Automorphism?

Question

If Aut G={e},then show either G is trivial group or G isomorphic to Z/2Z

Answer 1

danny -

x -> x g is not an automorphism unless g is the identity element. Otherwise, it moves the identity, something which automorphisms do not do.

You are thinking of the map x -> g^-1 x g, which is an automorphism (called an inner automorphism) for every element g of the group. Automorphisms not of this form are called outer automorphisms.

If for every x and y the map x -> y^-1 x y is the identity map, then for every x and y, x = y^-1 x y and so y x = x y and the group is Abelian.

fatal_flaw_death -

The map x -> x^-1 is always an automorphism for Abelian groups but is not for non-Abelian groups. If f(x) = x^-1 then an automorphism requires f(a b) = f(a) f(b). If a and b do not commute, then

f(a^-1 b^-1) = f(a^-1) f(b^-1) = (a^-1)^-1 (b^-1)^-1 = a b is what you want, but

f(a^-1 b^-1) = (a^-1 b^-1)^-1 = (b^-1)^-1 (a^-1)^-1 = b a is what you get.

Sometimes this is called an anti-automorphism, where f(xy) = f(y) f(x).

So you can't assume x -> x^-1 is an automorphism and use that to prove the group is Abelian, although above I showed it must be Abelian. Then yes, x -> x^-1 is then an automorphism, and so x = x^-1 for all x, or equivalently x^2 = e, the identity element of the group, for all x.

Now if a and b are distinct non-identity elements of G, then they generate a subgroup H = = { e, a, b, ab } isomorphic to (Z/2Z)^2. Take the factor group of right cosets G/H. Each coset has four elements: let the representative of the coset H itself be e, and choose a representative element of every coset not equal to H (if G is infinite, use of the axiom of choice may be necessary). For example, if g represents the coset B = Hg then B = Hg = { g, ga, gb, gab }. [Remember that despite using multiplicative notation that these all commute.] Let f(e) = e, f(a) = b, f(b) = a, f(ab) = ab, and for a coset Hx with representative element x let f(x) = x, f(xa) = xb, f(xb) = xa, f(xab) = xab. f is a nontrivial automorphism of G. So G has at most one non-identity element if Aut(G) is the one-element group.

Answer 2

Suppose G is not the trivial group.
f(x) = x^(-1) is always an isomorphism, so we must have x^(-1) = x for all group elements. Consider f,g in G
(fg)^(-1) = g^(-1)*f^(-1) = gf
but (fg)*(fg) = e, so fg^(-1) = fg.
thus fg = gf
So G is abelian.
Thus, G is a product of (Z/2Z)^n. If n > 1, then our elements are of the form
(a1,a2,a3,...,an)
and the morphism
f(a1,a2,a3,a4,a5,...,an)
= (a2,a1,a3,a4,a5,a6,a7,...,an) is an automorphism.
So n = 1.
Thus the only such groups are the trivial group and Z/2Z

Answer 3

take an element different than 0, g
and let the automorphism f(x)=xg
If there exist an element different than 0 and g ,say h, then
f(h)=e(h) or hg=h or g=0 contradiction