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\(\def\Z{\mathbb{Z}} \def\zn{\mathbb{Z}_n} \def\znc{\mathbb{Z}_n^\times} \def\R{\mathbb{R}} \def\Q{\mathbb{Q}} \def\C{\mathbb{C}} \def\N{\mathbb{N}} \def\M{\mathbb{M}} \def\G{\mathcal{G}} \def\0{\mathbf 0} \def\Gdot{\langle G, \cdot\,\rangle} \def\phibar{\overline{\phi}} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\Ker}{Ker} \def\siml{\sim_L} \def\simr{\sim_R} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

Section3.1Groups of small order

Let's start exploring groups in order of increasing, um, order. Before we do this, it will be helpful to introduce the notion of a group table (also known as a Cayley table 1 ) for a finite group. Given a finite group \(G\text{,}\) list its elements in some fixed order, say, \(a_1, a_2, \ldots, a_n\text{,}\) and then construct its group table by creating an array with exactly one row and exactly one column corresponding to each group element. We then put in the row \(i\) and column \(j\) the element \(a_ia_j\) of \(G\text{.}\) Note that a single group can have group tables that look different from one another, since reordering a group's elements will change its table.


Consider the group \(\Z_4\) under addition modulo 4. Ordering the elements of \(\Z_4\) as \(1,2,3,4\text{,}\) we have the following group table for \(\langle \Z_4,+\rangle\text{:}\)

\(+\) \(0\) \(1\) \(2\) \(3\)
\(0\) \(0\) \(1\) \(2\) \(3\)
\(1\) \(1\) \(2\) \(3\) \(0\)
\(2\) \(2\) \(3\) \(0\) \(1\)
\(3\) \(3\) \(0\) \(1\) \(2\)

Now, clearly, there is no group of order 0 (do you see why?). Is there a group of order 1? Well, suppose \(\langle G,*\rangle\) is such a group. Since \(G\) must contain an identity element \(e\text{,}\) we must have \(G=\{e\}\text{,}\) and since \(e\) is \(G\)'s identity element, we must have \(e*e=e\text{.}\) Clearly, in this case, the three group axioms hold. So \(G\) is a valid group, without much going on in it.


If \(G\) is a group with \(|G|=1\text{,}\) then \(G\) is called the trivial group.


A good question to ask here is why it's called “the” trivial group, rather than “a” trivial group. Indeed, there are infinitely many groups of order 1! (Do you see why?) But it turns out that all of these groups are structurally the same. Hence mathematicians end up thinking of them as various instantiations of one group, rather than separate groups. We will discuss this in more depth shortly, when we introduce the idea of isomorphism.

Next suppose that group \(\Gdot\) has order 2. Then \(G\) must contain an identity element, \(e\text{,}\) and a non-identity element, \(a\text{.}\) Since \(e\) is its own inverse, and inverses are unique, \(a\) must be its own inverse as well. So \(G\) must have the following table.

\(*\) \(e\) \(a\)
\(e\) \(e\) \(a\)
\(a\) \(a\) \(e\)

It is straightforward to show that such a structure does satisfy all the group axioms (the only one we really need to check is associativity).

Now, what if group \(\Gdot\) has order 3? Note that you can't have any entry appear more than once in the same row or same column (excluding of course the labels outside the grid we're filling in), given Theorem 2.5.17. Is there only one way of filling in the table for a group of order 3? (Hint: consider what element must be the second row, third column entry.)

\(*\) \(e\) \(a\) \(b\)
\(e\) \(\) \(\) \(\)
\(a\) \(\) \(\) \(\)
\(b\) \(\) \(\) \(\)

Finally, what if group \(\Gdot\) has order 4? It turns out in this case there are two valid ways of filling in a group table!

What we have been doing here is really getting into the idea of the structure of groups, and when we can consider groups to be essentially “the same” or fundamentally “different.” We approach this more formally via the concepts of homomorphism and isomorphism.