### William A. McWorter Jr.

Let ( which converts |

First, let's settle the fact that the operation * on **T/H** is well-defined. Let **H** + p = **H** + p'**H** + q = **H** + q'. Then **H**. We show that **H** + p + f(q - p) = **H** + p' + f(q' - p').

p + f(q - p) - (p' + f(q' - p')) = p - p' + f(q - q') - f(p - p').

Since f fixes **H** and p - p' and q - q' are in **H**, the right side of the above equation is a sum of elements of **H** and so is in **H**. Hence **H** + p + f(q - p) = **H** + p' + f(q' - p').

Next, we show that every equation **H** + p) * (**H** + x) = **H** + q**H** + x,**H** + p**H** + q**T/H**. Existence of **H** + x**T**. For uniqueness, let **H** + r'**H** + p) * (**H** + x) = **H** + q.**H** and **H**. So

p + f(r - p) - q - (p + f(r' - p) - q) = f(r - r') is in **H**.

Since f fixes **H**, the point r - r' is in **H**, and so **H** + r = **H** + r'.**H** + y) * (**H** + p) = **H** + q**H** + y**T/H** follows from a similar argument. Since **T**, **H** + s) * (**H** + p) = **H** + q.**H** + s'**H** + y) * (**H** + p) = **H** + q,**H**. So

s + f(p - s) - q - (s' + f(p - s') - q) = s - s' - f(s - s') = - f(f(s - s'))

is in **H**.

Since f fixes **H**, so does f^{ -1}, whence **H**, and so **H** + s = **H** + s'.

Finally, we show that **T/H** is anticommutative, that is, that **H** + p) * (**H** + q)≠(**H** + p) * (**H** + q)**H** + p≠**H** + q.**H** + p) * (**H** + q) = (**H** + q) * (**H** + p),**H** + p**H** + q,**H** + p≠**H** + q.**H**.

(**H** + p) * (**H** + q) = (**H** + q) * (**H** + p) implies that **H**.

Hence **H**.

Also, 2f(p - q - 2f(p - q)) = 2f(p - q) - 4f(f(p - q)) and

4(f(f(p - q)) - f(p - q) + p - q) = 4·0 = 0 are in **H**.

Thus, the sum

[2f(p - q) - 4f(f(p - q))] - [(p - q) - 2f(p - q)] + 4(f(f(p - q)) - f(p - q) + p - q) = 3(p - q)

**H**. Since p - q is not in **H**, this means that the factor group **T/H** contains the element **H** + p - q*a*^{2} + *ab* + *b*^{2}.*a*^{2} + *ab* + *b*^{2}*a*^{2} - 2*ab* + *b*^{2} = (*a* - *b*)^{2}*a* - *b*,*a* and *b*. The result is proved; **T/H** under * is anticommutative.

#### Freaky Links

- Fill the Grid
- Napoleon's Plane Tessellation
**T/H**is a Quasigroup

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

66328429