Group Theory for Beginners Trivia Quiz

Answer: Operator: Squaring

To make a group, you will need a set of objects (often numbers) and a binary operator - one that takes exactly two operands (input objects) and creates a result (output object). Squaring a number is not a binary operator - it takes only one operand.

Answer: A member of the set

While the result of any operation must always be unambiguously defined, this is an inherent property of a mathematical operator, so it would not serve as a criterion for a group. The first criterion for a group is that the result of applying the operator on any two operands in the set is part of the set again. As a counterexample, take the set of digits, 0 to 9, and use the operator "average" - while the average of 3 and 5 is 4 (and thus a member of the set), the average of 3 and 6 is 4.5, which is not a digit.

Any such combination of a set and operator that satisfies the closure criterion is called a magma or groupoid.

Answer: (a & b) & c = a & (b & c)

An operation is associative if it doesn't matter whether you evaluate an expression consisting of three operands and two operators by combining the left or the right two operands first. A counterexample would be the set of integers, using subtraction as the operator:

(7 - 4) - 2 = 3 - 2 = 1, but
7 - (4 - 2) = 7 - 2 = 5

This shows that the magma (integers; subtraction) is not a semigroup and thus can't be a group.

Answer: a & i = i & a = a, for all a

An identity element is one that leaves any other element unchanged when used in the operation. This must be satisfied regardless of whether you apply this from the left or from the right.

As a counterexample, let's use the set of (positive) whole numbers. Our operator "&" shall be writing the numbers next to each other, with no space, so 10 & 24 = 1024. This is both closed and associative, so we have a semigroup. 0 is also called a left identity element, because you can put a 0 before any number and not change its value (so 0 & a = a for all a) - however 0 is not a right identity element and thus not a true identity element: 1 & 0 = 10!

Answer: No

Interestingly enough, we do not need to state this requirement because we can prove that this is always the case, by using associativity:

Assume we found an x so that a & x = i.

(a & x) & a = i & a = a. Now we know that this needs to be the same as a & (x & a), so a & (x & a) = a as well, but that means that, by the definition of the identity element, x & a = i.

Congratulations. If all these criteria have been met, we now have constructed ourselves a group!

Answer: No, it is not invertible

So close and yet so far! Multiplication on real numbers is closed (multiplying two real numbers gives a real number), it is associative, it has the identity element 1 and all numbers have an inverse, 1/a - except for that pesky zero. There is no real number x so that 0 * x = 1 and thus our candidate fails the test, remaining a monoid.

Answer: Nowhere - it's not closed

If you check for associativity, identity and invertibility, this looks like a group, but it isn't, because not every sum of two irrational numbers is irrational. If, for example, you take a = pi and b = 4-pi, these two will add up to 4 which is rational and thus not part of the set I defined.

Answer: Yes, it's a group

This time we do have a group, no questions asked. All results are integers from 0 to 12, the operation is associative, the identity element is 0, the inverse of 0 is 0 and the inverse of any other element a is 13-a.

I wouldn't have to use 13 in this context - the equivalent structure for any other number of elements is a group as well.

Answer: True

Once again, we use associativity for the proof. We'll assume that there are two different inverses for a, which we shall call x and y. Again, the identity element is called i.

Now we look at (x & a) & y = i & y = y.
However, x & (a & y) = x & i = x.

Associativity requires that (x & a) & y = x & (a & y), thus x = y, contradicting our assumption that they are different.

Answer: Yes

Yes, there are such groups, but they are difficult to imagine. The simplest set giving one of these has six elements. You can visualize these elements as operations shuffling a three-letter string.

(Somewhat advanced content follows - this is just for the interested):

i is "do nothing". ABC remains ABC.
f is "swap the first two letters". ABC becomes BAC.
l is "swap the last two letters". ABC becomes ACB.
o is "swap the outer two letters". ABC becomes CBA.
x is "take the first letter and put it behind the last". ABC becomes BCA.
y is "take the last letter and put it before the first". ABC becomes CAB.

Now we take {i;f;l;o;x;y} as our set and the operation a & b is "perform a, then perform b". It can be proven by simple counting that this is a group. The identity element is i, the inverse of x is y and vice versa and all other elements are inverse to themselves.

However let's look at o & x versus x & o:

o & x first turns our ABC into CBA, then we shift the C right, resulting in BAC. That's the effect of l, so o & x = l.
x & o however first gives us BCA, then we swap the outer letters for ACB, the outcome of r. So x & o = r.

o & x = l, but x & o = r. Here's your non-commutative group, which, by the way, is called the dihedral group of order 6.