My question, in it's general formulation, is : is there a way to construct non trivial group endomorphism other than conjugation ?

Now for my specific needs, I wont to find some endomorphism other than conjugation of the following group :

$$G=\left\{ \left(\begin {array} {c c} a & b \\ 0 & 1 \end {array} \right ) \equiv [a,b] \,:\, a\in U(1) , ~ b\in \mathbb C \right\} \simeq U(1)\ltimes \mathbb C $$

This group is isomorphic to the special Euclidean group $ SE(2)= SO(2)\ltimes \mathbb C $, also called the rigid motions group, or displacement group which is the subgroup of Euclidean group that contains all orientation-preserving isometries (translations and rotations).

For information, the conjugation gives the following endomorphism

$ \mathfrak{h}=[\alpha_{\mathfrak{h}},\beta_{\mathfrak{h}}] $ be a fixed element in $ G $, $$\begin{equation} \rho_{\mathfrak{h}}(g):= \mathfrak{h} g\mathfrak{h}^{-1}=[a,(1-a)\beta_{\mathfrak{h}}+b\alpha_{\mathfrak{h}}]; \qquad g=[a,b]\in G \end{equation}$$

  • $\begingroup$ Do you mean group isomorphism? Or a homomorphism $G \to G$? $\endgroup$ – Travis Jun 24 '15 at 4:32
  • $\begingroup$ Also, I might avoid using Fraktur characters (e.g., $\mathfrak h$) for group elements, as they are often used for Lie algebras, including the ones associated to Lie groups like $G$. (The use of the bracket and comma in the notation for group elements makes this doubly confusing, and this notation is also used for the Lie bracket of a particular Lie algebra.) $\endgroup$ – Travis Jun 24 '15 at 4:35
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    $\begingroup$ @AymaneFihadi One option is the complex conjugation map, $[a, b] \mapsto [\bar{a}, \bar{b}]$, which by your last display equation cannot be given by (algebraic) conjugation by any element. $\endgroup$ – Travis Jun 24 '15 at 4:39
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    $\begingroup$ @AymaneFihadi Another seems to be the map $[a, b] \mapsto [a, 0]$. Note that this is just projection onto the quotient of $G$ by the normal subgroup $\{1\} \times \Bbb C$, which is isomorphic to $U(1)$, followed by inclusion back into the semidirect product. $\endgroup$ – Travis Jun 24 '15 at 4:44
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    $\begingroup$ @AymaneFihadi Since both examples are solutions to your second question, I've written them up in an answer in their own right. $\endgroup$ – Travis Jun 24 '15 at 5:07

For the specific case of the given group $G$, there seem to be at least a few interesting choices:

  1. As usual we can view $U(1)$ as the unit circle in $\Bbb C$, which in particular is preserved (as a set) under complex conjugation. In the bracket notation we can write the multiplication operation as $$[a, b] \cdot [c, d] := [ac, ad + b];$$ since complex conjugation is a field automorphism, we see immediately that the map $C: G \to G$ defined by $$C: [a, b] \mapsto [\bar{a}, \bar{b}]$$ is a group isomorphism.

  2. As indicated by the semidirect product notation, the group $N := \{1\} \times \Bbb C < G$ is normal, and the quotient map $\pi: G \to G / N \cong U(1)$ is a homomorphism. Postcomposing with the inclusion homomorphism $\iota$ back into the semidirect product gives the map $$\Pi := \iota \circ \pi$$ defined by $$\Pi : [a, b] \mapsto [a, 0],$$ which neither trivial nor an isomorphism.

  3. One can combine the above examples to produce the homomorphism $$\Pi \circ C = C \circ \Pi : [a, b] \mapsto [\bar{a}, 0],$$ which again is neither trivial nor an isomorphism.

  4. One can modify the examples in (2) and (3) by inserting in the composition the power map homomorphism $p_n : U(1) \to U(1)$ defined by $p_n(z) := z^n$ for any $n \in \Bbb Z$ to produce the maps $$P_n := \iota \circ p_n \circ \pi : [a, b] \mapsto [a^n, 0].$$ The map $p_0$ is trivial (and hence so is $P_0$), $p_1$ is the identity map (so $P_1 = \Pi$), and $p_{-1}$ is the conjugation map (so $P_{-1} = C \circ \Pi$), and hence this example subsumes (2) and (3) above. Note also that $P_{-k} = P_k \circ C = C \circ P_k$.

  • $\begingroup$ When I have seen your comment about [a,0] endomorphism, I thought about the power modification :) $\endgroup$ – Aymane Fihadi Jun 24 '15 at 5:32

In full generality, no. Any group homomorphism $\phi\colon G\to H$ yields a normal subgroup of $G$ via its kernel (and identifies the morphism up to an isomorphism of the image). This goes for group endomorphisms, in particular. So in a simple group the only endomorphisms are the trivial morphism and the automorphisms. Discarding the trivial morphism, we are left with automorphisms. If there is an outer automorphism, you are done. A simple group need not, in general, possess an outer automorphism, however.

So to solve your particular case, start considering the outer automorphisms, and then failing that the normal subgroups and whether $G$ contains a copy of the quotient by said subgroup. For normal subgroups, one should be staring you in the face from the groups you've listed it as isomorphic to.

Indeed, $SE(2)$ is normal (of index 2) in the full Euclidean group, so conjugation by a reflection acts as an order 2 isomorphism of $SE(2)$. Do you know if that is an inner isomorphism?

  • $\begingroup$ thank you. No I don't. but I will think about that later. another question : is there a way to identify Euclidean group with some 2x2 matrix group? because I am not used to manipulate Euclidean group. $\endgroup$ – Aymane Fihadi Jun 24 '15 at 5:26
  • $\begingroup$ @AymaneFihadi I think the answer is no, but that's worth asking as a separate question in its own right. If you do so, you should include a link in that question to this one. $\endgroup$ – Travis Jun 24 '15 at 5:42

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