# Symmetry group - symmetry types

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