Does this concept have a name?

[mnb96]

Hello,
Let's say I have a group $$(G,\ast,0)$$ and I define an endomorphism $$f:G \rightarrow G$$
I know there is a subset $$H \subseteq G$$ such that:

$$u \in H \\ \Rightarrow \\ f(u)=u$$

Essentially we have defined an endomorphism which, for a subset H, behaves like the identity function.
Is this a known concept which has a name? Does the subset H or the endomorphism f have a particular name?

 

[wsalem]

I'm not familiar with the name but this is actually a property shared with all endomorphisms. Since if you consider the trivial subgroup (of only the identity element), any endomorphism must preserve the identity. Hence $$e \in H = \{e\} \Rightarrow f(e)=e$$.

 

[mnb96]

You are right, but unfortunately I noticed that my question was a bit too vague (and incorrect).
Let's be more specific:

what is given are: the group G and a non-trivial subset $$H \subseteq G$$ which not necessarily contains the identity element;

what I want to find is an endmorphism f which satisfies the property $$f(u) = u$$ whenever $$u\in H$$.Let's assume that I find an f with that property. The question is: what can I say about it? Is it a known concept? Can I say that I found an XXXXXX endomorphism, or an XXXXXXX subset where I can replace the X's with a formal algebraic term?

 

[wsalem]

I actually never read about it, so I can't make any claims about it's significance.
Your last phrasing of the problem I believe doesn't help much. You are given a group G, and a non-trivial subset H, finding an f that is the identity when restricted to H is trivial. Just let f be the identity map!. But I'm not quite sure that is what you had in mind.

Now, let's try another way, let's say you have a group G, an endomorphism f and a non-trivial subset H such that x in H implies f(x) = x. Is H the only subset, or the largest subset with this property?

To answer this, we may investigate what the set of all such elements would look like?
Define H as a subset of G that contain the elements of G such that x in H implies f(x) = x.

H is non-empty and contains the identity, as shown earlier.

H is closed under the binary operation: assume there is a non trivial x that is preserved by f, i.e f(x) = x in H. Then we know that f(xx) = f(x)f(x) since f is a homomorphism. But we must have f(xx) = f(x)f(x) = xx. So xx also satisfy this property and are in H.

H contains the inverse of each element: $$1 = f(1) = f(x x^{-1}) = f(x) f(x^{-1}) = x f(x^{-1})$$, so that $$f(x^{-1})= x^{-1}$$

Hence you may conclude that H is a group (cyclic subgroup if we picked only x), and the restriction of f to H is a homomorphism(injective). It is also the smallest subgroup of G that has f(x) = x for some x in G.

If G is generated by a set X, and the endomorphism f preserve each element in that set, i.e f(x) = x for each x in X, then f is the identity map!

So you may conclude from this discussion that if H is a subset of G, where f(x) = x for all x in H, then there's a subgroup K of G generated by H such that f(K) = K.Does this help?

 

[mnb96]

your discussion was very interesting and, in fact, it also helped me to understand that I formulated the problem incorrectly.

I guess I was wrong in letting f be an endomorphism.
f is simply supposed to be a function (defined in all G) and with the property we mentioned in respect to H.

but then I guess not much can be further discussed(?)

 

[confinement]

http://en.wikipedia.org/wiki/Invariant_subspace

 

[wsalem]

The analog of invariant subspaces for groups is the the characteristic subgroup which is a subgroup H of a group G that is invariant under ALL automorphisms of G. If we let f be an arbitrary automorphism then for H to be a characteristic subgroup, f(x) is in H for every x in H. But it doesn't tell us that f(x)=x.

 

[mnb96]

uhm...I already stumbled upon the concept of characteristic subgroup, and thanks for the hint. However I agree with wsalem:

if H is to be a characteristic-subgroup, then it means that for all automorphisms f in G, f(x) is still in H but we don't necessarily have f(x)=x.

Moreover this definition apparently requires f to be an automorphism, which is not the case, because (as I corrected in my last post) f is simply a total function, perhaps surjective, whose restriction on H is an inclusion map.

Unfortunately I am not able to find a better way to formalize all this.

http://en.wikipedia.org/wiki/Restriction_(mathematics )
http://en.wikipedia.org/wiki/Inclusion_map

 

[wsalem]

mnb96, I believe dropping the homomorphism requirement is not the right direction, you'll lose plenty of information then.

Here's a couple of question for you.

Let G be a group, f an endomorphism, H the subgroup of all elements satisfying f(x) = x in G. Is the subset J where f(x) doesn't equal x (for all non trivial x in G) ever a subgroup? Do we need to impose any restrictions on G?

If J turns out to be a subgroup, then assuming that G is abelian, can we say that G is the direct sum of a subgroup that is preserved by f and another that isn't, i.e G is the direct sum of H and J?

What the quotient group G/H look like, how is it different from I am f/H.
Considering the canonical projection p: G -> G/H. How's pf and p are different?

 

[mnb96]

The group G must be abelian (again I forgot to mention it); and losing information outside H is not a big deal, actually it is likely to be desirable.

In fact, for what I want to do, it is ideal that the function f maps all the elements of H to themselves (like the identity function) and at the same time it maps all the remaining elements of G-H into the identity element.

Does this makes any difference?

 

[wsalem]

Wouldn't this make f a little bit trivial. What am I missing here? It doesn't look very interesting!

Which object are you trying to study? The group G, the subset H, or the map f? Also why G is a group and not just a set, what advantage does this give with f only a map and not a homomorphism?

 

[mnb96]

Hello,
please, keep in mind that this is not the whole problem I am studying.
The whole problem is essentially about two different algebraic-structures which consist both in an abelian group, plus this "trivial" operation f.

The idea is trying to find both an isomorphism g and the function f such that f is preserved by the isomorphism.

In this way one is able to have the following:

$$g(u+v)\bullet g(f(u+v))^{-1} =$$

$$= g(u)\bullet g(v)\bullet f(g(u+v))^{-1} =$$

$$= g(u)\bullet g(v)\bullet f(g(u) \bullet g(v))^{-1} =$$

$$= g(u)\bullet g(v)\bullet g(f(v))^{-1} \bullet g(f(u))^{-1} =$$

...assuming $$v \in H$$ we have f(v)=v, so

$$= g(u)\bullet g(v)\bullet g(v)^{-1} \bullet g(f(u))^{-1} =$$

$$= g(u)\bullet g(f(u))^{-1} =$$

...this quantity is interesting to me, because it is independent from v.

if ideally the function f mapped everything outside H into the indentity element we would have:

$$g(u) \bullet g(f(u))^{-1} = g(u)$$

and since g was an isomorphism I would be able to recover u.

This is the big picture: hopefully it makes more sense

 

[mnb96]

uhm... While checking what I've written in the previous post, I just noticed two important things:

1) I implicitly used the identity $$f(x\bullet y) = f(x) \bullet f(y)$$. If this is to be true, then one must admit that f is indeed an endomorphism.

2) if we want to get rid of the dependence on any $$v \in H$$ in the last equation, then we must admit that it is not possible to have $$\exists(v,v') : (v+v)' \notin H$$; if I am not wrong this is only possible when H is a subgroup of G.

If this is correct then we can agree that you made the best possible discussion in your second post.