Kernal of the left regular action

The left regular action is defined as follows: Let G be a group and A = G (as sets.) Define $\displaystyle \phi : G \times A \to A: g \cdot a \mapsto ga$ where g is any element of G and a is any element of A. ga is calculated using the group operation in G. The question is to find $\displaystyle ker( \phi )$. The kernal in my text is defined as $\displaystyle ker ( \phi ) = \{g \in G | g \cdot a = a~\forall a \in A \}$.

The only element I can come up with for the kernal is $\displaystyle 1_G$. That seems to little to me. Are there any others? A hint would be appreciated

Thanks.

-Dan

Re: Kernal of the left regular action

Dan,

I don't know what book(s) you are reading, so maybe you already know all this. If so, bear with me.

Given a set X a permutation of X is a one to one mapping from X onto X. A permutation group P is set of permutations of X that form a group under the operation of function composition. In particular, the identity permutation is in P and for any x in P, also x^{-1} is in P.

Now for any abstract group G, a permutation representation of G is a homomorphism f from G into a permutation group P. The representation is said to be faithful if Ker(f) = {1}, i.e. f is an injection.

A group action of a group G on a set A is as you describe. This group action is really just a permutation representation of G -- given the "group action" ga for g in G and a in A, the representation f is the mapping f where for any g in G, f(g) is the permutation of A defined by f(g)(a) = ga. The properties of action ensure that f indeed is a homomorphism. Conversely, given a permutation representation f of G where the underlying set of the permutation group is A defines an action by ga = f(g)(a). So group actions are really just permutation representations. The kernel (notice the spelling!) of the action is just the kernel of f. Namely, the kernel is the set {g in G : ga = a for all a in A}. That is, this is the set of elements of G that map to the identity permutation of A under f.

Now for your specific question. The left regular representation of any group G is as you describe. That is for any group G, associate with each g in G the permutation f(g) of the set G defined by f(g)(a) = ga. Now ask yourself: for what elements g of G is this associated permutation (mapping) of G the identity permutation? That is for what g in G is it true that ga = a for all a in G? I hope you see immediately that the only such g is the identity element of G.

As an old group theorist, the way I think of group actions is as permutation representations. So I can think about permutation group properties; example transitivity, orbits, primitivity, etc.

Re: Kernal of the left regular action

Quote:

Originally Posted by

**johng** The representation is said to be faithful if Ker(f) = {1}, i.e. f is an injection.

I knew most of what you said, except for the above comment about the "kernel." (Thanks for the spell check.) I knew about the injection, but I didn't know about ker(f) = {1}. I guess that's what the exercise was all about. Thanks for the explanation. (Bow)

-Dan

PS For the record I'm using "Abstract Algebra," 3rd ed. by Dummit and Foote.