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

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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.