Isomorphisms : abstract algebra?

Question

Let G be a group. How do you prove that the mapping alpha(g) = g^-1 for all g in G is an automorphism if and only if G is Abelian.

Answer 1

Note that the mapping α(g) = g⁻¹ has an inverse, namely itself. Therefore, α is automatically a bijection. Therefore, it suffices to show α is a homomorphism iff G is abelian.

Suppose G is abelian. Let a, b be any two elements of G. Then α(ab) = (ab)⁻¹ = b⁻¹a⁻¹ = a⁻¹b⁻¹ = α(a)α(b). So α is a homomorphism.

Conversely, suppose G is not abelian. Then there exist elements a and b such that ab≠ba. But if a⁻¹b⁻¹ = b⁻¹a⁻¹, then (a⁻¹b⁻¹)⁻¹ = (b⁻¹a⁻¹)⁻¹ and so ba=ab, a contradiction. Thus a⁻¹b⁻¹ ≠ b⁻¹a⁻¹, and so α(ab) = (ab)⁻¹ = b⁻¹a⁻¹ ≠ a⁻¹b⁻¹ = α(a)α(b), so α does not preserve the group operation and is not a homomorphism.

Answer 2

Let's use A for alpha.
Assume G is abelian:
A(g1g2) = (g1g2)^-1 = (g2g1)^-1 = g1^-1*g2^-1 = Ag1Ag2
Therefore a homomorph.
If Ag1 =Ag2 then g1^-1 = g2^-1 and g1 = g2
Therefore 1 - 1.
If g$ in G Ag$^-1 = (g$^-1)^-1 = g$
Therefore onto and an automorph
Now assume A is an automorph
Then and take g1, g2 in G
Ag1g2 = (g1g2)^-1 = g2^-1*g1^-1
Ag1g2 = Ag1Ag2 =g1^-1*g2^-1
Therefore g2^-1*g1^-1=g1^-1*g2^-1
or (g1g2)^-1 = (g2g1)^-1 or g1g2 =g2g1
and G is abelian.

Answer 3

ok, for the reason which you have already proved isomorphism, i visit in simple terms teach one-to-one and onto. enable ?(x) = ?(y). Then gxg^-a million = gyg^-a million. precise multiply the two components by using g. We get gx = gy. Left multiply the two components by using g^-a million. We get x = y. hence, ? is one-to-one. assume x is in G. concept: g^(-a million)xg is the preimage of x in G. all human beings comprehend g^(-a million)xg is in G by using the closure components of communities. ?(g^(-a million)xg) = g(g^(-a million)xg)g^-a million. employing associativity, inverse, and identity properties, we get x. hence, there's a preimage of all x in G. hence, ? is onto. hence ? is a bijection. by way of certainty it fairly is likewise a homomorphism, ? is an isomorphism (and bigger extraordinarily, an automorphism).