## MaBloWriMo 11: Examples of Groups

For reference, here’s the definition of a group again:

• a set $G$
• a special element $1 \in G$
• a binary operation $\cdot$ on $G$

such that

• $\cdot$ is associative, that is, $a \cdot (b \cdot c) = (a \cdot b) \cdot c$ whenever $a$, $b$, and $c$ are elements of $G$
• $1$ is the identity for $\cdot$, that is, $1 \cdot a = a \cdot 1 = a$ for every $a \in G$
• For every element $a \in G$, there is another element $a^{-1} \in G$, called the “inverse” of $a$, such that $a \cdot a^{-1} = a^{-1} \cdot a = 1$.

Today let’s just look at some examples (and non-examples).

• We’ve already seen one example from last time: take $G$ to be the set of all integers $\mathbb{Z}$, with the operation of addition. Adding two integers gives us another integer; addition is associative; and it has $0$ as an identity. For any $n \in \mathbb{Z}$, its negation $-n$ is the inverse: $n + (-n) = 0$.

• On the other hand, the integers with the operation of multiplication do not form a group. Multiplication is associative and has an identity (namely, $1$), but in general there are no inverses. The inverse of, say, $3$ would be some other integer $x$ such that $3x = 1$, which does not exist.

• Motivated by the above non-example, we can think about the set of all rational numbers $\mathbb{Q}$ with the operation of multiplication. Does this form a group? Multiplying two rational numbers yields a rational again; multiplication is associative; and it has $1 \in \mathbb{Q}$ as an identity. Given some $p/q$ its inverse is $q/p$ since $p/q \cdot q/p = 1$. Right? Well, not so fast! The problem is that every rational number has a multiplicative inverse except zero. (If $p/q = 0$ then $p = 0$, and $q/p$ is not well-defined.) So $\mathbb{Q}$ with the operation of multiplication is very close to being a group but not quite!

• We can fix this by simply excluding zero: $\mathbb{Q^\ast} = \mathbb{Q} - \{0\}$ is the set of all nonzero rational numbers. This does indeed form a group with the operation of multiplication: multiplying two nonzero rational numbers yields another nonzero rational; $1$ is a nonzero rational which serves as the identity; and any nonzero rational $p/q$ has $q/p$ (another well-defined nonzero rational) as its inverse.

• Consider the set $\mathbb{Z}_n = \{ 0, 1, 2, \dots, n-1 \}$ which has $n$ elements, consisting of the first $n$ nonnegative integers. Define the operation $+_n$ which works by adding and then taking the remainder $\pmod n$. For example, $1 +_8 2 = 3$, but $4 +_8 5 = 1$, since $4 + 5 = 9 \equiv 1 \pmod 8$. In other words, addition “wraps around” the end of the set. Does this form a group? The $+_n$ operation always gives us another element of $\mathbb{Z}_n$ as its output; it is associative; and $0$ is the identity. What about inverses? Well, the inverse of $k$ is $n-k$, since $k +_n (n-k) = 0$. For example, $1 +_8 7 = 0$, $2 +_8 6 = 0$, and so on. So this does make a well-defined group (it is sometimes called the cyclic group of order $n$).

• A group can’t have zero elements, since there has to be an identity element. Can we have a group with one element? Sure we can (it is called the trivial group). Let $G = \{e\}$ and define the binary operation as $e \odot e = e$. You can check that this is associative and has $e$ as its identity, and that $e$ is its own inverse. (You can also check that this is super boring.)

• How about a group with two elements? Let $G = \{e, a\}$ and suppose $e$ is the identity for the binary operation. That means $e \odot e = e$ and $e \odot a = a \odot e = a$. So the only thing left is to define what $a \odot a$ is. It might seem that we have two choices, but in fact we are not allowed to define $a \odot a = a$, because then $a$ would not have an inverse: if $a \odot e = a \odot a = a$, there would be nothing we can multiply $a$ by in order to to get $e$. So we are forced to choose $a \odot a = e$. But from there we can check that all the group properties do in fact hold. Note we have just proved that there is essentially only one group of size two (I say “essentially” because you could always call the elements something else, but that doesn’t really change the structure of the group). We saw above that $\mathbb{Z}_n$ is a group for any $n$ (actually, for $n \geq 1$), so $\mathbb{Z}_2$ must be the same as the $G = \{e,a\}$ we just defined, and indeed it is: $e$ is $0$ and $a$ is $1$. $1 +_2 1 = 0$, just like we defined $a \odot a = e$.

• You might like to think about groups with $3$ or $4$ elements. How many different possibilities are there?

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 5 Responses to MaBloWriMo 11: Examples of Groups

1. Pat Muchmore says:

This is great, I’m so glad to have you blogging again. So, is $\mathbb{R}$ without 0 also a group? What about $\mathbb{C}$ and $\mathbb{H}$? Intuitively, it seems like they must be as long as you exclude 0, but I might be missing other details. What really interests me though is whether $\mathbb{O}$ without 0 is a group. I know that associativity breaks down in the octonions, but is there some way to rescue it and make it a group? I’m assuming that $\mathbb{S}$ can’t be a group even if we exclude 0 because it has nonzero zero divisors…

• fegleynick says:

R is the set of reals, and C is the set of complex numbers, but what are H, O, and S?

• Brent says:

$\mathbb{H}$ is the set of quaternions (which are 4-dimensional numbers, in the same sense that the complex numbers are 2-dimensional), $\mathbb{O}$ are the octonions (8-dimensional), and $\mathbb{S}$ are sedenions (16-dimensional).

• Brent says:

Hi Pat, yes, $\mathbb{R}$ without zero is a group under multiplication, as are the complex numbers $\mathbb{C}$ and quaternions $\mathbb{H}$. You are right that the octonions $\mathbb{O}$ without 0 is not a group under multiplication, because of the lack of associativity. Off the top of my head, I do not know whether there is a way to “rescue” it — though there probably is some appropriate thing you can do to construct a group which is “as much like $\mathbb{O}$ as possible”.

Since $\mathbb{S}$ has zero divisors, excluding zero by itself does not make sense; the resulting set is then not even closed under multiplication. You would want to do something like remove zero, and then remove all the zero divisors, and then remove all the divisors of the things you just removed, … and so on. I do not know whether this results in anything sensible. And of course you would still have the problem with associativity.