Groups, Inner Automorphisms

Inner Automorphisms

The conjugate of an element w with respect to x is x*w/x. If w and x commute then the conjugate of w is w. Given a group G and an element x, replace every element w in G with its conjugate. In other words, f(w) = xw/x. This map commutes with *, and is an automorphism. Specifically, it is an inner automorphism. If G is abelian, xw/x = w, and every inner automorphism becomes the trivial automorphism. An automorphism that is not an inner automorphism is an outer automorphism. The composition of two inner automorphisms, derived from x and y, is the inner automorphism derived from yx. One can use x inverse to produce the inverse of a specific inner automorphism. Thus the inner automorphisms form a subgroup of all the automorphisms of G. If x produces an inner automorphism, and f is any other automorphism, having inverse j, apply f, then the x automorphism, then j. If w is a group element, we have j(x*f(w)/x). since j commutes with *, this is j(x)*j(f(w))/j(x), or j(x)*w/j(x). The composition gives another inner automorphism, hence the subgroup of inner automorphisms is normal inside the (possibly larger) group of automorphisms. This implies a quotient group, the automorphisms mod the inner automorphisms.

The center of G, i.e. those elements that commute with all of G, produce trivial inner automorphisms. If x commutes with everything then xG/x leaves G fixed. Conversely, if x is not in the center of G then xy/x is different from y for some y, and the resulting inner automorphism is nontrivial.

Map G onto the inner automorphisms of G by carrying x to the function xG/x. We must show that this is a homomorphism. Compose xG/x with yG/y and get yxG/x/y, the inner automorphism associated with yx. Technically, we were hoping for the inner automorphism associated with xy. This is not a show stopper though. Change the convention so that the composition of two automorphisms is the second followed by the first, rather than the first followed by the second. The automorphisms of G still form a group. Now the composition of xG/x and yG/y is xyG/y/x, as it should be, and the map from G onto its inner automorphisms is indeed a group homomorphism.

We showed that the kernel of this map is precisely the center of G. Thus the inner automorphisms of G, with function composition defined as above, form a group that is isomorphic to G mod its center.