MIT 18.703 Modern Algebra / Assignment 2

[E7-87-83]

Course: MIT 18.703 Modern Algebra
Document: ./MIT/Solutions/18.703/Assignment2/
Date: Early 2023

Ex6 ＃38, ＃55 okay

＃41 做咗＃40

＃46　比較詳細啲嘅寫法

(ab)^k = 1

a^{-1} 1 a = a^{-1} (ab)^k a = a^{-1}(ab)^k a = (ba)^k

呢道應該未教 Conjugacy class , 未講 identity 嘅 Conjugacy class 得一個element.
For arbitary x,y in G, x^-1 y x 未必等於 y .

仲有，應該補充證明 for 0 < n < k, (ba)^n != 1

＃48
如果提及 Every group with order >= 2 at least has one cyclic subgroup other than {e}，個證明就完整啦。
See also: https://math.stackexchange.com/questions/1310398/abstract-algebra-every-group-has-a-cyclic-subgroup

Ex7 straight-forward, ok

Ex8 ＃2, ＃8, ＃17, ＃23, ＃26, ＃30 ok

＃10 minor improvement

「Indeed, the isomorphism is $ F(x) = e^x $ that maps $ \mathbb{R} $ to $ \mathbb{R}^+ $.」

改為「Indeed, an isomorphism is $ F(x) = e^x $ that maps $ \mathbb{R} $ to $ \mathbb{R}^+ $.」

因為 isomorphism 可以唔至一個。

Ex9 ＃1, ＃10, ＃14 ok

Bonus ＃1, ＃2, ＃3, ＃5 ok

＃4

我估佢應該想問：「Show if two DISTINCT cycles are not disjoint, then they do not commute.」

第14行： we must have $ \sigma(i) = (i + k) \pmod n $ 未解釋k係咩

anyway 呢個相關resource解說得幾好： https://math.mit.edu/~mckernan/Teaching/12-13/Spring/18.703/l_6.pdf

"One reason why conjugation is so important, is because it measures how far the group G is from being abelian.
Another reason why conjugation is so important, is that really con- jugation is the same as translation."