for all \(x,y\) in \(G\text{.}\) A homomorphism that is both injective (one-to-one) and surjective (onto) is called an isomorphism of groups. If \(\phi\colon G\to H\) is an isomorphism, we say that \(G\) is isomorphic to \(H\text{,}\) and we write \(G\approx H\text{.}\)
Show that each of the following are homomorphisms.
\(GL(n,\R)\to \R^\ast\) given by \(M\to \det M\)
\(\Z\to \Z\) given by \(x\to mx\text{,}\) some fixed \(m\in \Z\)
\(G\to G\text{,}\)\(G\) any group, given by \(x\to
axa^{-1}\text{,}\) some fixed \(a\in G\)
\(\C^\ast\to\C^\ast\) given by \(z\to z^2\)
Show that each of the following are not homomorphisms. In each case, demonstrate what fails.
\(\Z\to \Z\) given by \(x\to x+3\)
\(\Z\to \Z\) given by \(x\to x^2\)
\(D_4\to D_4\) given by \(g\to g^2\)
Instructor’s solution for Checkpoint 2.4.2.
Given two \(n\times n\) matrices \(M,N\text{,}\) we have \(\det(MN)=\det(M)\det(N)\text{,}\) so the determinant map is a homomorphism on \(GL_n\text{.}\) Given \(a,b\in \Z\text{,}\) the distributive law \(m(a+b)=ma+mb\) says that multiplication by \(m\) is a homomorphism. Given a group \(G\) and elements \(a,x,y\in G\text{,}\) we have
so conjugation by \(a\) is a homomorphism. Given \(z,w\in
\C^\ast\text{,}\) we have \((zw)^2=zwzw=zzww=z^2w^2\text{,}\) so the squaring map is a homomorphism.
Proposition2.4.5.Basic properties of homomorphisms.
Let \(\phi\colon G\to H\) be a homomorphism of groups. Let \(e_G,e_H\) denote the identity elements of \(G,H\text{,}\) respectively. We have the following.
(identity goes to identity) \(\phi(e_G) = e_H\)
(inverses go to inverses) \(\phi\left(g^{-1}\right) = \left(\phi(g)\right)^{-1}\) for all \(g\in G\)
\(\ker(\phi)\) is a subgroup of \(G\)
\(\phi(G)\) is a subgroup of \(H\)
(preimage sets are cosets of the kernel) \(\phi(x)=y\) if and only if \(\phi^{-1}(y) =
x\ker(\phi)\)
\(\phi(a)=\phi(b)\) if and only if \(a\ker(\phi)=b\ker(\phi)\)
\(\phi\) is one-to-one if and only if \(\ker(\phi)=\{e_G\}\)
Proposition2.4.6.\(G/K\) is a group if and only if \(K\) is a kernel.
Let \(K\) be a subgroup of a group \(G\text{.}\) The set \(G/K\) of cosets of \(K\) forms a group, called a quotient group (or factor group), under the operation
if and only if \(K\) is the kernel of a homomorphism \(G\to G'\) for some group \(G'\text{.}\)
Corollary2.4.7.(First Isomorphism Theorem).
Let \(\phi\colon G\to H\) be a homomorphism of groups. Then \(G/\ker(\phi)\) is isomorphic to \(\phi(G)\) via the map \(g\ker(\phi) \to \phi(g)\text{.}\)
Definition2.4.8.Normal subgroup.
A subgroup \(H\) of a group \(G\) is called normal if \(ghg^{-1}\in H\) for every \(g\in G\text{,}\)\(h\in H\text{.}\) We write \(H\trianglelefteq G\) to indicate that \(H\) is a normal subgroup of \(G\text{.}\)
Proposition2.4.9.Characterization of normal subgroups.
Let \(K\) be a subgroup of a group \(G\text{.}\) The following are equivalent.
\(K\) is the kernel of some group homomorphism \(\phi\colon G\to H\)
\(G/K\) is a group with multiplication given by Equation (2.4.1)
Instructor’s solution for Exercise 2.4.2.1 (proof of Proposition 2.4.5, part 1).
For Property 1, apply \(\phi\) to both sides of \(e_G=e_Ge_G\) to get \(\phi(e_G)=(\phi(e_G))^2\text{.}\) Now multiply both sides by \((\phi(e_G))^{-1}\) to get \(e_H=\phi(e_G)\text{.}\) For Property 2, apply \(\phi\) to both sides of \(gg^{-1}=e_G\) to get \(\phi(g)\phi(g^{-1})=e_H\) (using Property 1), so that we have \(\phi(g^{-1})=(\phi(g))^{-1}\text{.}\)
Instructor’s solution for Exercise 2.4.2.2 (proof of Proposition 2.4.5, part 2).
For Property 3, let \(a,b\) be elements of \(\ker(\phi)\text{.}\) Then \(\phi(ab)=\phi(a)\phi(b)=e_H\text{,}\) so \(ab\) is in \(\ker(\phi)\text{.}\) We also have \(e_H=(\phi(a))^{-1}=\phi(a^{-1})\) (by Property 2) so \(\ker(\phi)\) is closed under taking inverses. Also, \(\ker(\phi)\) is not empty (\(\ker(\phi)\) contains \(e_G\) by Property 1), so \(\ker(\phi)\) is a subgroup by the 2-step subgroup test. For Property 4, if \(\phi(x_1)=y_1\) and \(\phi(x_2)=y_2\) are in \(\phi(G)\text{,}\) then \(y_1y_2 = \phi(x_1)\phi(x_2)=
\phi(x_1x_2)\text{,}\) so \(\phi(G)\) is closed under multiplication. Also, \(y_1^{-1}=(\phi(x_1))^{-1}=\phi(x_1^{-1})\) (by Property 2), so \(\phi(G)\) is closed under taking inverses. Finally, \(\phi(G)\) is not empty because it contains \(e_H\) (by Property 1), so \(\phi(G)\) passes the 2-step subgroup test.
Instructor’s solution for Exercise 2.4.2.3 (proof of Proposition 2.4.5, part 3).
For Property 5, suppose \(\phi(x)=y\text{.}\) If \(k\in
\ker(\phi)\text{,}\) then \(\phi(xk)=
\phi(x)\phi(k)=\phi(x)=y\) so \(\phi^{-1}(y)\) contains \(x\ker(\phi)\text{.}\) Conversely, if \(a\in \phi^{-1}(y)\text{,}\) then \(\phi(x^{-1}a)=\phi(x)^{-1}\phi(a)=y^{-1}y=e_H\text{,}\) so \(x^{-1}a\in \ker(\phi)\text{.}\) This implies \(a\in
x\ker(\phi)\) by Proposition 2.3.9. For Property 6, apply Property 5 to conclude that \(\phi(a)=\phi(b)=c\) if and only if \(a\ker(\phi)=\phi^{-1}(c)=b\ker(\phi)\text{.}\) For Property 7, observe that Property 6 implies that the kernel of \(\phi\) has more than one element if and only if \(\phi\) is not one-to-one.
4.
Show that the inverse of an isomorphism is an isomorphism.
Instructor’s solution for Exercise 2.4.2.4.
An isomorphism is a bijection, so the inverse mapping is a bijection. We need to prove the inverse map is operation preserving. Let \(\phi\colon G\to H\) be an isomorphism of groups. Let \(a,b\) be elements in \(H\text{,}\) and let \(x=\phi^{-1}(a), y=\phi^{-1}(b)\text{.}\) Then
First, suppose \(K=\ker(\phi)\) for some homomorphism \(\phi\colon G \to G'\text{.}\) Explain why Item 6 of Proposition 2.4.5 can be rephrased to say that there is a one-to-one correspondence \(G/K
\leftrightarrow \phi(G)\) given by \(gK\leftrightarrow
\phi(g)\text{.}\) Now use the bijection \(G/K\leftrightarrow
\phi(G)\) to impose the group structure of \(\phi(G)\) (Item 4 of Proposition 2.4.5) on \(G/K\text{.}\) Conversely, if \(G/K\) is a group with the group operation (2.4.1), define \(\phi\colon G\to
G/K\) by \(\phi(g)=gK\text{,}\) then check that \(\phi\) is a homomorphism and that \(\ker(\phi)=K\text{.}\)
Instructor’s solution for Exercise 2.4.2.5 (proof of Proposition 2.4.6).
For the "if" direction, let \(K=\ker(\phi)\) for some homomorphism \(\phi \colon G\to G'\) for some group \(G'\text{.}\) By Item 6 of Proposition 2.4.5, we have a one-to-one correspondence \(G/K\leftrightarrow \phi(G)\text{.}\) The group structure of \(\phi(G)\) (see Item 4) induces the group multiplication (2.4.1) on \(G/K\text{.}\) For the "only if" direction, suppose \(G/K\) is a group under the operation (2.4.1). Let \(\phi\colon G\to G/K\) be given by \(g\to gK\text{.}\) Then it is easy to check that \(K=\ker(\phi)\text{.}\)
7.
Let \(n,a\) be relatively prime positive integers. Show that the map \(\Z_n\to \Z_n\) given by \(x\to ax\) is an isomorphism.
Use the fact that \(\gcd(m,n)\) is the least positive integer of the form \(sm+tn\) over all integers \(s,t\) (see Exercise 2.3.2.4). Use this to solve \(ax=1 \pmod{n}\) when \(a,n\) are relatively prime.
Instructor’s solution for Exercise 2.4.2.7.
Suppose \(\gcd(a,n)=1\text{.}\) Then there exist integers \(s,t\) such that \(sa+tn=1\text{.}\) Then we have \(as = 1-tn = 1\pmod{n}\text{,}\) so \(x=s \pmod{n}\) is a solution to \(ax=1 \pmod{n}\text{.}\) This give us an inverse mapping.
More comprehensive version of the problem: Show that \(\mu_a\colon \Z_n\to \Z_n\) given by \(\mu_a(x)=ax\) is an isomorphism if and only if \(n,a\) are relatively prime.
Solution to the more comprehensive version of the problem. We claim the following.
Claim2.4.10.
\(\ker(\mu_a)=\left\langle
\frac{\lcm(a,n)}{a}\right \rangle\text{.}\) Here is how result follows from the Claim. If \(n,a\) are relatively prime, then \(\lcm(n,a)=na\text{,}\) so \(\lcm(n,a)/a=n\text{,}\) so \(\ker(\mu_a)=\{0\}\text{.}\) Now apply Property 7 of Proposition 2.4.6, together with the fact that for a mapping from a finite set to itself, the mapping is one-to-one if and only if the mapping is onto. If \(n,a\) are not relatively prime, then \(\lcm(n,a)\lt
na\text{,}\) and \(\lcm(n,a)/a\leq n\text{.}\) Then \(\ker(\mu_a)\) is not trivial. Now apply Property 7 of Proposition 2.4.6 to conclude that \(\mu_a\) is not one-to-one, and therefore is not an isomorphism.
Now we prove the Claim. Let \(b=\langle \lcm(a,n)/a\rangle\text{.}\) We have
so \(b\in \ker(\mu_a)\text{.}\) If \(c\in \ker(\mu_a)\text{,}\) then \(\mu_a(c)=ac=kn\) for some integer \(k\text{.}\) Since \(kn\) is divisible by \(a\text{,}\) we have \(kn\geq \lcm(a,n)\text{,}\) so \(c\geq
b\text{.}\) This shows \(b\) is the smallest positive element in \(\ker(\mu_a)\) (or \(b=0\)). Applying Exercise 2.3.2.4, we conclude the statement in the Claim.
Note: It is not necessary to prove the Claim to solve the given exercise. It is enough simply to note that \(b\in \ker(\mu_a)\text{.}\)
8.Another construction of \(\Z_n\).
Let \(n\geq 1\) be an integer and let \(\omega=e^{i2\pi/n}\text{.}\) Let \(\phi\colon
\Z\to S^1\) be given by \(k\to \omega^k\text{.}\)
Show that the the image of \(\phi\) is the group \(C_n\) of \(n\)th roots of unity.
Show that \(\phi\) is a homomorphism, and that the kernel of \(\phi\) is the set \(n\Z=\{nk\colon k\in \Z\}\text{.}\)
Conclude that \(\Z/\!(n\Z)\) is isomorphic to the group of \(n\)-th roots of unity.
Instructor’s solution for Exercise 2.4.2.8.
By Checkpoint 2.1.10, we have \(C_n=\{\omega^0,\ldots,\omega^{n-1}\}\text{,}\) so it is clear that \(C_n\subseteq \img(\phi)\text{.}\) Conversely, given \(k\in Z\text{,}\) write \(k=qn+r\) for some \(q\in \Z\) and \(0\leq r\leq n-1\text{.}\) Then we have \(\omega^k=\omega^r\text{.}\) Thus we have \(\img(\phi)\subseteq
C_n\text{.}\) We conclude that \(\img(\phi)=C_n\text{,}\) as desired.
That \(\phi\) is a homomorphism is clear: we have \(\phi(a+b) =
\omega^{a+b}=\omega^a\omega^b=\phi(a)\phi(b)\text{.}\) We have \(\omega^{kn}=\omega^0=1\) for all \(k\in \Z\text{,}\) so \(n\Z\subseteq \ker(\phi)\) Conversely, let \(k\in \Z\) and write \(k=nq+r\) as in the previous part, so that \(\omega^k=\omega^r\text{.}\) Since \(\omega^r=1\) if and only if \(r=0\) for \(0\leq r\leq n-1\text{,}\) we have \(\ker(\phi)\subseteq n\Z\text{.}\) We conclude that \(\ker(\phi)=n\Z\text{,}\) as desired.
By the First Isomorphism Theorem, we have \(\Z/(n\Z)\approx C_n\text{.}\)
9.Isomorphic images of generators are generators.
Let \(S\) be a subset of a group \(G\text{.}\) Let \(\phi\colon G\to H\) be an isomorphism of groups, and let \(\phi(S)=\{\phi(s)\colon s\in S\}\text{.}\) Show that \(\phi(\langle S\rangle)=\langle \phi(S)\rangle\text{.}\)
Instructor’s solution for Exercise 2.4.2.9.
Elements of \(\phi(\langle S\rangle)\) have the form \(\phi(s_1s_2\cdots s_k)\) for some \(s_i\in S\cup
S^{-1}\text{.}\) Elements of \(\langle \phi(S) \rangle\) have the form \(\phi(s_1)\phi(s_2)\cdots \phi(s_k)\text{.}\) The result follows from the fact that \(\phi(s_1s_2\cdots s_k) = \phi(s_1)\phi(s_2)\cdots \phi(s_k)\text{.}\)
10.Conjugation.
Let \(G\) be a group, let \(a\) be an element of \(G\text{,}\) and let \(C_a\colon G\to G\) be given by \(C_a(g)=aga^{-1}\text{.}\) The map \(C_a\) is called conjugation by the element \(a\) and the elements \(g,aga^{-1}\) are said to be conjugate to one another.
Show that \(C_a\) is an isomorphism of \(G\) with itself.
Show that "is conjugate to" is an equivalence relation. That is, consider the relation on \(G\) given by \(x\sim y\) if \(y=C_a(x)\) for some \(a\text{.}\) Show that this is an equivalence relation.
Instructor’s solution for Exercise 2.4.2.10.
Since we’ve already shown that conjugation is a homomorphism (see Checkpoint 2.4.2), it is enough to show that \(C_a\) is bijective. We claim that \(C_{a^{-1}}=(C_a)^{-1}\text{.}\) Indeed, we have
so \(C_{a^{-1}}\circ C_a=\Id\text{.}\) Similarly, we have \(C_a\circ
C_{a^{-1}}=\Id\text{,}\) so we conclude that \(C_{a^{-1}}=(C_a)^{-1}\text{.}\)
For reflexivity, we have \(C_{e}(x) = exe^{-1}=x\text{,}\) so \(x\sim
x\) for all \(x\text{.}\) For symmetry, if \(y=C_a(x)=axa^{-1}\text{,}\) then \(C_{a^{-1}}(y)=a^{-1}ya=x\text{,}\) so \(x\sim y\) implies \(y\sim x\text{.}\) For transitivity, if \(C_a(x)=y\) and \(C_b(y)=z\text{,}\) then \(C_{ba}(x)=baxa^{-1}b^{-1}=byb^{-1}=z\text{,}\) so if \(x\sim
y\) and \(y\sim z\text{,}\) then \(x\sim z\text{.}\)
11.Isomorphism induces an equivalence relation.
Prove that "is isomorphic to" is an equivalence relation on groups. That is, consider the relation \(\approx\) on the set of all groups, given by \(G\approx H\) if there exists a group isomorphism \(\phi\colon G\to H\text{.}\) Show that this is an equivalence relation.
Instructor’s solution for Exercise 2.4.2.11.
For reflexivity, notice that \(\Id\colon G\to G\) is an isomorphism for any group \(G\text{.}\) For symmetry, suppose \(\phi\colon G\to H\) is an isomorphism. By Exercise 2.4.2.4, \(\phi^{-1}\colon
H\to G\) is also an isomorphism. For transitivity, suppose \(\phi\colon G\to H\text{,}\)\(\psi\colon H\to K\) are isomorphisms. For \(g,g'\in G\text{,}\) we have
so \(\psi\) is a homomorphism. Since \(\phi,\psi\) are bijective, so is \(\psi\circ \phi\text{.}\) Thus we have \(G\approx
K\) by the isomorphism \(\psi\circ \phi\text{.}\)
Show that Item 3 implies Item 2. The messy part of this proof is to show that multiplication of cosets is well-defined. This means you start by supposing that \(xK=x'K\) and \(yK=y'K\text{,}\) then show that \(xyK=x'y'K\text{.}\)
14.Further characterizations of normal subgroups.
Show that Item 3 is equivalent to the following conditions.
\(gKg^{-1}= K\) for all \(g\in G\)
\(gK = Kg\) for all \(g\in G\)
Instructor’s solution for Exercise 2.4.2.12.
Let \(K\) be the kernel of \(\phi\colon G\to
H\text{.}\) Let \(g\in G\) and let \(k\in K\text{.}\) Then we have
Suppose that \(K\) is a normal subgroup of \(G\text{.}\) We just need to show that the multiplication given by Equation (2.4.1) is well-defined. Suppose that \(xK=x'K\) and \(yK=y'K\text{.}\) First write \(x'=xk_1\) and \(y'=yk_2\) for some \(k_1,k_2\in K\text{.}\) Finally, let \(k_4=y^{-1}k_1y\text{,}\) and note that \(k_4\in K\) by the assumption of normality. Now let \(x'y'k_3\) be an element of \(x'y'K\text{.}\) Then
which is in \(xyK\text{,}\) as required. This proves \(x'y'K\subseteq
xyK\text{.}\) A similar argument shows containment in the other direction.
Instructor’s solution for Exercise 2.4.2.14.
Let \(g\in G\text{.}\) We showed in Exercise 2.4.2.11 that \(C_g\colon
G\to G\) is an isomorphism. By the definition of normality, \(C_g\) restricts to a map from \(K\) to itself. This restriction must also be an isomorphism (with inverse \(C_{g^{-1}}\)), so \(C_g(K)=K\text{,}\) or, said another way, \(gKg^{-1}=K\text{.}\)
Part (a) says that for all \(k\in K\text{,}\) we have \(gkg^{-1}=k'\) for some \(k'\in K\text{.}\) This is the same as \(gk=k'g\text{.}\) That this holds for all \(k\in
K\) means that \(gK\subseteq Kg\text{.}\) A similar argument shows that \(Kg\subseteq gK\text{.}\) We conclude that \(gK=Kg\text{.}\)
15.Automorphisms.
Let \(G\) be a group. An automorphism of \(G\) is an isomorphism from \(G\) to itself. The set of all automorphisms of \(G\) is denoted \(\Aut(G)\).
Show that \(\Aut(G)\) is a group under the operation of function composition.
is a subgroup of \(\Aut(G)\text{.}\) (The group \(\Inn(G)\) is called the group of inner automorphisms of \(G\text{.}\))
Find an example of an automorphism of a group that is not an inner automorphism.
Instructor’s solution for Exercise 2.4.2.15.
We showed in Exercise 2.4.2.11 that the composition of isomorphisms is an isomorphism, so the group operation is well defined. The identity is an isomorphism, and we showed in Exercise 2.4.2.4 that the inverse of an isomorphism is an isomorphism. Associativity is a property of function composition. All of the requirements of a group hold for \(\Aut(G)\text{.}\)
We will use the 2-step subgroup test. First, \(\Inn(G)\) contains the identity map \(C_e\colon G\to G\text{,}\) so \(\Inn(G)\) is not empty. Given \(g,h\) in \(G\text{,}\) we have
so \(\Inn(G)\) is closed under composition. We showed in Exercise 2.4.2.11 that \((C_g)^{-1}=C_{g^{-1}}\text{,}\) so \(\Inn(G)\) is closed under inversion. This completes the 2-step subgroup test, and we conclude that \(\Inn(G)\) is a subgroup of \(\Aut(G)\text{.}\)
One example is complex conjugation on \(\C^\ast\text{.}\) Any nontrivial automorphism of an Abelian group is an example, because conjugation by any element is the identity. So, for example, \(\mu_2\colon \Z_5\to \Z_5\) given by \(\mu_2(x)=2x\) works.
