Groups: the mathematics of symmetry
Before this sheet: Matrices that move the plane and Fermat's little theorem.
Tonight's idea
Label the corners of an equilateral triangle A, B, C. There are exactly six ways to put it back in its outline: do nothing (\(e\)); rotate by 120° (\(r\)) or 240° (\(r^2\)); or flip it about the axis through corner A (\(a\)), B (\(b\)) or C (\(c\)). These are the symmetries of the triangle.
Flip about A, then rotate by 120°: track the corners in the figure and the result is the single flip \(c\). We write \(r \circ a = c\), reading right to left. The other order gives \(a \circ r = b\). Order matters.
In the table, the entry in row \(g\), column \(h\) is \(g \circ h\): do \(h\) first, then \(g\). So row \(r\), column \(a\) gives \(c\).
| ∘ | e | r | r² | a | b | c |
|---|---|---|---|---|---|---|
| e | e | r | r² | a | b | c |
| r | r | r² | e | c | a | b |
| r² | r² | e | r | b | c | a |
| a | a | b | c | e | r | r² |
| b | b | c | a | r² | e | r |
| c | c | a | b | r | r² | e |
Four things to notice. (1) Combining two symmetries gives a symmetry: no gaps. (2) There is a do-nothing element \(e\). (3) Every element can be undone: every row contains \(e\). (4) \((g \circ h) \circ k = g \circ (h \circ k)\). Any set with an operation obeying these four rules is a group. This one is \(D_3\), of order 6. The numbers \(\{0, \ldots, 5\}\) under addition mod 6 are another group, \(\mathbb{Z}_6\). A theorem proved once applies to all of them.
Lagrange's theorem. If a group has \(n\) elements and a subgroup (a subset that is itself a group) has \(m\) elements, then \(m\) divides \(n\).
Watch one: why Lagrange is true
Take the subgroup \(H = \{e, a\}\) inside \(D_3\). For each element \(g\), form the coset \(gH = \{g \circ e,\ g \circ a\}\) from the table:
\(eH = aH = \{e, a\}\), \(rH = cH = \{r, c\}\), \(r^2H = bH = \{r^2, b\}\).
Only three different cosets appear. They do not overlap, each has exactly 2 elements, and together they cover all 6. So \(6 = 3 \times 2\). Cosets always cut a group into pieces the size of the subgroup: that is the proof. One consequence: the order of an element \(g\), the smallest \(k\) with \(g^k = e\), is the size of the subgroup of its powers, so it divides the order of the group. In \(D_3\) the orders are 1, 3, 3, 2, 2, 2.
Your turn
- Use the table to compute \(r \circ b\), \(b \circ r\), and \((r \circ a)^2\). Find the inverse of every element.
- The symmetries of a non-square rectangle: list them and write the 4 × 4 Cayley table. Could you relabel its elements to make its table the table of \(\mathbb{Z}_4\) (addition mod 4)? Compare the orders of the elements before you decide.
- In \(\mathbb{Z}_{12}\) (addition mod 12), find the order of 8, of 9 and of 5. Then list every subgroup of \(\mathbb{Z}_{12}\), using Lagrange to limit the search.
- The numbers \(1, 2, \ldots, p - 1\), multiplied mod a prime \(p\), form a group of order \(p - 1\). Use "the order of an element divides the order of the group" to prove Fermat's little theorem, \(a^{p-1} \equiv 1 \pmod p\), in three lines. (Compare the Fermat sheet.)
Stretch: the rotations of a cube
The top face of a cube can land in 6 positions and then be spun 4 ways, so the cube has \(6 \times 4 = 24\) rotations. The rotations that keep one particular corner fixed form a subgroup. How many are there? How many cosets does Lagrange predict, and what do they mean geometrically?
Answers
- \(r \circ b = a\), \(b \circ r = c\) (row \(b\), column \(r\)). \((r \circ a)^2 = c \circ c = e\). Inverses: \(e^{-1} = e\), \(r^{-1} = r^2\), \((r^2)^{-1} = r\), and \(a, b, c\) are each their own inverse.
- Four symmetries: \(e\), the half-turn \(h\), the horizontal flip \(x\), the vertical flip \(y\). Every element squared is \(e\), and the product of any two different non-identity elements is the third. It is not \(\mathbb{Z}_4\): there the element 1 has order 4, while here every element has order 1 or 2. This is the Klein four-group.
- Order of 8: \(8, 4, 0\), so 3. Order of 9: \(9, 6, 3, 0\), so 4. Order of 5: 12. Subgroup sizes divide 12, and there is one of each size: \(\{0\}\), \(\{0, 6\}\), \(\{0, 4, 8\}\), \(\{0, 3, 6, 9\}\), \(\{0, 2, 4, 6, 8, 10\}\), \(\mathbb{Z}_{12}\). Six subgroups.
- Let \(k\) be the order of \(a\). By Lagrange, \(k\) divides \(p - 1\), say \(p - 1 = km\). Then \(a^{p-1} = (a^k)^m = 1^m = 1\). Fermat's theorem is Lagrange's theorem applied to one group.
Stretch. The rotations fixing a corner are the identity and the rotations by 120° and 240° about the long diagonal through it: a subgroup of order 3. Lagrange gives \(24/3 = 8\) cosets, one for each of the 8 corners the chosen corner can be sent to. Each coset is "all the rotations that send our corner there".