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Retraction math help 
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#1
Nov306, 10:20 AM

P: 867

A subgroup H of a group G is a retract of G if there exists a homomorphism [itex]q\,:\,G \rightarrow H[/itex] such that [itex]q(h) = h[/itex] for all [itex]h \in H[/itex]. 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, [tex]\mbox{ker}(q) = K < G[/tex] 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 [itex]G = KH[/itex] and [itex]K \cap H = \{e_H\}[/itex] (from wikipedia). The group G, then, should be the semidirection product of K and H. Is this right? Now, since H acts on K by conjugation: [tex]k \mapsto hkh^{1}[/tex] this defines a group homomorphism [tex]p\,:\,H \rightarrow \mbox{Aut}(K)[/tex] 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 semidirect product [itex]K \rtimes [/itex] is a group consisting of pairs [itex]hk[/itex] with multiplication [tex](h_1k_1)\cdot (h_2k_2) = (h_1 h_2)(p(h_1)(k_1)k_2)[/tex] and we also get [tex]hkh^{1} = p(h)(k)[/tex] Now, my main question (and the reason why I brought the semidirect 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 semidirect product [itex]K \rtimes H[/itex] 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 semidirect product came pretty close. 


#2
Nov306, 08:34 PM

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Z_{4} is a nonsimple group with only trivial retracts. Double check this.



#4
Nov506, 10:40 PM

P: 867

Retraction math help
Wow, how did you guys come up with such easylooking answers!?
Well, [itex]4\mathbb{Z}[/itex] is a subgroup of [itex]\mathbb{Z}_4[/itex] is it not? (Also, this is a normal subrgoup, no?) and [itex]4\mathbb{Z}[/itex] is certainly not a trivial subgroup. So, [itex]\mathbb{Z}_4[/itex] is nonsimple because it contains normal subgroups which are not trivial. Suppose that [itex]H[/itex] is a subgroup of [itex]\mathbb{Z}_4[/itex]. H is a retract if there is a homomorphism [itex]p\,:\,\mathbb{Z}_4 \rightarrow H[/itex] such that [itex]p(h) = h[/itex] for all h in H. So do I have to check every subgroup of [itex]\mathbb{Z}_4[/itex] and see if there is a homomorphism? 


#5
Nov506, 10:44 PM

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


#6
Nov506, 10:52 PM

P: 867

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



#7
Nov506, 10:53 PM

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



#8
Nov506, 11:06 PM

P: 867

{0,1,2,3} {0,2} {0} These are the only subgroups of Z_4 that I could find. Does this look right? 


#9
Nov506, 11:22 PM

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#10
Nov506, 11:49 PM

P: 867

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


#11
Nov606, 07:27 AM

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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
Nov606, 08:20 PM

P: 867

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
Nov606, 09:13 PM

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#14
Nov606, 09:44 PM

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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
Nov706, 01:42 AM

P: 867

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 [itex]\mathbb{Z}_4[/itex] is 0 mod 4 = {...,8,4,0,4,8,...} and the homomorphism h(x) = x which takes an element from [itex]\mathbb{Z}_4[/itex] 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
Nov706, 01:47 AM

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#17
Nov706, 08:13 AM

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#18
Nov706, 01:57 PM

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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.) 


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