In each case, we want to "move" the rotation, which is a motion itself, to some new location by means of a second isometry. The way this is done is called conjugation. You have seen conjugation many times, but you may not have articulated it formally, as we will do now.

In a new sketch, draw a point P, a point Q and a point A. Then let the isometry P45 be rotation with center P by 45 degrees; let Q45 be rotation with center Q by 45 degrees; let T be translation by vector PQ and S be the inverse of T, translation by vector QP.

• Rotate the point A with center P by 45 degrees to get A’. Relabel A’ as A1. Select points A and A1 and choose Define Transform from the Transform menu. Label this P45.
• Mark the vector PQ and translate A and A1 by this vector to get A’ and A1’. Select points A and A’ and choose Define Transform from the Transform menu. Label this T.

Then A1’ is related to A’ by rotation by 45 degrees by center Q.

Thus a first guess as to how to form the isometry Q45 would be that Q45 is P45 followed by T.

• So define this transform: Draw a point B and apply R45 to B to get B’ and then apply T to get B’’. Hide B’. Select points B and B’’ and choose Define Transform from the Transform menu. Label this T P45.

• Now apply this transform T P45 to B’’ to get a new point B’’’’ and then again to this new point to get another new point and continue until you have several points and a pattern emerges.
• Do these points seem to come from a rotation of B? Is the angle of rotation still apparently 45 degrees? Is the center Q?
• Construct the center K as the circumcenter of BB’B’’ and observe where K is located. Form the triangle KPQ. What kind of triangle is this? What is the vertex angle at K? What is the image of the point P by the isometry T P45? Assuming this isometry is in fact a rotation by 45 degrees, does knowing the image of P explain the shape of KPQ?

We discovered something about the nature of T P45, but it is not the rotation around Q that we seek. So we will try another candidate.

Let’s reexamine the original figure. What we want to do is start with point A’ and end up with point A1’. How do we do this? The first step is to move from A’ to A (this can be done by S); then rotate by P45 to get A1, and move from A1 to A1’ by T. Thus the process is a three-step procedure, namely T P45 S. Let’s call this triple composition U.

• Before starting, hide all the points except P and Q. (Select all points but P and Q and create a Hide/Show button.)
• To investigate U in our sketch, we need to define S. Draw any point C and mark the vector QP. Translate C by this vector to get C’. Select C and C’ and Define Transform, naming this S.
• Now continue. Apply P45 to C’ to get C’’ and then apply T to C’’ to get C’’’.
• Let’s look at this figure. Construct segments to form the quadrilateral CC’C’’C’’’. Drag C around and observe the shape. What kind of quadrilateral is this figure? Is it clear why it is this shape? Measure angle C’PC’’ and also angle CQC’’’. Can you explain why the angles are the same?
• Now, let’s iterate isometry U. First we must define it.
• To reduce clutter, hide C’ and C’’ and the sides of the quadrilateral, except segment CC’’’. Then select C and C’’’ and Define Transform. Call it U.

• Now we can experiment with U as we did before with T P45. We have C and C’’’. Apply U to C’’’ to get C’’’’’’ and then keep applying U to the new points. Does U appear to be a rotation? What is the center of rotation this time? Also apply U repeatedly to the segment CC’’’ to get a symmetric pattern.

Note on Hiding Labels: If the labels are cluttering up the landscape, you can hide them all. First, select all these objects. Then go to the Display menu. If it only says Show Labels, do not despair. Choose that. Now immediately go back to the menu, without changing the selection. Now it will say Hide Labels. So choose that.

The isometry U is an example of a general principle. Let F and J be any isometries. Let K be the inverse of J. Then let G = JFK.

Then G is an isometry, since it is a composition of isometries. But G has the same nature as F. So if F is a rotation, so is G. If F is a reflection, so is G.

One way to see this is to look at a point A and its images, B = F(A) and C = F(C). Then if A’ = J(A), B’ = J(B), C’ = J(C). Then G(A’) = JFK(J(A)) = JF(A) = J(B) = B’. Likewise, G(B’) = C’.

Thus, for example, if A is a fixed point of F, then A’ is a fixed point of G. A consequence of this is that if F has exactly one fixed point (which happens if and only if F is a rotation), so does G. So G is a rotation if and only if F is a rotation.

F has two fixed points M and N, if and only if F is reflection in the line MN. Thus G is then reflection in line J(M) J(N).

Likewise, if A, B, C are collinear, then A’, B’, C’ is collinear. But A, B, C is collinear for any choice of A only when F is a translation, so F is a translation if and only if G is a translation.

This only leaves glide reflections. We can conclude that F is a glide reflection if and only if G is a glide reflection by process of elimination, or we can observe a feature that is preserved. For instance, F is a glide reflection if and only if there is exactly one invariant line m that is mapped into itself by F. We can show that J(m) is mapped by G into itself.

Now that we have stepped through two experiments, we will outline the next experiments and let you fill in the details.

This time you can vary the angle of rotation by dragging F, so you can experiment with different angles. If the angle of R is t, what is the angle of V? How does this differ from what happened with U? Can you see a reason for this?

What kind of isometry is the composition WT? How is it related to TW? What can you say about the direction of these isometries?

What kind of isometry is the composition W^(-1)T? [We use W^(-1) for the inverse of W.] How is it related to T^(-1)W? What can you say about the direction of these isometries?

Note: This particular combination is a key element in understanding wallpaper groups than contain line reflections or glide reflections. It is discussed as a theorem in Bix.

There are many more possibilities to investigate. Here are some to try if you have time. Try to guess what you will get before trying the experiment.

• Conjugate a line reflection by a rotation.
• Conjugate a line or glide reflection by a translation. (What if the translation is parallel to the mirror line of the reflection?)