# Retraction math help

1. Nov 3, 2006

### Oxymoron

A subgroup H of a group G is a retract of G if there exists a homomorphism $q\,:\,G \rightarrow H$ such that $q(h) = h$ for all $h \in H$. This map, q, is called the retraction from G onto H.

If my definition of a retract is correct then could I form a subgroup, K, of G that consists of the kernel of the retraction? That is,

$$\mbox{ker}(q) = K < G$$

So K is the subgroup consisting of all elements in G that get mapped to the identity of H. Obviously, this is a normal subgroup of G and we have $G = KH$ and $K \cap H = \{e_H\}$ (from wikipedia). The group G, then, should be the semi-direction product of K and H. Is this right?

Now, since H acts on K by conjugation:

$$k \mapsto hkh^{-1}$$

this defines a group homomorphism

$$p\,:\,H \rightarrow \mbox{Aut}(K)$$

In other words, given a group G, and a subgroup H, one can set K to be the subgroup of G consisting of elements of G that get mapped to the identity of H under the retraction. This set K is then normal, and one then has a homomorphism, p, from H into the automorphism group of K. Then the semi-direct product $K \rtimes$ is a group consisting of pairs $hk$ with multiplication

$$(h_1k_1)\cdot (h_2k_2) = (h_1 h_2)(p(h_1)(k_1)k_2)$$

and we also get

$$hkh^{-1} = p(h)(k)$$

Now, my main question (and the reason why I brought the semi-direct product up) is this: Is the existence of a group G, which is not simple (that is, a group whose normal subgroups are not necessarily the trivial group and the group itself) and whose only only retracts are G itself, and the trivial subgroup, possible?

I figured that such a group did exist. It had to be the following things:

1] It has to be a group.
2] It must not be simple.
3] It must have trivial retracts.

I figured that the semi-direct product $K \rtimes H$ was not quite what I wanted. It is a group (by definition), it is not simple because it contains K and H as subgroups (at least), but it has H as a retract! Therefore it fails #3 of the 3 restraints. Is this all correct so far?

Does anyone know of a group which satisfied all three conditions? I thought the semi-direct product came pretty close.

Last edited: Nov 3, 2006
2. Nov 3, 2006

### AKG

Z4 is a non-simple group with only trivial retracts. Double check this.

3. Nov 5, 2006

### mathwonk

as is Z/p^n for all primes p, all n > 1.

4. Nov 5, 2006

### Oxymoron

Wow, how did you guys come up with such easy-looking answers!?

Well, $4\mathbb{Z}$ is a subgroup of $\mathbb{Z}_4$ is it not? (Also, this is a normal subrgoup, no?) and $4\mathbb{Z}$ is certainly not a trivial subgroup. So, $\mathbb{Z}_4$ is non-simple because it contains normal subgroups which are not trivial.

Suppose that $H$ is a subgroup of $\mathbb{Z}_4$. H is a retract if there is a homomorphism $p\,:\,\mathbb{Z}_4 \rightarrow H$ such that $p(h) = h$ for all h in H.

So do I have to check every subgroup of $\mathbb{Z}_4$ and see if there is a homomorphism?

Last edited: Nov 5, 2006
5. Nov 5, 2006

### Hurkyl

Staff Emeritus
Nope. No element of $4\mathbb{Z}$ is an element of $\mathbb{Z}_4$, so it can't be a subgroup.

There is a canonical map $\phi: \mathbb{Z} \rightarrow \mathbb{Z}_4$, but $\phi(4\mathbb{Z})$ is a trivial subgroup of $\mathbb{Z}_4$.

6. Nov 5, 2006

### Oxymoron

I verified that there is indeed a homomorphism, p, from the group $\mathbb{Z}_4$ into the two trivial subgroups $\mathbb{Z}_4$ and $\{e\}$ satisfying the condition.

How do you check other subgroups?

Last edited: Nov 5, 2006
7. Nov 5, 2006

### Hurkyl

Staff Emeritus
Directly. There aren't many subgroups or automorphisms of Z_4 (or of Z_p^n).

8. Nov 5, 2006

### Oxymoron

Of course. I dont know what I was thinking! Is $\mbox{ker}(\phi)$ a subgroup of $\mathbb{Z}_4$?

Ok, so subgroups of $\mathbb{Z}_4$ have to contain 0 and have to be closed under addition.

{0,1,2,3}
{0,2}
{0}

These are the only subgroups of Z_4 that I could find. Does this look right?

Last edited: Nov 5, 2006
9. Nov 5, 2006

### Hurkyl

Staff Emeritus
Nope. The kernel is a subgroup of Z. The image is a subgroup of Z_4.

Yes.

Last edited: Nov 5, 2006
10. Nov 5, 2006

### Oxymoron

Well, {0,1,2,3} is a retract since there exists a homomorphism (namely the trivial homomorphism), p, such that p(0)-0, p(1)=1, p(2)=2, p(3)=3 for 0,1,2,3 in the subgroup.

Also, {0} is a retract because the trivial homomorphism p(0) =0.

But if $\mathbb{Z}_4$ then {0,2} mustn't be a retract. That means that there must not exist a homomorphism $p\,:\,\mathbb{Z}_4 \rightarrow \{0,2\}$ such that $p(h) = h$ for all h in {0,2}. But surely there is a homomorphism.

11. Nov 6, 2006

### Hurkyl

Staff Emeritus
1 generates Z_4. Any homomorphism is completely determined by what it does to the generators. Where can a homomorphism Z_4 -> {your subgroup} map 1?

12. Nov 6, 2006

### Oxymoron

If it is a homomorphism, must it map 1 to 1? That is, must it map the generator of the group to the same element in the subgroup? But if the subgroup does not have 1, will the homomorphism break down?

13. Nov 6, 2006

### Hurkyl

Staff Emeritus
What does the definition of homomorphism say? (answer: no)

14. Nov 6, 2006

### mathwonk

the answer to how did i come up with my answer is: i am teaching abstarct algebra, amnd so i know the classification of abelian groups, and i know that no group of form Z/p^n is a direct sum oif any other pair of groups, as they have different numbers of elements of order p^n.

15. Nov 7, 2006

### Oxymoron

A homomorphism must map the generator somewhere, but the only choices I have are 0 and 2. I dont see anything that forces me to choose either element.

Wikipedia (which says nothing about homomorphisms having anything to do with the generator) says that a homomorphism is simply a mapping which preserves the identity element. The identity element in $\mathbb{Z}_4$ is 0 mod 4 = {...,-8,-4,0,4,8,...} and the homomorphism h(x) = x which takes an element from $\mathbb{Z}_4$ to the corresponding element in {0,2} is a mapping which preserves the identity: h(0) = 0 which is in {0,2}. So this map is a homomorphism because it preserves the identity. What is wrong with this argument.

16. Nov 7, 2006

### Oxymoron

Still, to make that connection is pretty impressive. And not groups of the form $\mathbb{Z}_p$ but $\mathbb{Z}_{p^2}$! That takes some thinking. Unfortunately I cannot see the relationship between simple groups with only trivial retracts and abelian groups which are not the direct sum of any other pair of groups. Wait..If you have an abelian group which is not the direct sum of any other pair of groups then is the main reason why you only get trivial retracts? In other words, do you get non-trivial retracts when a group is able to be expressed as a direct product of other groups?

Last edited: Nov 7, 2006
17. Nov 7, 2006

### Hurkyl

Staff Emeritus
So, there are potentially two homomorphisms from Z_4 to {0, 2}; one that sends 1 to 0, and one that sends 1 to 2. (And, in this case, both possibilities do give rise to homomorphisms)

Last edited: Nov 7, 2006
18. Nov 7, 2006

### mathwonk

as observed above, a group with a retract has a normal subgroup, namely the kernel of the retract, which has a "complement", n amely the image of the retract. and any such group is a semi direct product of those two subgroups.

but an abelain group has only trivial semi direct products, so if an abelian group has a retract, then it is the direct product of those two subgroups,

hence any abelian grouop which is not a direct product, does not have a retract. now cyclic groups are good candidates, but the chienese remainder theorem tells us that a cyclic group whose order invovkles mroe than one prime, isa prouct, e.g. Z/15 = Z/5 x Z/3.

but a cyclic group of prime power order, cannot be a product, because then the order of one factor would always divide the order of the other afctor. this means that the order of the smaller factor would annihilate the group, hence it could noit anylonger be cyclic.

(in a cyclic group the annihilator cannot be a proepr factor of the order of the group.)

Last edited: Nov 7, 2006
19. Nov 7, 2006

### mathwonk

but the secret was entirely contained in the answer Z/4 = Z/2^2. i just generalized it to Z/p^n.

20. Nov 7, 2006

### Hurkyl

Staff Emeritus
Have you tried constructing one?

Theorem: Let G and H be nontrivial groups. Then GxH has a nontrivial retract.

I'm unhappy with the wording... I think what you said is wrong for nonabelian groups. But for abelian groups:

Theorem: Suppose an abelian group G has a nontrivial retract H. Then G is a direct sum of nontrivial groups.

(I think you should be capable of proving both of these. The first one is much easier, though)

21. Nov 7, 2006

### mathwonk

direct sum is perhaps the wrong term, except in the cvategory of abelin groups. ie AxB does not have the mapping property of a direct sum of groups.

22. Nov 7, 2006

### Oxymoron

Ok, so there is a homomorphism $p\,:\,\mathbb{Z}_4 \rightarrow \{0,2\}$ but it certainly doesn't satisfy $p(h) = h$ for all $h \in \{0,2\}$. So this means {0,2} is not a retract, right?

23. Nov 7, 2006

### AKG

There's a homomorphism from any group to any group, so in particular, yes, there's one from Z4 to {0,2}. There are two homomorphisms so saying "it doesn't satisfy p(h) = h for all h in {0,2}" doesn't really make sense. But you can prove that "they don't satisfy p(h) = h for all h in {0,2}". And by definition of "retract", you can answer the last question yourself. We've spent a large number of posts on a relatively simple thing. I think what you need to do is actually try to make a retract from Z4 to {0,2}, and figure out for yourself what goes wrong. If you do this, it should become entirely obvious why Zpk works.

The following explains it, but see if you can figure it out for yourself first:

$\mathbb{Z}_{p^k}$ is cyclic, hence abelian, hence all its subgroups are central, hence normal. What do it's proper subgroups look like? For each m in {1, 2, ..., k-1}, the subset {pm, 2pm, ..., (pk-m)pm} forms a subgroup, and these are the only subgroups. Suppose one of them is a retract, then we get a homomorphism f which maps pm to pm. What does it map 1 to? It maps it to apm for some a in {1, 2, ..., pk-m}. So on the one hand:

f(pm) = pm

but on the other:

f(pm)
= f(1 + 1 + ... + 1) [adding pm times]
= f(1) + ... + f(1) [since f is a homomorphism]
= apm + ... + apm
= ap2m

So if f is to be well-defined, we require:

pm = ap2m (mod pk)
pm(apm - 1) = 0 (mod pk) (*)
pm(apm - 1) | pk
apm - 1 | pk-m

But the only numbers which divide a power of p like pk-m are other (non-bigger) powers of p. On the other hand, apm-1 is clearly not a power of p, for if it were, it would either be congruent to 1 (mod pk), in which case line (*) would be a contradiction, or p would divide it, in which case there would be some integer j such that:

pj = apm - 1
p(apm-1 - j) = 1
p | 1

But of course, no prime divides 1, so again we get a contradiction.

24. Nov 8, 2006

### Oxymoron

AKG, a perfect explanation! Incredible! What Hurkyl and Mathwonk said now make so much sense it isn't funny. Also if I had realised all this, I may have come up with the same idea as Hurkyl and Mathwonk did. I cant believe this came down to prime numbers! Amazing.

This also means that if there are homomorphisms from $\mathbb{Z}_4$ to {0,2} then they are not well-defined. Therefore the only homomorphisms are the ones from {0,1,2,3} -> {0,1,2,3} ({0,1,2,3} is a trivial retract) and from {0,1,2,3} -> {0} ({0} is a trivial retract).

One last question:

Is it possible to have a countable group with countably many retracts?

I was thinking I would start with finite groups, because they are finitely generated. And every finitely generated group is countable. Then if I want to take a quotient I know that it will be finitely generated as well but Im not sure if I want to do this, but it seems handy to have.

Does this sound like the right place to start?

25. Nov 8, 2006

### Hurkyl

Staff Emeritus
When you say countable, I presume you mean countably infinite?

I suspect you will need to make use of an infinite direct sum. (an infinite product would be uncountable!)

Actually... you already know a particular example of an infinite direct sum: the positive rational numbers under multiplication!