# Transfer-free proof that the power map is a homomorphism

For a central subgroup $C$ of finite index of a group $G$ it is well-known that the power map $$G \to C,\;g \mapsto g^{(G:C)}$$ is a group homomorphism. This is commonly proved by help of the transfer map (see for example the group theory book of Robinson (10.1.3) or Isaacs (5.6)). I wonder, however, if there is a direct proof that avoids using the transfer ?

-
One can write down a direct proof which basically encodes the construction and the homomorphism property of the transfer map; which will be a quite long calculation with coset representatives. I wonder if there is a completely different approach (probably not). It cannot be a purely formal algebraic manipulation since the statement is wrong for monoids. –  Martin Brandenburg Jan 5 '13 at 2:05
@Martin: Yes, my question is about such a completely different approach. Thanks for clarifying. I don't understand why the failure in monoids rules out a proof based on formal manipulations: Can't such manipulations make use of the existence of inverses ? –  Todd Leason Jan 5 '13 at 2:56
What statement is wrong for monoids? How would you define $(G:C)$? –  Tom Goodwillie Jan 5 '13 at 3:06
@Martin and Todd: there is no good notion of index in monoids, so I don't really see which analogous statement would make sense. Actually the notion of index itself in the statement makes a "purely formal algebraic" approach quite unclear (at least to me). –  YCor Jan 5 '13 at 3:09
In order to stress what Martin and Yves are saying, let $n$ be an integer such that $g^n$ is central for all $g$. Then, in general, the $n$th power map is no hom (take $G$ the dihedral group of order 8 and $n=2$). Hence, "formal manipulations" alone, like $(xy)^n=x(yx)^nx^{-1}=(yx)^n$ don't suffice to show that your map is a hom. –  Ralph Jan 5 '13 at 11:07