- #1

Oxymoron

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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 semi-direction 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 semi-direct 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 semi-direct product up) is this: Is the existence of a group G, which is

I figured that such a group

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 [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 semi-direct product came pretty close.

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 semi-direction 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 semi-direct 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 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 [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 semi-direct product came pretty close.

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