If G is a finite group and H is a subgroup of G then we say that H is a Frobenius subgroup of G if xHx-'nH5 (1) if and only if xeH. Concerning such Frobenius subgroups there is a celebrated theorem due to Frobenius [1] which asserts: let T= {aEGjaEfxHx-1 all xEG or a= 1 }; then T is a normal subgroup of G. We call it the complementary subgroup of H. Suppose now that a finite group G has an automorphism cb leaving only 1 fixed. Thus the elements x-'b(x) are all distinct as x ranges over G, so they must fill out all of G. Let p be a prime which divides the order of G and let Sp be a p-Sylow subgroup of G. Then ?)(Sp) is also a p-Sylow subgroup of G and so cb(S,) =ySpyfor some yEEG. Pick xCG so that y-1 =x-l(x). Then as can immediately be verified, c(xSx-1) =xSx-1. That is, for each prime p there is a p-Sylow subgroup Sp, say, of G left set-wise invariant by c. We claim that Sp is unique in this regard. For if N(Sp) = { x G j xSx-1 = S, } then it is clear that ?)(N(Sp)) = N(Sp). Since N(Sp) is invariant under Xb, c induces an automorphism on N(Sp) leaving only 1 fixed, whence every element in N(Sp) can be written in the form n-10(n) with nCN(Sp). Thus if a-1l(a) CN(Sp), then a-lc(a) =n-lb(n) for some nEN and so a = n from which, of course, we have that aEN. Now suppose cb(Sp) = S, and ?)(Sp') =Sp' for some other p-Sylow subgroup. Since Sp' = aSpa-1 for some a E G, aSpa-1 = Sp' = ?)(Sp) = q5(aSpa-') =-c(a)Spq5(a)-1, leading to a-10(a) EN(Sp), and so aCN(Sp) and finally to Sp' = aSpa-1 =Sp. Thus Sp is indeed unique with respect to being left invariant by 4. If Vf is an automorphism of G which commutes with cb then since cb(Sp) = Sp, ciI'(Sp) =+ iP(Sp) = f (Sp), and so, since t'(Sp) is a p-Sylow subgroup left fixed by Xb, Vf(Sp) = Sp. Thus if W is an Abelian group of