RE: Cryptography 101 (An Interactive Class) : An Introduction to Group Theory - Week 1
The set of even integers is an additive group. The identity is 0, since 0 + g = g for any integer g (even or otherwise). Addition is associative: if g1, g2, and g3 are even integers, then (g1 + g2) + g3 = g1 + (g2 + g3). Given an even integer g, its inverse is just -g. g + (-g) = 0. These seem pretty obvious. There's nothing else we need to say to show associativity and inverses, right?
For closure, suppose you've got two even integers, g1 and g2. Since they're even, you can factor out a 2 from each: g1 = 2m and g2 = 2 for some integers m and n. So, g1 + g2 = 2m + 2n = 2(m + n), and m + n is the sum of two integers, so it is also an integer, meaning the sum is an even integer.The group permutation that moves {1, 2, 3} to {3, 2, 1} is (1 3).
1) Associativity can be taken from the integers, since we know the integers are associative, a subset of them will be as well. Inverses we do not automatically obtain, but it is obvious that if g = 2k in 2Z, for k in Z, then -g = -2k is also in 2Z. But, we need to explicitly state that and verify that it's the case.
Closure is done correctly. Good job!
2) Correct.
95/100, since you didn't explicitly state how the inverse of an even integer is also even.Oh, ok. Thanks