2.5: Group Conventions and Properties
Before we discuss more examples, we present a theorem and look at some conventions we follow and notation we use when discussing groups in general; we also discuss some properties of groups.
2.5.1: Some Group Conventions
Theorem \(\PageIndex{1}\)
The identity element of a group is unique (by Theorem 2.1.9), and given any element \(a\) of a group \(G\text{,}\) the inverse of \(a\) in \(G\) is unique (by Theorem \(2.1.2\)).
Note
We will generally use \(e\) or \(e_G\) as our default notation for an identity element of group, but be aware that many mathematicians denote a group's identity element by 1. We denote the inverse of element \(a\) in \(G\) by \(a^{-1}\text{.}\)
Note
Although it is written in what we call multiplicative notation , do not assume \(a^{-1}\) is what we usually think of as a multiplicative inverse for \(a\text{;}\) remember, we don't even know if elements of a group are numbers! The type of inverse that \(a^{-1}\) is (a multiplicative inverse for a real number? an additive inverse for a real number? a multiplicative inverse for a matrix? an inverse function for a function from \(\mathbb{R}\) to \(\mathbb{R}\text{?}\)) depends on both \(G\)'s elements and its operation.
Note
-
We usually don't use the notation \(*\) when describing group operations. Instead, we use the multiplication symbol \(\cdot\) for the operation in an arbitrary group, and call applying the operation “multiplying”—even though the operation may not be “multiplication” in the non-abstract, traditional sense! It may actually be addition of real numbers, composition of functions, etc. Moreover, when actually operating in a group \(\langle G, \cdot\,\rangle\text{,}\) we typically omit the \(\cdot\) . That is, for \(a,b\in G\text{,}\) we write the product \(a\cdot b\) as \(ab\text{.}\) We call this the “product” of \(a\) and \(b\text{.}\)
For every element \(a\) in a group \(\langle G, \cdot \rangle\) and \(n\in \mathbb{Z}^+\text{,}\) we use the expression \(a^n\) to denote the product
\begin{equation*} a \cdot a \cdot \cdots \cdot a \end{equation*}
of \(n\) copies of \(a\text{,}\) and \(a^{-n}\) to denote \((a^{-1})^n\) (that is, the product of \(n\) copies of \(a^{-1}\)). Finally, we define \(a^0\) to be \(e\text{.}\) Note that our “usual” rules for exponents then hold in an arbitrary group: that is, if \(a\) is in group \(\langle G, \cdot\,\rangle\) and \(m,n\in \mathbb{Z}\text{,}\) then \(a^m a^n = a^{m+n}\) and \((a^m)^n=a^{mn}=(a^n)^m\text{.}\)
However:
-
When we know our operation is commutative, we typically use additive notation, denoting the group operation by \(+\text{,}\) calling the group operation “addition,” and denoting the inverse of an element \(a\) by \(-a\text{.}\) When we use additive notation, we do not omit the \(+\) when operating in a group \(\langle G,+\rangle\text{,}\) and we call \(a+b\) a sum rather than a product. Also, when working with an operation that is known to be commutative, the identity element may be denoted by 0 rather than by \(e\text{,}\) \(e_G\text{,}\) or 1, and for \(n\in \mathbb{N}\text{,}\) we write \(na\) instead of \(a^n\text{.}\)
Finally, note that \((-n)a=n(-a)=-(na)\) (where \(-a\) and \(-(na)\) indicate the additive inverses of \(a\) and \(na\text{,}\) respectively); we can therefore unambiguously use the notation \(-na\) for this element. Using this notation, note that for \(m\in \mathbb{Z}\text{,}\) \(na+ma=(n+m)a\) and \(n(ma)=(nm)a\text{.}\)
Note
Be careful to always know where an element you are working with lives! For instance, if, as above, \(n\in \mathbb{Z}^+\) and \(a\) is a group element, \(-n\) and \(-a\) look similar but may mean very different things. While \(-n\) is a negative integer, \(-a\) may be the additive inverse of a matrix in \(\mathbb{M}_2(\mathbb{R})\text{,}\) the additive inverse \(2\) of the number \(4\) in \(\mathbb{Z}_6\text{,}\) or even something completely unrelated to numbers.
Remark
Multiplicative notation can be used when working with any arbitrary group, while additive notation should be used only when working with a group whose binary operation is commutative.
We summarize multiplicative versus addition notation in the following table, where \(a,b\) are elements of a group \(G\text{.}\)
| Table 2.5.1: Summary of Multiplicative and Addition Notation in Groups. | ||
|---|---|---|
| Multiplicative Notation | Additive Notation | |
| Operation Notation | \(\cdot\) | \(+\) |
| \(a\) operated with \(b\) | \(ab\) | \(a+b\) |
| Identity Element | \(e\) or \(e_G\) (or \(1\)) | \(e\) or \(e_G\) (or \(0\)) |
| Inverse of \(a\) | \(a^{-1}\) | \(-a\) |
Note
We do use the notation \(*\) when using multiplicative or additive notation would lead to confusion. For instance, if we want to define an operation on \(\mathbb{Q}^*\) that assigns to pair \((a,b)\) the quantity \(ab/2\text{,}\) it would be unwise to use multiplicative or additive notation for this operation since we already have conventional meanings of \(ab\) and \(a+b\text{.}\) Similarly, we would not denote the identity element of \(\mathbb{Q}^*\) under this operation by \(0\) or \(1\text{,}\) since the identity element in this group is the rational number \(2\text{,}\) and writing \(0=2\) or \(1=2\) would look weird.
Finally:
Note
If there is a default notation for a particular operation (say, \(\circ\) for composition of functions) or identity element (say, \(I_n\) in \(GL(n,\mathbb{R})\)) we usually use that notation instead.
2.5.2: Some Group Properties
While we don't need to worry about “order” when multiplying a group element \(a\) by itself, we do need to worry about it in general.
Note
Group operations need not be commutative!
Definition: Abelian and Nonabelian
A group \(\langle G, \cdot\,\rangle\) is said to be abelian if \(ab=ba\) for all \(a,b\in G\text{.}\) Otherwise, \(G\) is nonabelian . (The word “abelian” derives from the surname of mathematician Niels Henrik Abel.)
Remark
If we know that a binary operation \(\cdot\) on a set \(G\) is commutative, then in checking to see if axioms \(\mathcal{G}_2\) and \(\mathcal{G}_3\) hold we need only verify that there exists \(e\in G\) such that \(ae=a\) (we don't need to check that \(ea=a\)) for all \(a\in G\) and that for each \(a\in G\) there exists \(b\in G\) such that \(ab=e\) (we don't need to check that \(ba=e\)).
Remark
If \(G\) is not known to be abelian, we must be careful when multiplying elements of \(G\) by one another: multiplying on the left is, in general, not the same as multiplying on the right!
Definition: Order, Finite Group, and Infinite Group
If \(G\) is a group, then the cardinality \(|G|\) of \(G\) is called the order of \(G\) . If \(|G|\) is finite, then \(G\) is said to be a finite group ; otherwise, it's an infinite group .
Example \(\PageIndex{1}\)
Of the groups we've discussed, which are abelian? Which are infinite/finite?
We have already seen that identity elements of groups are unique, and that each element \(a\) of a group \(G\) has a unique inverse \(a^{-1}\in G\text{.}\) Here are some other basic properties of groups.
Theorem \(\PageIndex{2}\)
If \(\langle G, \cdot \rangle\) is a group, then left and right cancellation laws hold in \(G\text{.}\) That is, if \(a,b,c\in G\text{,}\) then
- If \(ab=ac\text{,}\) we have \(b=c\) (the left cancellation law); and
- If \(ba=ca\text{,}\) we have \(b=c\) (the right cancellation law).
- Proof
-
Let \(a,b,c∈G\) and assume that \(ab=ac\). Multiplying both equation sides on the left by \(a^{−1}\), we obtain
\(\begin{array}& &2 & &a^{−1}(ab)=a^{−1}(ac) \\ & &\Rightarrow &(a^{−1}a)b=(a^{−1}a)c \\ & &\Rightarrow &eb=ec \\ & &\Rightarrow &b=c \end{array}\)
This proves that the left cancellation law holds. A similar proof shows that the right cancellation law holds.
Theorem \(\PageIndex{3}\)
Let \(\langle G, \cdot \rangle\) be a group and let \(a,b\in G\text{.}\) Then there exist unique elements \(x,y\in G\) such that \(ax=b\) and \(ya=b\text{.}\)
- Proof
-
If \(x=a^{−1}b\) and \(y=ba^{−1}\), then \(ax=a(a^{-1}b)=(aa^{−1})b=eb=b\) and \(ya=(ba^{−1})a=b(a^{−1}a)=be=b\). So such elements \(x\)and \(y\) exist. The fact that they are unique follows from the cancellation laws: if \(ax=b\) and \(ax′=b\) then \(x=x′\) by left cancellation, and if \(ya=b\) and \(y′a=b\) then \(y=y′\) by right cancellation.
Note
We only of necessity have \((ab)^{-1}=a^{-1}b^{-1}\) if \(G\) is known to be abelian!
However, we do have the following:
Theorem \(\PageIndex{4}\)
If \(a\) and \(b\) are elements of a group \(\langle G, \cdot \rangle\text{,}\) then
\begin{equation*} (ab)^{-1}=b^{-1}a^{-1}. \end{equation*}
- Proof
-
We have that
\((ab)(b^{−1}a^{−1})=a(bb^{−1})a^{−1}=aea^{−1}=aa^{−1}=e\)
Similarly, (b−1a−1)(ab)=e.