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Writing a finite group out in full, and reading the identity, the inverses, commutativity and cancellation straight off the grid.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to build the Cayley table of a small group, complete a partly filled table using the rule that no element repeats in a row or a column, find the identity and each element's inverse by looking, decide from the table whether a group is abelian, tell the two groups of order four apart by their diagonals, and say which axiom a table cannot confirm.
The last lesson checked the axioms on sets described in words: the integers, the symmetries of a polygon, the invertible matrices. For a finite group there is a better option — write down every product at once. A group with $n$ elements has $n^2$ products, and for small $n$ that is a table that fits on a page.
The Cayley table of a finite group lists the elements along the top and down the side; the cell in row $x$ and column $y$ holds the product $xy$, in that order. A Latin square is a square grid in which each symbol appears exactly once in every row and every column. The Klein four-group is the group of order four in which every element is its own inverse; the other group of order four is cyclic, generated by a single element.
Write the elements across the top and down the left, with the identity first. Fill the cell in row $x$, column $y$ with $xy$. Order matters: the row element is on the left.
Four things are then visible at a glance.
The identity. Its row copies the heading row and its column copies the heading column. Nothing else does.
Inverses. The inverse of $x$ is the element heading the column in which the identity appears in row $x$. It appears exactly once per row, so each element has exactly one inverse.
Commutativity. The group is abelian exactly when the table is symmetric about the main diagonal, because that symmetry says $xy = yx$ for every pair.
Cancellation. No element appears twice in a row, or twice in a column: a repeat would give $xa = xb$ with $a \ne b$, and multiplying by $x^{-1}$ would say $a = b$. So every row and column is a rearrangement of the elements, and the table is a Latin square.
One thing is invisible: associativity. A Latin square can be written down that is no group's table, and the only way to tell is to check $(xy)z = x(yz)$ for every triple. So a table confirms three axioms cheaply and the fourth not at all.
Another way: picture
Think of the table as a filled-in grid you are solving like a puzzle. The identity row and column come free. Each remaining row must contain every symbol once, like a Sudoku row, and so must each column. For a group with three elements those two rules alone finish the grid — there is no choice left anywhere — which is the reason there is exactly one group of order three.
Another way: steps
To build the table of a small group: 1. List the elements with the identity first, along the top and down the side. 2. Fill the identity's row and column by copying the headings. 3. Work out the remaining products from the operation itself. 4. Check each row and each column holds every element exactly once; a repeat means an arithmetic slip. 5. Read off the inverses, and check the diagonal symmetry to decide whether the group is abelian.
Four elements, identity $e$ first. Cancellation forces most of the table, but not all of it: one genuine choice remains, and it produces the only two groups of order four there are.
If some element $a$ has order four, the elements are $e, a, a^2, a^3$ and the group is cyclic. Its diagonal reads $e, a^2, e, a^2$ — only two elements square to the identity.
If no element has order four, every non-identity element must have order two, so $x^2 = e$ for all $x$. That is the Klein four-group, and its diagonal is all identity. It is the symmetry group of a non-square rectangle: do nothing, flip horizontally, flip vertically, rotate a half turn.
Both are abelian, so the diagonal symmetry does not separate them; the diagonal entries do. And an order-four group can be recognised instantly by asking a single question of the table: how many cells on the main diagonal hold the identity? Two means cyclic, four means Klein.
This is the smallest case of a pattern that runs through the whole course: counting elements of each order is often enough to name a group, without ever writing an isomorphism down.
Reading the product in the wrong order. The cell in row $x$, column $y$ is $xy$, not $yx$. In an abelian group it makes no difference, which is exactly why the habit goes unnoticed until it matters.
Thinking a Latin square is a group. It is not. Associativity is invisible in the pattern, and there are Latin squares of order five that no group produces.
Taking symmetry about the diagonal for granted. Most interesting groups are not abelian, so most tables are not symmetric. Assuming symmetry while filling one in will produce a table that is not any group's.
Looking for the inverse in the wrong place. It is the column heading where the identity sits in that element's row, not the row heading.
Believing the table proves the set is a group. It shows closure, an identity and inverses. The fourth axiom still has to be argued, usually by saying the operation is really composition of functions, which is associative for free.
Elements $0, 1, 2, 3$; the cell in row $x$, column $y$ is the remainder of $x + y$ on division by $4$.
The operation.
Row $1$ reads $1, 2, 3, 0$ and row $2$ reads $2, 3, 0, 1$: each is a rearrangement, as it must be.
The Latin square property.
The identity $0$ sits in row $1$ under column $3$, so $3$ inverts $1$. The diagonal reads $0, 2, 0, 2$, so only $2$ has order two and the group is cyclic.
Inverses and order, read off.
Elements $1, 3, 5, 7$ under multiplication modulo $8$. $3 \times 3 = 9 \equiv 1$, $5 \times 5 = 25 \equiv 1$, $7 \times 7 = 49 \equiv 1$.
The diagonal is all $1$.
So every element is its own inverse, and nothing has order four.
No generator.
This is the Klein four-group, not the cyclic one — even though it is written multiplicatively and the cyclic example was written additively. Notation is not structure.
Two groups of order four, and this is the second.
A diagonal cell holds $x^2$, so the question is how many elements satisfy $x^2 = e$ — that is, how many are their own inverse.
Translate the picture into algebra.
Pair each non-identity element with its inverse. A group of order five has four non-identity elements, and if any were self-inverse the remaining three could not pair up evenly.
Count in pairs.
So only the identity is self-inverse: exactly one cell. The same counting argument shows a group of even order always has an element of order two.
Here is $\{1, 3, 9\}$ under multiplication modulo $13$, with identity $1$. Complete its table: the cell in row $x$ and column $y$ holds $xy$.
| 1 | 3 | 9 | |
|---|---|---|---|
| 1 | 1 | 3 | 9 |
| 3 | 3 | ||
| 9 | 9 |
Here is $U(5) = \{1, 2, 3, 4\}$ under multiplication modulo $5$, with identity $1$. Complete the nine products that are missing.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| 1 | 1 | 2 | 3 | 4 |
| 2 | 2 | |||
| 3 | 3 | |||
| 4 | 4 |
Match each thing you can see in a Cayley table to the algebraic fact it records.
| The group is abelian | That row belongs to the identity element | Every element has an inverse | Cancellation holds | The operation is associative | |
|---|---|---|---|---|---|
| A row that copies the heading row | |||||
| The identity appears in every row | |||||
| The table is symmetric about its main diagonal | |||||
| Each element appears exactly once in each row |
Select every statement that is true of the Cayley table of a finite group.
This task has no paper form; do it on a device.
Look at $U(12) = \{1, 5, 7, 11\}$ under multiplication modulo $12$, a group with four elements. Which group is it?
Build the proof that no element appears twice in one row of a group's Cayley table.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Here is $\{1, 4, 7\}$ under multiplication modulo $9$, with identity $1$. Complete its table: the cell in row $x$ and column $y$ holds $xy$.
| 1 | 4 | 7 | |
|---|---|---|---|
| 1 | 1 | 4 | 7 |
| 4 | 4 | ||
| 7 | 7 |
You can complete a Cayley table and read the identity, the inverses and commutativity off it. Say in your own words why no element can appear twice in one row, and what a table cannot tell you. Next: the facts that follow from the axioms alone, proved rather than seen.
9. Your turn: in a group of order five, how many cells of the main diagonal hold the identity?, step 3