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Suppose I have an some plane and an identity mapping on the points of the plane. I see that the identity can be expressed as a product of an even number of reflections, since any reflection has itself as its own inverse. But why is it impossible to ever express the identity as the product of an odd number of reflections? Thanks.

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Hint: Look at the orientation (determinant). – t.b. Mar 22 '11 at 9:50
It might be interesting for you: The product of two reflections is a rotation (look up in Wikipedia). – shuhalo Mar 22 '11 at 10:03
@Theo: perhaps add your comment as an answer? – lhf Mar 22 '11 at 13:07
up vote 6 down vote accepted

On lhf's request I make my comment into an answer:

A reflection reverses the orientation of the plane: If you look what happens to a triangle whose corners are $ABC$ (oriented counterclockwise) then after a reflection the corners will be ordered as $ACB$ (oriented counterclockwise). So after applying an odd number of reflections the corners of $ABC$ will again be arranged as $ACB$ counterclockwise, so you can't possibly have the identity.

If you want to make this into a rigorous proof, you can do this by introducing the determinant (of the linear part) of an isometry and show that it is multiplicative: That is to say $\det(ST) = \det(S)\det(T)$. Now a reflection has determinant $-1$, so the determinant of the composition of $n$ reflections will be $(-1)^{n}$. If $n$ is odd then the determinant will be $(-1)^n = (-1)^{2k+1} = -1$. On the other hand the determinant of the identity transformation is $1$, so only an even number of reflections can give the identity.

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For something essentially the same, but more folksy: reflections are rigitd motions, so take an object which is red on one side and green on the other. A reflection in a line turns it over and changes the visible colour. The identity mapping does not change the visible colour (nor do rotations in the plane or translations). So to get the object back to its original colour you need an even number of colour changes, i.e. an even number of reflections. – Henry Mar 22 '11 at 13:27
@Henry: Thanks! this is a much nicer way to put it. – t.b. Mar 22 '11 at 13:30

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