that the elements of U_n is the ones of the form x^n = 1. So all x satisfying this must have |x| = 1. Hence if a^n = 1 for a = e^{iv} for some v then a' = e^{-iv} and this is the conjugate of a. Ie a' = \bar{a}. It can be prove that if a is a root of a polynom with real coefficients then \bar{a} 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.
