For congruence classes, what does $b = g^{–1}ag$ mean?

Question

Under Wikipedia's definition of congruence classes, (which I got to via studying space groups, point groups and crystallography) it says:

In mathematics, especially group theory, two elements $a$ and $b$ of a group are conjugate if there is an element g in the group such that $b = g^{–1}ag$. This is an equivalence relation whose equivalence classes are called conjugacy classes.

That looks like matrix similarity.... How does that translate to group symmetry?

Answer 1

Let me explain conjugacy in the setting of a planar crystallographic group (which is a little simpler than space groups, and the concepts are pretty much identical).

The key idea of conjugacy in this setting is that if two rigid motions in the plane are conjugate then they are symmetries of "the same type", where by "same type" I am referring (somewhat vaguely) to the classification of planar rigid motions according to their geometric properties: rotation, translation, reflection, or glide reflection; and more numerical properties such as rotation angle and translation distance. Let me put some meat on it by going through the classification of planar rigid motions.

For the proof, let $a,b$ be planar rigid motions that are conjugate to each other. It follows that there exists a planar rigid motion $g$ such that $b = g^{-1} a g$.

• If $a$ is a rotation around some point then $b$ is also a rotation around some point. Furthermore, the rotation angles of $a$ and $b$ are equal.

Here's the detailed proof.

First, let $p$ be the point around which $a$ rotates. It follows that $a(p)=p$, i.e. the point $p$ is fixed by $a$. From that, it follows that the point $q = g^{-1}(p)$ is fixed by $b$, i.e. $b(q)=q$, because $$b(q) = g^{-1} a g(q) = g^{-1} a (p) = g^{-1}(p) = q $$ That example pretty much captures the whole strategy: one keeps using the conjugacy equation over and over again, to prove that the "classification" properties of $a$ and of $b$ are the same.

Let's move on to the rotation angles of $a$ and $b$. Let $\theta$ be the rotation angle of $a$, which means that if $\overrightarrow{pr}$ is any ray based at $p$, then $\theta$ is equal to the angle between the ray $\overrightarrow{pr}$ and the image ray $a(\overrightarrow{pr})$. Since $g^{-1}$ is a rigid motion, it follows that $\theta$ is equal to the angle between the rays $g^{-1}(\overrightarrow{pr})$ and $g^{-1}(a(\overrightarrow{pr}))$. Now let's re-express those last two rays. \begin{align*} g^{-1}(\overrightarrow{pr}) &= \overrightarrow{g^{-1}(p) g^{-1}(r)} \\ &= \overrightarrow{q g^{-1}(r)} \\ &= \overrightarrow{qs} \end{align*} where we set $s = g^{-1}(r)$.

Also, \begin{align*} g^{-1}(a(\overrightarrow{pr})) &= (g^{-1} a)(\overrightarrow{pr}) \\ &= (b g^{-1})(\overrightarrow{pr}) \\ &= b \left(\overrightarrow{g^{-1}(p) g^{-1}(r)}\right) \\ &= b(\overrightarrow{q s}) \end{align*} Thus we have shown that $\theta$ is also the angle between the the ray $\overrightarrow{qs}$ and its image ray $b(\overrightarrow{qs})$ under $b$, and therefore $b$ also has rotation angle $\theta$.

That's probably enough detailed proof so you can get the idea, but here's the rest of the rigid motion classification, with proof outlines:

• If $a$ is a reflection across a line then $b$ is also a reflection across a line.

Proof outline: Let $L$ be the line that $a$ reflects across; then $g^{-1}(L)$ is the line that $b$ reflects across.

• If $a$ is a translation, with translation distance $d$, then $b$ is also a translation with translation distance $d$.

Proof outline: For any point $p$, $d$ equals the distance from $x$ to $a(p)$. Letting $q = g^{-1}(p)$, it follows that $d$ equals the distance from $q=b(p)$ to $b(q)=b(g^{-1}(p)=g^{-1}(a(p))=g^{-1}(q)$.

• If $a$ is a glide reflector along a line $L$, translating that line along itself by distance $d$, then $b$ is a glide reflector along the line $g^{-1}(L)$, translating the line along itself by distance $d$.