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I am reading about symmetries in Wikipedia and I am trying to understand this statement:

Two geometric figures are considered to be of the same symmetry type if their symmetry groups are conjugate subgroups of the Euclidean group $E(n)$ (the isometry group of $\mathbb R^n$), where two subgroups $H_1$, $H_2$ of a group $G$ are conjugate, if there exists $g \in G$ such that $H_1=g^{−1}H_2g$.

What does conjugacy class have to do with this ? If I consider matrices I understand what is the meaning of being in the same conjugacy class, but in this example I don't...

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Matrices do not suffice to describe the group $E(n)$ of isometries of $\Bbb R^n$. The orthogonal group $O(n)$ of orthogonal matrices is the subgroup of $E(n)$ that fix the origin $0\in\Bbb R^n$.

In order to get all of $E(n)$ you have to "add" (in the group theoretic sense) all the translations, i.e. the transformations $P\mapsto P+\vec v$ where $\vec v$ is an arbitrary fixed vector in $\Bbb R^n$.

The concept of conjugation, and of conjugacy class, can be readily extended to just any group $G$: if $g\in G$, "conjugation by $g$" is the transformation $x\mapsto gxg^{-1}$ (actually an automorphism).

For example, let $n=2$ and let $\cal C$ and $\cal C^\prime$ two circles centered at points $C$ and $C^\prime$ respectively. The subgroup of symmetries of $\cal C$ is the subgroup $H$ of $E(n)$ whose elements are the rotations and reflexions around $C$. The same, with respect to $C^\prime$, for the subgroup $H^\prime$ of the symmetries of $\cal C^\prime$.

It's not difficult to see that $H^\prime=tHt^{-1}$ where $t\in E(2)$ is the translation defined by the vector $\vec{CC^\prime}$, i.e. $H$ and $H^\prime$ are conjugate inside $E(2)$.

Mind that if $\cal C$ and $\cal C^\prime$ have different radii there's no isometric transformation that changes one into the other, yet their groups of symmetries are conjugate, hence isomorphic (I would say that this is the point in introducing the concept of symmetry type).

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great example, really clears some problems I had. thank you (+1) –  Belgi Mar 24 '12 at 11:18

The two figures may have different orientations and/or positions. Just as in the definition of congruence we demand not that the figures are identical but that they can be made identical by an isometry, so in the definition of "same symmetry type" we demand not that the figures have the same symmetry group but that their symmetry groups can be made identical by an isometry. Conjugation is just how transformations act on groups: To compare the symmetry groups of two figures, you apply an isometry that brings one figure into an orientation and position corresponding to that of the second figure, then apply a symmetry operation of the second figure, then "undo" the isometry by applying its inverse, and then check whether the combined result of these operations corresponds to a symmetry operation of the first figure. That's just what conjugation is.

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"we demand not that the figures have the same symmetry group but that their symmetry groups can be made identical by an isometry" - does this mean that their symmetry groups are isomorphic ? do you mean that this is done by the same isometry ? –  Belgi Mar 24 '12 at 9:15
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@Belgi: This means more than that the symmetry groups are isomorphic. They might be abstractly isomorphic but not via conjugation with an isometry. For instance, one figure might have inversion as its only non-trivial symmetry operation, and the other might have reflection in some plane. These two symmetry groups are abstractly isomorphic (as all groups with two elements are), but they represent different types of spatial operations and are not conjugates of each other as subgroups of the Euclidean group of isometries of Euclidean space. What do you mean by "this is done by the same isometry"? –  joriki Mar 24 '12 at 9:25
    
you wrote "can be made identical by an isometry", what I asked is if you mean that there exist an isometry phi s.t. applying phi to each element of one symmetry group turns that symmetry group to the second figure symmetry group ? –  Belgi Mar 24 '12 at 9:33
    
@Belgi: If by "applying the isometry to an element of the symmetry group" you mean "conjugating the element of the symmetry group with the isometry", then yes. It has to be the same isometry for all elements of the symmetry group, but I can't think of an example where that would make a difference, i.e. where you could conjugate the symmetry operations with different isometries to get the other symmetry group, and yet couldn't also do it with the same isometry for all of them. –  joriki Mar 24 '12 at 9:59
    
thank you for your help! –  Belgi Mar 24 '12 at 11:18

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