Homomorphism • Definition ○ Let G,H be groups ○ A function f:G→H is a homomorphism if ○ f(g_1 g_2 )=f(g_1 )f(g_2 ), ∀g_1,g_2∈G ○ One says f “respects”, or “preserves” the group operation • Trivial Examples ○ Let G be a group ○ The identity map of G→G is a homomorphism § f(g_1 )f(g_2 )=1⋅1=1=f(g_1 g_2 ) ○ The map G→G given by g↦1 is a homomorphism § This only works if we send every element of G to 1 § If x∈G∖{1}, and f:G→G is given by g↦x,∀g § f(g_1 g_2 )=f(g_1 )(g_2 ) § i.e. x=x^2 § Thus x=1 § This is impossible since x∈G∖{1} • Example 1 ○ Let f:R→R∗ be given by f(x)=e^x ○ f is a homomorphism ○ f(x_1+x_2 )=e^(x_1+x_2 )=e^(x_1 ) e^(x_2 )=f(x_1 )f(x_2 ) • Example 2 ○ Let G be a group, and let x∈G ○ The map f:G→G, g↦xgx^(−1) is a homomorphism ○ f(g_1 g_2 )=xg_1 g_2 x^(−1)=xg_(1 ) x^(−1) xg_2 x^(−1)=f(g_1 )f(g_2 ) ○ This homomorphism is called conjugation by x • Example 3 ○ Let n∈Z ○ Is f:Z→Z, x↦x+n a homomorphism? ○ Answer: Only when n=0 ○ f(0+0)=f(0) ○ i.e. n+n=n ○ Thus n=0 • Example 4 ○ Let n∈{0,1,2,3,…} ○ Is α:Z→Z, x↦x^n a homomorphism? ○ Answer: Only when n=1 ○ When x=0 § α(x)=1 § 1 is not the identity (0 is) § So this doesn’t work ○ For n≥2 § α(x_1+x_2 )=α(x_1 )+α(x_2 ) § (x_1+x_2 )n=x_1n+x_2^n § This is not always true § For instance, when x_1=x_2=1, 2^n≠2 for n≥2 • Example 5 ○ Let n∈Z ○ Is β:Z→Z, x↦nx a homomorphism? ○ Answer: Yes ○ β(x_1+x_2 )=n(x_1+x_2 )=nx_1+nx_2=β(x_1 )+β(x_2 ) • Example 6 ○ The previous examples is a special case of the following: ○ Let G be a group, and n∈Z ○ Define β:G→G, g↦g^n ○ Then β is homomorphism ∀n∈Z iff G is abelian ○ Proof: homomorphism ⇒ abelian § Say n=−1 § Let g_1,g_2∈G § β(g_1,g_2 )=β(g_1 )β(g_2 ) § (g_1 g_2 )(−1)=g_1(−1) g_2^(−1) § g_2^(−1) g_1(−1)=g_1(−1) g_2^(−1) § (g_2^(−1) g_1^(−1) )(−1)=(g_1(−1) g_2^(−1) )^(−1) § (g_1^(−1) )^(−1) (g_2^(−1) )(−1)=(g_2(−1) )^(−1) (g_1^(−1) )^(−1) § g_1 g_2=g_2 g_1 § Thus G is abelian ○ Proof: abelian ⇒ homomorphism § (Homework) Isomorphism • Definition ○ Let G,H be groups ○ A homomorphism α:G→H is a isomorphism if ○ there is a homomorphism β:H→G s.t. ○ αβ=id_H and βα=id_G ○ In this case, we say G and H are isomorphic • Fact ○ L:G→H is an isomorphism iff it is a bijective homomorphism ○ Proof: isomorphism ⇒ bijective homomorphism § Clear ○ Proof: bijective homomorphism ⇒ isomorphism § Need to show α^(−1) is a homormorphism § (Homework) • Example ○ R(0)≔{r∈Rr0} is a group under multiplication ○ Define f:R→R(0) where f(x)=e^x ○ Then f is a homomorphism ○ Moreover, f is an isomorphism ○ The inverse of f is ln • Observation ○ If G,H are isomorphic groups, then |G|=|H| Proposition 16 • Statement ○ If f:G→H is an isomorphism, then ○ G is abelian iff H is abelian ○ ∀g∈G, |g|=|f(g)| ○ Note |g|=inf{n∈Z(0)│g^n=1}
