First look at the left picture below. Top left, top right, bottom right and bottom left are fixed positions on the table, and the centre stays fixed too. A is gold, $B$ green, $C$ blue and $D$ pink. The letters and their colours belong to the card and move with it; each corner keeps its own letter.

Turn the card a quarter-turn clockwise: A goes from top left to top right, $B$ goes to bottom right, $C$ to bottom left and $D$ to top left. The outline is still the same square, but the four letters are in different positions. A record that only says “still a square” loses the information we need.

We should distinguish two things here. Where the letters are now is an arrangement, or state. “Turn a quarter-turn clockwise” is a move that changes a state. We can perform the same move from different starting states. We will write $R$ for the instruction, not just for the picture with A at the top right.

Here is why we want a theory of these moves: we want to calculate with a long instruction list. For example, which single move has the same effect as $R$ followed by $R$? Re-enacting every long list is awkward. Instead, we can work out how a small collection of moves combine and reuse those rules.

We measure a full turn as 360°, so a quarter-turn is $360^\circ \div 4 = 90^\circ$. Two quarter-turns give $90^\circ + 90^\circ = 180^\circ$; three give $180^\circ + 90^\circ = 270^\circ$; four give $270^\circ + 90^\circ = 360^\circ$. At that point, every corner is back where it started.

Press “Next move” to perform $R$ once. Follow A as it moves, and check the other three corners too. The left picture keeps the starting arrangement while the right one holds the current result. “Play steps” performs four moves and stops. You can pause, reset or select a completed step; it will not advance the lesson.

Write $R^{2}$ for two successive $R$ moves and $R^{3}$ for three. The superscript counts repetitions; we are not squaring an angle. Write $e$ for doing nothing. Turning through a full circle has a different physical history from doing nothing, but it sends every point to exactly the same final position. As rules for final positions, $R^{4}$ and $e$ are the same object.

Why keep just four rotations? The card must finish exactly on its original outline. A can finish at one of four corners. If we only rotate, choosing where A finishes fixes where $B$, $C$ and $D$ finish as well. The moves $e, R, R^{2}$ and $R^{3}$ already use all four possible destinations of A. A 45° turn leaves corners and edges outside the outline, so it is not in our collection.

We use the notation $a \star b$ to mean perform $b$ after $a$. Throughout this page, moves are carried out from left to right. This is not ordinary multiplication of numbers; it is our notation for combining two moves.

The first expression joins one turn to two turns, giving three. The second joins three to two, giving five. The first four restore the starting position and the fifth turns 90° again, so the result is $R$. We obtained this by following the moves, not by memorising a table. Below, set the selectors to $R^{3}$ and $R^{2}$ and check each corner through the starting, intermediate and final pictures.

Rotating does not reverse the order $A, B, C$, $D$ around the boundary. But turning the card over across its vertical middle line exchanges left and right: top left swaps with top right, and bottom left swaps with bottom right. The order $A, B, C$, $D$ that was clockwise is now anticlockwise. No rotation produces that reversal, so we need a new move.

A reflection sends each point to the other side of a line, the same perpendicular distance from that line. Choose the vertical middle line and call this move $F$. Assume the corner letters are readable from both sides of the card. The pictures record corner positions rather than asking you to imagine a three-dimensional flip. The dashed line stays fixed on the table, and the second picture shows the exchanged destinations.

Performing $F$ twice exchanges each pair back, so $F \star F = e$. We can also perform $F$ after a rotation. Turns of 0°, 90°, 180° and 270° give four different reflected results. Choose the rotation below. The middle picture keeps its result; exchanging its left and right positions across the fixed vertical line gives the last picture.

These four results are also reflections across four different lines. Check the destinations: $F$ exchanges left and right. $R \star F$ fixes $A$ and $C$ but exchanges $B$ and $D$, so it reflects across the diagonal from top left to bottom right. $R^{2} \star F$ exchanges top and bottom. $R^{3} \star F$ fixes $B$ and $D$ but exchanges $A$ and $C$, so it reflects across the other diagonal. The dashed lines below show those equivalent reflection lines, with the final arrangements.

Why are there exactly eight moves, with none missing? A has four possible destinations. $B$ starts next to A, and a rigid move preserves the side length, so $B$ must finish at one of the two corners next to A’s destination, not at the opposite corner. Once $A$ and $B$ are placed, $C$ must be the unused neighbour of $B$, and $D$ takes the final corner. There are at most $4 \times 2 = 8$ possibilities. Our four rotations and four reflections are different and already give eight, so they exhaust the possibilities.

We distinguish the moves by their final point positions. A move that preserves the whole outline this way is a symmetry of the square. The four rotations are commonly called $C_{4}$. On this page we call the collection of eight symmetries $D_{4}$: the 4 refers to the four sides, not to the number of moves. Next we will check why these collections deserve the name “group”.

With rotations, turning 90° and then 180° gives 270°, as does the reverse order. That might tempt us to think all moves can be swapped. With reflections, we need to check again. Choose $R$ first and $F$ second below. The three pictures preserve the start and both steps, and the lower pair compares the two final results in opposite orders.

Let’s follow A. The move $R$ sends it from top left to top right, and the following $F$ sends it back to top left. In reverse order, $F$ sends A to top right, but the following $R$ carries it to bottom right. A ends at different corners, which proves that the combined moves differ. So these moves do not satisfy the commutative law: we cannot claim $a \star b = b \star a$ for every pair. This does not mean every pair differs; $e$, for example, can be swapped with any move.

Now we can organise all combinations in a table. For row $R$ and column $F$, we already followed A. $B$ goes from top right to bottom right, then to bottom left. $C$ goes from bottom right to bottom left, then to bottom right. $D$ goes from bottom left to top left, then to top right. Reading the final letters clockwise from top left gives $A, D, C$, $B$, exactly the picture for $R \star F$. Every cell is worked out this way: feed each destination of the first move into the second move.

Counting how many rotations and reflections occur is not enough. $R$ then $F$ and $F$ then $R$ each contain one rotation and one reflection, yet their results differ. We need to record the order and how the moves combine. The table keeps the information that a count alone loses.

The table gives results, but we also want to know why we can keep using it. The next four properties describe what makes combining and undoing these moves work.

1. Combining two moves stays inside the collection

Each move preserves distances and returns the square to its outline. After the first move, distances are unchanged; after the second, they are still unchanged and the outline is still the original one. The combined move is therefore another symmetry of the square. We proved that our eight moves exhaust those symmetries, so the result must be one of them. When combining any two objects in a set always gives another object in that set, we call this closure.

2. We can change the brackets, without changing the order

Suppose the execution order is $a$, $b$, $c$. We could look up the combined effect of $a$ and $b$ and put $c$ after it. Or we could look up the combined effect of $b$ and $c$ and put that after $a$. A few trials do not prove that these groupings always agree. We can check an arbitrary point $P$ instead.

Write $a(P)$ for the position of $P$ after $a$. Then $b(a(P))$ means apply $b$ to that new position. Through all three moves, the route is $P$ → $a(P)$ → $b(a(P))$ → $c(b(a(P)))$. Whether we group the first two moves or the last two, every point finishes at $c(b(a(P)))$. At no stage did we exchange the order of $a$, $b$ and $c$.

This is associativity. Brackets decide which part we calculate first, not which move happens first. It lets us combine adjacent pieces of a long list. It does not contradict the fact that $R$ and $F$ cannot be swapped.

3. We have a move that changes nothing

Doing nothing and then performing $a$ has the same effect as $a$ alone, as does performing $a$ and then doing nothing. Thus $e \star a = a$ and $a \star e = a$ for every $a$. We call $e$ the identity. It is not the number 1; it plays the role of leaving the result unchanged under this operation. The starting arrangement is a state, whereas $e$ is a move that leaves any starting state unchanged.

4. Each move can be undone

After $R$, performing $R^{3}$ gives four quarter-turns and restores the starting position; $R^{3}$ followed by $R$ does the same. We call $R^{3}$ the inverse of $R$. Performing $R^{2}$ twice gives $180^\circ + 180^\circ = 360^\circ$, so $R^{2}$ is its own inverse. The inverse of $e$ is $e$. Reflecting across the same line twice sends each point back from the opposite side, so each of our four reflections is its own inverse.

Write $a^{-1}$ for the move that undoes $a$; it does not mean dividing a number by a move. It satisfies both $a \star a^{-1} = e$ and $a^{-1} \star a = e$. Below is the calculation for the reflection selected earlier; you can also check it in the table.

Now give this structure a name

A set is a collection of objects; it may be finite or infinite. An operation takes two objects and gives a result by a definite rule. Put a set $G$ together with an operation $\star$. If they satisfy the following four conditions, we call $(G, \star )$ a group. These conditions define the word “group”; we are not claiming that all operations satisfy them.

The square itself is not the group. The group is the eight moves together with their composition operation. The four rotations alone also form a group: combining rotations stays among those four, the same point-by-point proof gives associativity, $e$ belongs to the collection, and their inverses are $e, R^{3}, R^{2}$ and $R$, all still in it. Commutativity is not on the list. A group may allow all pairs to commute, as the rotations do, or may not, as with all eight moves.

Here is a specific list: $R, F, R, F, R$, $R$ — turn, reflect, turn, reflect, turn, turn. We want a shorter list that leaves every corner in the same final position. We also want an instruction that restores the starting arrangement. Use the buttons to step through the list and compare each position with the calculation underneath.

The first pair is $R \star F$, and the third and fourth moves form the same pair. We saw that $R \star F$ is a reflection across a diagonal. Repeating that reflection twice restores the starting arrangement, so the first four moves combine to $e$. The two remaining $R$ moves combine to $R^{2}$. The whole six-move list therefore has the same effect as one 180° clockwise turn.

Each step has a reason: associativity lets us group adjacent moves, the reflection is its own inverse, and the identity leaves the remaining $R^{2}$ unchanged. We did not exchange $R$ and $F$ in order to cancel them. To restore the original, perform $R^{2}$ again, since $R^{2} \star R^{2} = e$. The six moves leave $C, D, A$, $B$ clockwise from top left; after the extra half-turn we read $A, B, C$, $D$ again.

Even without finding a shortcut, we can undo the list by taking the inverse of each move in reverse order. The inverse of the final $R$ is $R^{3}$; the preceding $R$ also has inverse $R^{3}$. The remaining inverse moves are $F, R^{3}$, $F$ and $R^{3}$. Why reverse the order? Undoing the last move brings us back to the state before it happened. Repeating this for the earlier moves takes us back to the start.

This is the use we have completed here: replacing six instructions with one, proving that two different lists have the same effect, and finding how to undo them. The group conditions collect the reasons that make this possible. Other objects satisfying the same four conditions can be combined and undone using the same reasoning. The rules do not solve every problem for us; they tell us why these particular manipulations are justified.

The definition did not require rotations or a finite collection. Let’s check familiar objects: the integers …, $-2, -1, 0, 1, 2$, …, usually written $\mathbb Z$, with ordinary addition. We can read $+3$ as moving three places right on a number line and $-2$ as moving two places left. Completing both moves leaves us one place to the right of the start: $3 + (-2) = 1$.

Closure follows because adding whole-number steps still ends at an integer position. For associativity, consider the same list of unit steps left or right. Total the first two sections first or total the last two first: no step is added, removed or altered, so the total displacement is the same. This gives $(a + b) + c = a + (b + c)$. We are using the rules of integer addition, not assuming this for every new operation.

Adding 0 means taking no steps, so $0 + a = a + 0 = a$: the identity is 0. After the steps for $a$, take the opposite steps for $-a$. Each step is paired with an opposite one, giving $a + (-a) = (-a) + a = 0$. The inverse of $a$ is $-a$, still an integer. Thus $(\mathbb Z, +)$ is a group. It has infinitely many objects, unlike our eight square moves, but satisfies the same four conditions.

In this comparison, 0 plays the role of $e$, adding an integer plays the role of appending a move, and adding its negative undoes it. We use the same word because the combination rules share the four properties, not because an integer is literally a square.

Not every collection or operation works

Keep only 0, 1, 2, 3, … and still use addition. We retain the identity 0, associativity and closure. But undoing $+3$ requires a $b$ with $3 + b = 0$. Subtracting 3 from both sides gives $b = -3$, which we excluded. Missing even one required inverse breaks the definition, so nonnegative integers under addition are not a group.

Now restore all integers but change the operation to subtraction. We get $(5 - 3) - 1 = 2 - 1 = 1$, whereas $5 - (3 - 1) = 5 - 2 = 3$. The groupings differ, so associativity fails and integers under subtraction are not a group. We must specify both the set and the operation; saying only “integers form a group” leaves out important information.

Use the diagrams and table above if needed. Pause to write your reasoning, then open the answer. The answers include how we obtain the result, not only the move name.

We began by wanting to place a square correctly, shorten a list and undo it. Following those concrete questions led us to composition, closure, associativity, identity and inverses. Only then did we give a name to a set and operation satisfying all four conditions: a group. We have finished a complete small problem here; we do not need another group theory lesson to justify what we have done.