Martin Exercise Solution(updating)

Groups

Laws of Composition

1.1

For all \(a,b,c\in S\), we have \((ab)c = a = a(bc)\), therefore law of composition in \(S\) is associative.

Suppose \(e\) is the identity, then for all \(a\in S\), we have \(ae=a=ea=e\), therefore \(S=\{e\}\)

1.2

1.3

For \(n\in \mathbb{N}\), since \(s\circ s^{-1}=s(s^{-1}(n))=n(n\ge2)\), we have \(s^{-1}(n)=n-1(n\ge2)\), however \(s(s^{-1}(1))=1\), assume \(s^{-1}(1)=x\), then \(s(x)=x+1=1\)\(\to x=0\) which is a contradiction to \(x\in\mathbb{N}\).

\(s^{-1}\circ s=s^{-1}(s(n))=s^{-1}(n+1)=n(n\ge 1)\), therefore \(s^{-1}(n)=n-1(n\ge 2)\), in addtion, we can define that \(s^{-1}(1)\) equals to any number in \(\mathbb{N}\).

Groups and Subgroups

2.2

• closure, apparently(law of composition)
• inverse, \(\forall a\in S,\exists a^{-1}\) satisfys that \(a^{-1}a=1\)

2.3
(a)\(y=x^{-1}w^{-1}z\)
(b)yes, no

2.4
(a)yes
(b)yes
(c)no
(d)yes
(e)no

2.5

Suppose \(e_{H}\neq e_{G}\), for all \(a\in H\), \(a\in G\) also, \(aa^{-1}=e_{H}\neq e_G\), it's impossible, therefore \(e_H=e_G\).