Problem:
Solution:
(Part a)
The closure, associativity, identity properties are obvious. The only thing unclear is why such an operation has multiplicative inverse.
Note that for a fixed $ a \neq 0 $, if $ x \neq y $, then $ ax \neq ay $ because otherwise $ a(x - y) = 0 $ and that is impossible if both of them are non-zero.
When $ x $ varies from 1 to $ p $, there are $ p $ different $ ax $ values ranging from 1 to $ p $. By the pigeonhole principle one of them must be 1.
(Part b)
Thanks for the hint for using the Lagrange's theorem, we know that the subgroup generated by $ a $ have an order $ s $ which is a factor of $ p - 1 $. Now we can write $ a^{p-1} = a^{ks} = (a^s)^k = 1^k = 1 $.
For myself in the future, I need to explain more why $ a^s = 1 $. Consider the following sequence that represent the cyclic subgroup.
$ a^0, a^1 \cdots a^{s-1} $.
Now it is obvious why $ a^s = 1 $
(Part c)
Again, thanks for the hint. It is obvious now if $ a = 0, a^p = 0 = a $. If $ a \neq 0 $, part b applies and just multiply both side by $ a $.
(Part d)
What else would that be? $ a^p - a $, that has to be 0.
Solution:
(Part a)
The closure, associativity, identity properties are obvious. The only thing unclear is why such an operation has multiplicative inverse.
Note that for a fixed $ a \neq 0 $, if $ x \neq y $, then $ ax \neq ay $ because otherwise $ a(x - y) = 0 $ and that is impossible if both of them are non-zero.
When $ x $ varies from 1 to $ p $, there are $ p $ different $ ax $ values ranging from 1 to $ p $. By the pigeonhole principle one of them must be 1.
(Part b)
Thanks for the hint for using the Lagrange's theorem, we know that the subgroup generated by $ a $ have an order $ s $ which is a factor of $ p - 1 $. Now we can write $ a^{p-1} = a^{ks} = (a^s)^k = 1^k = 1 $.
For myself in the future, I need to explain more why $ a^s = 1 $. Consider the following sequence that represent the cyclic subgroup.
$ a^0, a^1 \cdots a^{s-1} $.
Now it is obvious why $ a^s = 1 $
(Part c)
Again, thanks for the hint. It is obvious now if $ a = 0, a^p = 0 = a $. If $ a \neq 0 $, part b applies and just multiply both side by $ a $.
(Part d)
What else would that be? $ a^p - a $, that has to be 0.
No comments:
Post a Comment