In our Modern Algebra class, the teacher mentioned that, up to isomorphism, there are only two distinct groups of order four, and similarly, only two groups of order six. It was also mentioned that this is a common topic for graduate entrance examinations in algebra. Hearing this, I investigated the matter with great interest. I found that the fact that there are only two non-isomorphic groups of order four is almost trivial—it felt a bit too simple for a graduate exam. I then analyzed the case for order six, which proved to be more complex due to the increased number of elements. Today, during a Real Analysis class, I thought of a simplification technique and successfully proved that there are only two non-isomorphic groups of order six. I am posting the results and the research process here to share with everyone.
Two Groups of Order Four
Whether we are considering groups of order four or order six, they are all finite groups. A characteristic of finite groups is that their multiplication tables can be explicitly written out (provided one is willing to take the trouble). To determine the number of groups of order four, we simply need to list the possible multiplication tables. Let the group of order four be G_4 = \{e, a, b, c\}, where e is the identity element. Based on this, we can at least write part of the multiplication table:
\begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & & & \\ b & b & & & \\ c & c & & & \end{array}
Next, consider the result of a^2. It can only be e or b (if it were c, it would be equivalent to being b by renaming). Let us consider these cases separately.
If a^2 = e, we fill in e at the corresponding position. At the same time, we must note: In a multiplication table, each element must appear exactly once in every row and every column (the Latin Square property of finite groups). This makes the construction of the multiplication table similar to solving a Sudoku puzzle! Therefore, after filling in a^2 = e, we can deduce the table as follows:
\begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & & \\ c & c & b & & \end{array}
The remaining four empty cells must be either \begin{array}{cc} a & e \\ e & a \end{array} or \begin{array}{cc} e & a \\ a & e \end{array}. It is easy to prove that both cases constitute valid group multiplication tables. Thus, we have successfully constructed two "multiplication tables" for groups of order four:
\begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \end{array} \quad \text{and} \quad \begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & a & e \\ c & c & b & e & a \end{array}
These two groups are the Klein four-group and the cyclic group of order four, respectively. The other case is a^2 = b. In this case, the table is completed automatically (once b is filled, the rest are determined):
\begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & b & & \\ b & b & & & \\ c & c & & & \end{array} \quad \to \quad \begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & b & c & e \\ b & b & c & & \\ c & c & e & & \end{array} \quad \to \quad \begin{array}{c|cccc} \cdot & e & a & b & c \\ \hline e & e & a & b & c \\ a & a & b & c & e \\ b & b & c & e & a \\ c & c & e & a & b \end{array}
This is simply another way of writing the cyclic group of order four. Therefore, there are only these two cases.
Two Groups of Order Six
When we reach order six, our "Sudoku" method becomes less efficient because the number of elements increases while the initial information remains relatively sparse. As Sudoku players know, puzzles with less initial information are often more difficult. We can use some properties of groups to simplify our reasoning.
Since it is a finite group, every element in a group of order six has a finite order. According to Lagrange’s Theorem, the order of an element must be a divisor of the group’s order (6). Now, let a be an element of maximal order in the group. Then |a| can be 6, 3, or 2. If |a| = 6, the group is the cyclic group of order six. This is the first case, and it is fully considered. (Indeed, changing the perspective makes it much faster!)
In the second case, |a| = 3, meaning a^3 = e. We then know three elements of the group: e, a, a^2. Let the fourth element be c. According to our assumption, the order of c can only be 3 or 2 (it cannot be greater than 3). However, |c| = 3 is impossible. If |c| = 3, the group would contain two more elements: c, c^2, giving us five known elements. What about ac and ca? Neither of them can be any of e, a, a^2, c, c^2. Thus, they would be new elements. However, the group is of order 6, so we must have ac = ca. This would imply (ac)^3 = a^3 c^3 = e. But what about (ac)^2? It clearly cannot be any of e, a, a^2, c, c^2, ac, which would mean the order of the group is greater than 6, a contradiction.
Consequently, we must have |c| = 2. This adds the element c to the group. Now consider ac and ca. Neither is among e, a, a^2, c, so they are new elements. Furthermore, ac \neq ca, because if ac = ca, then (ac)^6 = a^6 c^6 = e, and since (ac)^3 \neq e and (ac)^2 \neq e, the order of ac would be 6. This contradicts the assumption that the maximal order is 3. Thus, ac \neq ca, and all six elements of the group are identified: e, a, a^2, c, ac, ca. We can then list the multiplication table:
\begin{array}{c|cccccc} \cdot & e & a & a^2 & c & ac & ca \\ \hline e & e & a & a^2 & c & ac & ca \\ a & a & a^2 & e & ac & ca & c \\ a^2 & a^2 & e & a & ca & c & ac \\ c & c & ca & ac & e & a & a^2 \\ ac & ac & c & ca & a^2 & e & a \\ ca & ca & ac & c & a & a^2 & e \end{array}
The final case is |a| = 2. Is this case possible? If |a| = 2, we only know two elements e, a. Consider a third element b; its order must also be 2. Now we have e, a, b. Consider ab and ba. If they are not equal, we would have five elements. But what about the sixth element? If we add a new element c, it would generate further elements like ac, leading to a group order greater than 6. Even if ab = ba, we would have four elements \{e, a, b, ab\}. Adding a new element c would result in ac and abc being distinct new elements, again making the group order greater than 6. Both scenarios lead to a contradiction.
Thus, there are only two non-isomorphic groups of order six.
Order n?
The analysis for order six is already quite lengthy. If a reader wishes to analyze the number of groups of order eight, it would require significant courage and perseverance (the answer is 5). Interestingly, analyzing groups of order nine or ten is actually simpler; the answer is 2 for both. The complexity depends largely on the number of factors of the order n. Some results can be found here:
https://en.wikipedia.org/wiki/List_of_small_groups
http://www.douban.com/note/245316892/
Although the analysis above is somewhat long and perhaps a bit clumsy, the truth is that even with other methods, the results obtained are not much more extensive. Determining the number of non-isomorphic groups of order n and their specific constructions is a very difficult problem. To this day, we do not have a general formula to calculate the number of non-isomorphic groups for any given n. Data from baike.com indicates that even considering how many non-isomorphic groups of order p^k exist is a tough problem; it has only been solved for cases where p is an odd prime and k \le 6, or when p=2 and k \le 7.
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