# Group action on a subgroup

1. Nov 28, 2007

### lion8172

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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
$$\mbox{ker} \sigma(g, *) = \{ x| \sigma(g, x) = g \ \forall x \in H \}$$?

Last edited: Nov 28, 2007
2. Nov 28, 2007

### morphism

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.

3. Nov 28, 2007

### 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.

Last edited: Nov 29, 2007