2.3: The Definition of a Group
At this point you may be asking yourself, why do we care? We've covered a lot of definitions and proved some theorems, but what is the goal of all this? Well, there are actually many goals that we can achieve using such material. Consider the following as an example. Suppose we want to solve the equation \(5+x=2\text{.}\) We can probably solve this quite easily almost just by looking at it (\(x=-3\)), but what facts are we actually using there? If we break down the reasoning leading to this answer, we may obtain something like the following, where each line is equivalent to the one preceding it.
\(\begin{array} &5+x& =2 \\ -5+(5+x)& =-5+2\\ (-5+5)+x & =-3\\ 0+x& =-3\\ x& =-3. \end{array}\)
In the second line we add the inverse of \(5\) in \(\langle \mathbb{Z}, +\rangle\) to each side of the equation; in the third line we use associativity of \(+\) in \(\mathbb{Z}\) (along with computation); and in the fourth line we use the fact that \(-5\) is the additive inverse of \(5\) (that is, the inverse of \(5\) in \(\mathbb{Z}\) under \(+\)). Finally, we use the fact that \(0\) is an additive identity element in \(\mathbb{Z}\) (that is, the identity element in \(\mathbb{Z}\) under \(+\)).
In summary, we used associativity , identity elements , and inverses in \(\mathbb{Z}\) to solve the given equation. This perhaps suggests that these would be useful traits for a binary structure and/or its operation to have. They are in fact so useful that a binary structure displaying these characteristics is given a special name. We note that these axioms are rather strong; “most” binary structures aren't groups.
Definition: Group Axioms
A group is a set \(G\text{,}\) equipped with a binary operation \(*\text{,}\) that satisfies the following three group axioms :
- \(*\) is associative on \(G\text{.}\) (Axiom \(\mathcal{G}_1\))
- There exists an identity element for \(*\) in \(G\text{.}\) (Axiom \(\mathcal{G}_2\))
- Every element \(a\in G\) has an inverse in \(G\text{.}\) (Axiom \(\mathcal{G}_3\))
We denote group \(G\) under \(*\) by the binary structure notation \(\langle G,*\rangle\text{,}\) or simply by \(G\) if the operation \(*\) is known from context (or need not be known in the current situation).
Remark
When proving/disproving that a set \(G\) is/is not a group under an (apparent) operation \(*\text{:}\)
- The first thing you should do is check to make sure that \(\langle G,*\rangle\) is a binary structure: that is, you should make sure that \(a*b\) is well-defined and in \(G\) for every \(a, b \in G\text{.}\) If this is not the case, it doesn't make sense to check to see if axioms \(\mathcal{G}_1\)–\(\mathcal{G}_3\) hold). (For instance: \(\mathbb{Z}^*\) has no chance of being a group under \(\div\text{,}\) since, e.g., \(3,4\in \mathbb{Z}^*\) but \(3 \div 4 \not\in\mathbb{Z}^*\text{.}\))
- You should never check \(\mathcal{G}_3\) before confirming \(\mathcal{G}_2\) holds, because it makes no sense to look for inverses if you haven't confirmed that \(G\) contains an identity element under \(*\text{.}\)