Isomorphic Subgroup Action on a Group: Proving the Isomorphism between H^g and H

[lion8172]

Homework Statement

I'm trying to prove that, if H is a subgroup of an arbitrary group G, then H^g, the action of a given element in G on H, is isomorphic to H.

Homework Equations

The Attempt at a Solution

Let \sigma denote a given action of G on H. We are considering the map \sigma(g, *) : H -> H^g (where g is fixed and * denotes a variable element of H). I think that the kernel of this map is the set of all elements x in H such that \sigma(g, x) = g (this is the part I'm not sure of). Since x lies in H and H is a subgroup of G, it follows that x must be equal to e, the identity element of G. Thus, ker{\sigma(g, x)} is trivial, and, as a result,
H^g \cong G/ \mbox{ker}(\sigma(g, *)) = G,
by the first isomorphism theorem.
Does this seem correct? If so, can you please justify the statement that
\mbox{ker} \sigma(g, *) = \{ x| \sigma(g, x) = g \ \forall x \in H \}?

 

[morphism]

H^g, the action of a given element in G on H, is isomorphic to H.

This doesn't make sense - an action is not a group. To have any chance of solving this problem correctly, you should write down what H^g actually is.

 

[lion8172]

H^g would be a group in certain cases (i.e. conjugation), but not in others. So the proof above certainly does not apply in general.