Brian Bi
$\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal}$

## Section 2.2. Groups and Subgroups

Exercise 2.2.1 Using Artin's notation,

 1 $$x$$ $$x^2$$ $$y$$ $$xy$$ $$x^2y$$ 1 1 $$x$$ $$x^2$$ $$y$$ $$xy$$ $$x^2y$$ $$x$$ $$x$$ $$x^2$$ 1 $$xy$$ $$x^2y$$ $$y$$ $$x^2$$ $$x^2$$ 1 $$x$$ $$x^2y$$ $$y$$ $$xy$$ $$y$$ $$y$$ $$x^2y$$ $$xy$$ 1 $$x^2$$ $$x$$ $$xy$$ $$xy$$ $$y$$ $$x^2y$$ $$x$$ 1 $$x^2$$ $$x^2y$$ $$x^2y$$ $$xy$$ $$y$$ $$x^2$$ $$x$$ 1

Exercise 2.2.2 Clearly 1 is invertible since $$1\cdot 1 = 1$$. If $$a \in S$$ is invertible with inverse $$a^{-1}$$, then $$a$$ is also the inverse of $$a^{-1}$$, so $$a^{-1}$$ is invertible too. And if $$a, b \in S$$ are invertible, then $$ab$$ is invertible with inverse $$b^{-1} a^{-1}$$. This establishes that the set of invertible elements of $$S$$ with law of composition inherited from $$S$$ is a group.

Exercise 2.2.3

1. Multiply both sides by $$x^{-1}$$ on the left and $$w^{-1}z$$ on the right to obtain $$y = x^{-1}w^{-1}z$$.
2. Yes, $$yzx = x^{-1}(xyz)x = x^{-1}(1)x = 1$$. No, it isn't necessarily the case that $$yxz = 1$$. For example, in $$S_3$$, we have $$x \cdot y \cdot xy = 1$$, but $$y \cdot x \cdot xy = x \neq 1$$.

Exercise 2.2.4

1. Yes. $$H$$ is itself a group under the same operation as $$G$$ and is a subset of $$G$$, therefore $$H$$ is a subgroup of $$G$$.
2. Yes. $$H$$ contains the identity element 1; the elements 1 and -1 are their own inverses; and $$H$$ is closed under multiplication.
3. No. $$H$$ doesn't contain the identity element 0.
4. Yes. $$H$$ contains the identity element 1; the inverse of a positive real is another positive real; and the positive reals are closed under multiplication.
5. No. $$H$$ doesn't contain the identity matrix.

Exercise 2.2.5 If $$1_H$$ is the identity of $$H$$, then it must be the case that $$1_H 1_H = 1_H$$. When this is regarded as an equality in $$G$$, we can cancel $$1_H$$ from both sides to obtain $$1_H = 1_G$$. Analogously, if $$a \in H$$ and $$a^*$$ denotes the inverse of $$a$$ in $$H$$, then $$aa^* = 1_H = 1_G$$. Left-multiplying both sides by the element $$a^{-1}$$ (the inverse of $$a$$ in $$G$$) yields $$a^* = a^{-1}$$.

Exercise 2.2.6 The element 1 is the identity in $$G^\circ$$ since $$1*a = a1 = a$$ and $$a*1 = 1a = a$$ for all $$a \in G^\circ$$. If $$a \in G^\circ$$ then the inverse element $$a^{-1}$$ in $$G$$ is also the inverse in $$G^\circ$$ since $$a*a^{-1} = a^{-1}a = 1$$ and $$a^{-1}*a = aa^{-1} = 1$$. If $$a, b, c \in G^\circ$$ then $$a*(b*c) = (b*c)a = (cb)a = c(ba) = (ba)*c = (a*b)*c$$, so $$*$$ is associative. We conclude that $$(G^\circ, *)$$ is a group.