Let $G$ be a group acting on a set $X$. For all $ g\in G $, we consider the application \begin{align*} \varphi_{g}: &~ X \longrightarrow X \\ &~ x \longrightarrow g.x \end{align*} It is clear that $\varphi_{gh}=\varphi_g\circ\varphi_h$, $\forall~ g,h \in G,$ $\varphi_e=\text{Id}_X$ ( $e$ the neutral element of $G$ ), and $\varphi_g \circ \varphi_{g^{-1}}=\varphi_{g^{-1}} \circ \varphi_g $, so $\varphi_g$ is bijective, for all $g\in G$. i.e: $\varphi \in \mathcal S(X)$, $\forall~ g\in G$, and the application: \begin{align*}\Phi :&~ G \longrightarrow \mathcal S(X) \\ &~g\longrightarrow \varphi_g\end{align*} is a group homomorphism that we call a representation of $G$ in $\mathcal S(X).$

I didn't understand what this definition is for, I'm looking for the intuition or the idea behind it. Why do we call the application $\Phi$ by this name: "representation", it is only a homomorphism? if anyone has any ideas or comments that he can add, I will be very grateful.

  • 3
    $\begingroup$ en.wikipedia.org/wiki/Group_representation may give you more information on the term. $\endgroup$ – freakish Aug 11 at 9:53
  • $\begingroup$ Explore the action of $D_8$ (dihedral group on 4 vertices) on the vertex set $V = \{ 1, 2, 3, 4\}$. Write down the group table by hand. This will give you a good intuition. $\endgroup$ – abcd123 Aug 11 at 9:56
  • 2
    $\begingroup$ The original purpose of groups was to describe symmetry as a collection of transformations. It was later abstracted into, well, the definition of an abstract group by a bunch of axioms that makes no reference to transformations. As a result, an abstract group can be represented by transformations of various other mathematical objects (or even other things besides transformations, like numbers or loops) in more than one way. For instance, $S_4$ could be represented by permutations of $4$ objects, or it could be represented by 3D rotations that preserve a cube. $\endgroup$ – runway44 Aug 11 at 9:57
  • $\begingroup$ The definition of a group as we know it goes back to 1882 (von Dyck). Sylow's theorem was proved in 1872. $\endgroup$ – David A. Craven Aug 11 at 10:42
  • 2
    $\begingroup$ See also the discussion here. $\endgroup$ – Arturo Magidin Aug 11 at 18:49

Put in general terms, a representation of an abstract object (group, ring, etc.) is a way of thinking about that object acting on a certain other thing we can describe in concrete terms. The most natural candidates for the other thing are sets (so that group elements act as permutations of those objects) or vector spaces (so that group elements act as matrices).

It's quite hard to understand dihedral groups abstractly, just using muliplication tables, but easy to understand them as symmetries of an $n$-gon via a matrix representation (i.e., as $2\times 2$ matrices over $\mathbb{R}$), or as permutations of the vertices of the $n$-gon.

So one aspect of group actions is that they can be used to more easily study the group. They can also be used to study the object being acted upon. For example, a cubic equation with real coefficients must have a real root. Why? Because complex conjugation permutes the roots, and it therefore swaps them in pairs. But there are an odd number of roots, so (at least) one is fixed. That's a trivial application of group actions.

| cite | improve this answer | |
  • $\begingroup$ (Of course, you can use the intermediate value theorem for the cubic equation, but you try proving the intermediate value theorem from first principles. I think it's a little easier to use group actions, and it generalizes to certain discontinuous functions.) $\endgroup$ – David A. Craven Aug 11 at 9:56
  • $\begingroup$ Thank you very much, your explanation clarified the ideas for me. another question please: what is the action used to model the cubic equation problem? do you have a course or a book that explains this to master the actions of a group? $\endgroup$ – M-S Aug 11 at 10:43
  • $\begingroup$ Complex conjugation does not affect the equation $ax^3+bx^2+cx+d=0$ (if $a,b,c,d\in\mathbb{R}$), so cannot send roots of the equation to non-roots. Therefore it permutes them. $\endgroup$ – David A. Craven Aug 11 at 10:44
  • $\begingroup$ Can you please elaborate more ? i'm a beginner in group theory so i need more explanation to understand better. $\endgroup$ – M-S Aug 11 at 10:48
  • $\begingroup$ For the level of detail you seem to need, I would recommend talking to your course instructor. $\endgroup$ – David A. Craven Aug 11 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.