(It is a line reflection if the center of the half-turn is on the mirror line.)
In this figure are two points A and B, the line AB = k, and the midpoint C of AB. Also shown are the lines m through C and n through A, both perpendicular to k.
Let G denote the glide reflection GAB with invariant line k and translation vector AB. For any point P, let Q = G(P).
We will show three ways to construct Q as a product of isometries. This will demonstrate that G is the product of a half-turn and a line reflection.
Let
T = TAB, the translation that takes A to B and
K, M, N = reflections in lines k, m, n
H = halfturn HA with center A
By definition, G = KT (Brown, p. 34). Let Q = G(P). We now study the figure to see 3 ways of constructing Q.
(1) Use the
definition of glide reflection, which states G = KT.
Translate P to get T(P). Reflect T (P) in k to get Q = KT (P). By definition Q = G (P).
(2) Change the order
of the translation and the glide to get Q = TK(P).
Reflect P in k to get P' = K(P). Then we claim Q = TP'). The reason this is true is that PP'QT (P) is a parallelogram (actually a rectangle). Since the sides PP' and T(P)Q are congruent and parallel, the quadrilateral is a parallelogram. Since PP'QT (P) is a parallelogram, the same translation T maps P to T(P) and also maps P' to Q. So Q = TK (P). (Since the angle at P is a right angle, the parallelogram is also a rectangle. If P is actually on line AB, a simple extra argument is needed for the special case.)
(3) Show Q = MH (P)
First, apply H to P to get H(P). Notice that H(P) = N(P') = NK(P). Then reflect in m to get MH (P) = MNK (P). Since MN = T (product of two parallel line reflections), then MH (P) = Q also.
Since P was any point, since MH (P) = G (P) for all P, then MH = G.
This shows that any glide reflection G is a product MH as explained above.
The reasoning above did not use anything about P. While it is not necessary, it may be helpful to look at the same figure for a different P. Here is an example.
Proof of the Big
Theorem about Half-turns and Line Reflections
Theorem: Let m be a line and let A be a point. Denote by A' the reflection of A in m.
Let k be the line through A perpendicular to m. Let n be the line through A parallel to m.
a) If A is on m, then Rm HA = HA Rm = Rk.
b) If A is not on m, then Rm HA = GAA', a glide reflection with invariant line k that maps A to A'.
c) If A is not on m, then HA Rm = GA'A, a glide reflection with invariant line k that maps A' to A.
Proof: The proof is essentially in the figure and discussion above, but starting with H and N and producing G rather than starting with G. Here is a short algebraic argument, without the figure and geometrical motivation.
Proof of (a)
If A is on m, then HA = Rm Rk
=
Therefore, Rm
HA =
Likewise, Rk Rm = Rk Rm Rm = Rk.
Proof of (b)
The halfturn HA = Rn Rk, so Rm HA = Rm (Rn Rk) = (Rm Rn )Rk = TAA' Rk .
We see that Rm Rn is a translation T, since m and n are parallel. Since T(A) = Rm (Rn(A)) = Rm (A) = A', T = TAA'.
Proof of (c)
This is very similar to the proof of (b). This time we use HA = Rk Rn. Then
HA Rm = (Rk Rn)Rm = Rk (RnRm) = Rk TA'A.
Again RnRm is a translation, but it is the inverse of the translation in (b), since the order of the reflections is reversed, so A' is mapped to A.
QED