One rule for many symmetries

Mathematics · Abstract Algebra · ages 21-22

Name ______________________ Date ____________

Which set with the given operation forms a group?
- Positive integers under addition
- Integers under addition
- Integers under subtraction
How many elements are in the planar symmetry group of a square?
- 4
- 6
- 8
Why do positive integers fail to form a group under addition?
- They exclude 0 and additive inverses.
- Adding two positive integers can give a negative integer.
- Regrouping three positive integers changes their sum.
Use multiplication modulo 6. Which candidate set forms a group?
- The classes {0, 1}
- The classes {1, 5}
- The classes {1, 2}
- The classes {0, 5}
Which clockwise rotation is the inverse of a clockwise 90 degree rotation of a square? All rotations are about its center.
- A clockwise rotation through 90 degrees
- A clockwise rotation through 180 degrees
- A clockwise rotation through 270 degrees
- A clockwise rotation through 0 degrees
Integers under subtraction form a group because differences of integers stay integers.
Circle one: True False
Under addition modulo 6, Nina says H = {0, 2, 4} fails closure because 2 + 4 = 6 and 6 is not listed in H. Is her conclusion valid?
Circle one: True False
Lee says that the class 1 is its own inverse under addition modulo 6. Is this correct?
Circle one: True False
A student claims that all six congruence classes modulo 6 form a group under both addition and multiplication. Which check corrects the claim?
- Both operations pass because reducing results proves every axiom.
- Addition passes, but multiplication gives 0 no inverse.
- Multiplication passes, but addition has no identity class.
In the planar symmetry group of a square, what is the order of a clockwise 90 degree rotation about its center?
- 4
- 2
- 8
- 1

Answer key

For grown-ups. Fold this page away before handing over the rest.

One rule for many symmetries W1-mt_HFAhcaGo2L-s1

Integers under addition · Integer addition is closed and associative, has identity 0, and gives each n the inverse -n. The other candidates fail group axioms.
8 · There are four rotations about the center and four reflections, giving eight distinct motions.
They exclude 0 and additive inverses. · The additive identity would be 0, which is not positive. The additive inverse of a positive integer is negative and is also outside the set.
The classes {1, 5} · In {1, 5}, products reduce to 1 or 5, the identity is 1, and 5 times 5 gives remainder 1. Each element has an inverse. The other candidates fail.
A clockwise rotation through 270 degrees · The inverse must return every vertex to its starting position. A further clockwise 270 degrees makes a total of 360 degrees, the identity motion.
False · Integer subtraction is closed but not associative. For example, (5 - 2) - 1 = 2, while 5 - (2 - 1) = 4.
False · The operation includes reduction modulo 6. The sum 2 + 4 therefore gives class 0, which is in H. Every pairwise sum reduces to 0, 2 or 4.
False · An additive inverse must give remainder 0. Since 1 + 1 has remainder 2, 1 is not its own inverse. Its inverse is 5 because 1 + 5 has remainder 0.
Addition passes, but multiplication gives 0 no inverse. · Addition modulo 6 has identity 0 and an inverse for every class. Multiplication has identity 1, but 0 times any class is 0, so 0 has no multiplicative inverse.
4 · One, two and three quarter-turns move vertices away from their own starting positions. Four quarter-turns total 360 degrees and give the identity, so the order is 4.