That field is the Galois field of order 4, denoted GF(2^2) or F_{2^2}. The operations there are addition modulo 4 and multiplication modulo 4 (I think this is what you meant by clock arithmetic). We can form finite fields GF(p^r) or F_{p^r} of order p^r where p is a prime.
meaning of "vanishes outside a set of finite measure"
What does it mean when we say that a function vanishes outside a set of finite measure? As in the definition of the integral as a prelude to the Lebesgue integral in Royden's Real Analysis, 3rd ed. (p. 77). It said:
If \phi vanishes...
How do you explain Littlewood's three principles in simpler terms? What does "nearly" mean (as in nearly a finite union of intervals, nearly continuous, and nearly uniformly convergent)?
And why are these important if I'm going to study the Lebesgue integral?
I'm learning this on my own so...
My teacher returned my assignment to me. She pointed out I made a mistake with the proof regarding my idea that the subgroup of G containing all elements of order 2 together with the identity has 2^k elements. From how I see it now, this idea could not be true. (Or is it?)
Dick has suggested...
Ah... I see where this should be going.
I decided to scrap the idea of using two cases. Instead, I have proven that if G is of order 2n, then G has an element of order 2 by partitioning G into two -- one partite cell containing the elements whose inverse is itself; and the other, those elements...
But I think we are not allowed to use that concept yet. The only concepts available to me, as far as the problem is concerned, are the definition of groups, homomorphisms and subgroups, cyclic groups, and cosets and counting (where this problem is part of).
So what I did was to use two cases...
The order of a group G, denoted |G| is the number of elements in G.
The order of an element a in G, denoted |a| is the least integer n such that a^{n}=e where e is the identity element in G.
Yes, you are correct, Z_3 has no elements of order 2 since 3 cannot be expressed as 2n for an integer...
Homework Statement
Let G be a group of order 2n. Show that G contains an element of order 2. If n is odd and G is abelian, then there is only one element of order 2
Homework Equations
Theorem (Lagrange):
If H is a subgroup of a group G, then |G|=[G:H]|H|. In particular, if G is finite...
I've not heard of such a morphism. But note that I'm not a seasoned mathematician. I'm just curious about what you would like to show. Of course, the identity mapping restricted to S would be an example of the kind of mapping that you want to construct.
Are you trying to make an analogue of...
I'm looking at the exercises of Hungerfod's Algebra. Some looks easy but it seems the proofs are not so obvious. Here's one I'm particularly having a hard time solving:
Let G be an abelian group of order pq with (p,q)=1. Assume that there exists elements a and b in G such that |a|= p and |b|...
I found this problem in a book for secondary students. It says it's from the Australian Mathematical Olympiad in 1982.
The sequence a_1,a_2,... is defined as follows:
a_1 = 2 and for n \geq 2, a_n is the largest prime divisor of a_1a_2...a_{n-1}+1.
Prove that 5 is not a member of this...