No explanations, only problems. Work through them in order and you'll go from the axioms to the reason there's no formula for the quintic, having proved every step yourself.
A group is a set $G$ with a way of combining two elements into a third, $ab$, such that the combining is associative, there's an identity $e$ with $ea = ae = a$, and every $a$ has an inverse $a^{-1}$ with $aa^{-1} = a^{-1}a = e$. That's the only definition you're given. The rest you build.
Every term is defined in a shaded box before the problems that use it. Each problem has hints, and a solution sketch for when you're properly stuck. The scratchpad on the right is for experimenting. Plenty of these problems go faster once you've played with an example. Mark a problem solved to find out what you proved.
New to proofs? The start-here page has a quick check and a free book to warm up with.
Moves
What the axioms already force.
A square has eight symmetries: four rotations and four flips. Label its corners 1 to 4. Write each symmetry as a permutation of the corners, check that doing one symmetry after another always gives one of the eight, and find two symmetries $a, b$ with $ab \neq ba$.
try r and s in the scratchpad
Let $r$ be a quarter turn and $s$ a flip. Every symmetry is $r^k$ or $r^ks$. Now compare $rs$ with $sr$.
With $r = (1\,2\,3\,4)$ and $s = (2\,4)$, the eight symmetries are $e, r, r^2, r^3, s, rs, r^2s, r^3s$. A symmetry is fixed by where corner 1 goes (4 choices) and whether corner 2 follows it clockwise or anticlockwise (2 choices), which is why there are exactly 8. Doing two symmetries in a row still moves the square onto itself, so it's one of the 8. Reading right to left, $rs = (1\,2)(3\,4)$ but $sr = (1\,4)(2\,3)$.
You've met your first non-abelian group, the dihedral group $D_4$. The order you do things in matters, and that's the whole reason group theory is interesting.
Show that a group has only one identity element, and that each element has only one inverse.
If $e$ and $e'$ are both identities, what is $ee'$?
If $b$ and $c$ are both inverses of $a$, look at $b(ac)$ and $(ba)c$.
If $e$ and $e'$ are both identities, then $ee' = e'$ (because $e$ is an identity) and $ee' = e$ (because $e'$ is), so $e = e'$. If $b$ and $c$ are both inverses of $a$, then $b = be = b(ac) = (ba)c = ec = c$.
Uniqueness of identity and inverses. It's why we're allowed to write $e$ and $a^{-1}$ as if there's only one of each.
Show that $(ab)^{-1} = b^{-1}a^{-1}$, and that $ab = ac$ implies $b = c$. Deduce that in the multiplication table of a finite group, every row and every column contains each element exactly once.
Multiply $ab$ by $b^{-1}a^{-1}$ and regroup. For cancellation, multiply both sides on the left by $a^{-1}$.
$(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = e$, and the same in the other order, so by problem 2 it's the inverse. If $ab = ac$, multiply on the left by $a^{-1}$ to get $b = c$. Row $a$ of the table lists $ag$ for every $g$. By cancellation no two entries are equal, so all $|G|$ elements appear once each. Columns work the same way, cancelling on the right.
This is cancellation, and its corollary says every group table is a Latin square, which is why the field guide's plates look the way they do. The first fact is the second most viewed group theory question on Math Stack Exchange.
Suppose $g^2 = e$ for every $g$ in $G$. Show that $G$ is abelian: $ab = ba$ for all $a, b$.
If $g^2 = e$ then $g = g^{-1}$. Apply that to $a$, $b$ and $ab$.
Your first abelian-forcing condition. Proving that a group has to be abelian is the single most common kind of group theory question people ask about online.
Order
How long until an element comes back to $e$?
Let $G$ be finite and $g \in G$. Show there's some $n \ge 1$ with $g^n = e$. The smallest such $n$ is the order of $g$. Show that $g^m = e$ exactly when the order of $g$ divides $m$.
The powers $g, g^2, g^3, \dots$ can't all be different. If $g^i = g^j$ with $i < j$, cancel.
For the second part, divide $m$ by the order $n$: $m = qn + r$ with $0 \le r < n$.
The powers $g, g^2, \dots, g^{|G|+1}$ can't all be different, so $g^i = g^j$ for some $i < j$. Cancel $g^i$ to get $g^{j-i} = e$. For the second part, write $m = qn + r$ with $0 \le r < n$. Then $g^m = (g^n)^qg^r = g^r$, which is $e$ exactly when $r = 0$, since $n$ is the smallest positive power giving $e$.
The order of an element. It's the first thing to compute about any element, and the first fingerprint the field guide uses.
Suppose $g$ has order $n$. Show that $g^k$ has order $n / \gcd(n, k)$. Deduce that the cyclic group $\{e, g, \dots, g^{n-1}\}$ has exactly $\varphi(n)$ generators, where $\varphi(n)$ counts the numbers from $1$ to $n$ that share no factor with $n$.
$(g^k)^m = e$ exactly when $n$ divides $km$, by problem 5. What's the smallest such $m$?
Let $d = \gcd(n, k)$. By problem 5, $(g^k)^m = e$ exactly when $n$ divides $km$, which happens exactly when $n/d$ divides $(k/d)m$. Since $n/d$ and $k/d$ share no factor, that's when $n/d$ divides $m$. So the order is $n/d$. $g^k$ generates the whole group exactly when its order is $n$, meaning $d = 1$, and there are $\varphi(n)$ such $k$ between $1$ and $n$.
This answers the most viewed group theory question on Math Stack Exchange: how to find a generator of a cyclic group. $g^k$ generates exactly when $\gcd(n, k) = 1$.
Show that every subgroup of a cyclic group is cyclic.
If $H \ne \{e\}$ is a subgroup of $\langle g\rangle$, let $g^d$ be the element of $H$ with the smallest positive exponent. Show every element of $H$ is a power of $g^d$ by dividing exponents by $d$.
Let $H \neq \{e\}$ be a subgroup of $\langle g\rangle$. It contains some $g^k$ with $k > 0$ (if it contains $g^{-k}$, it contains the inverse). Let $d$ be the smallest positive exponent with $g^d \in H$. For any $g^m \in H$, write $m = qd + r$ with $0 \le r < d$. Then $g^r = g^m(g^d)^{-q} \in H$, so $r = 0$ by the choice of $d$. So $H = \langle g^d\rangle$.
Subgroups of cyclic groups are cyclic, and in fact there's exactly one of each order dividing $n$.
Show that a finite group with an even number of elements has an element of order 2.
Pair each element with its inverse. Which elements are paired with themselves?
Pair each element with its inverse. Elements with $g \neq g^{-1}$ come in pairs of two. What's left are the elements with $g = g^{-1}$, that is $g^2 = e$. Since $|G|$ is even, the number left over is even. $e$ is one of them, so there's at least one more, and it has order 2.
This is the $p = 2$ case of Cauchy's theorem. Problem 28 proves it for every prime, with a much better trick.
Cosets
A subgroup chops the group into equal pieces.
Let $H$ be a subgroup of $G$. For $g \in G$, the left coset $gH$ is $\{gh : h \in H\}$. Show that two left cosets are either equal or don't overlap at all, and that every left coset has exactly $|H|$ elements.
If $x \in aH \cap bH$, then $x = ah = bh'$. Show $aH = bH$.
For the size: $h \mapsto gh$ is one-to-one by cancellation.
If $x \in aH \cap bH$, then $x = ah = bh'$, so $a = bh'h^{-1}$. Every $ah''$ is then $b(h'h^{-1}h'') \in bH$, so $aH \subseteq bH$, and the same argument the other way round gives equality. $h \mapsto gh$ is onto $gH$ by definition and one-to-one by cancellation, so $|gH| = |H|$. Every $g$ lies in its own coset $gH$, so the cosets cover $G$.
Cosets partition the group. This picture is the one to keep in your head for the rest of the subject.
Show that if $G$ is finite, the size of every subgroup divides $|G|$. Deduce that the order of every element divides $|G|$, that $g^{|G|} = e$, and that a group with a prime number of elements is cyclic.
Problem 9 cuts $G$ into pieces of size $|H|$. For elements, apply the first part to $\{e, g, g^2, \dots\}$.
The cosets cut $G$ into $[G : H]$ pieces of size $|H|$, so $|G| = [G : H]\,|H|$. The subgroup $\{e, g, \dots, g^{n-1}\}$ has exactly $n = $ the order of $g$ elements, so the order of $g$ divides $|G|$, and then $g^{|G|} = e$ by problem 5. If $|G| = p$ is prime, any $g \neq e$ has order dividing $p$ and bigger than 1. So its order is $p$, and $g$ generates everything.
Lagrange's theorem. The converse is false: problem 19 finds a group of order 12 with no subgroup of order 6.
Let $p$ be prime. Show that the numbers $1, 2, \dots, p - 1$ form a group under multiplication mod $p$. Deduce that $a^{p-1} \equiv 1 \pmod p$ whenever $p$ doesn't divide $a$.
For inverses: $x \mapsto ax \bmod p$ is one-to-one on $\{1, \dots, p-1\}$, so something maps to $1$. Then use problem 10.
If $p$ divides neither $a$ nor $b$, it doesn't divide $ab$, so the set is closed. Associativity and the identity $1$ are inherited from ordinary multiplication. For inverses: if $ax \equiv ay$, then $p$ divides $a(x - y)$, so $x = y$. So $x \mapsto ax$ is one-to-one on a finite set, hence onto, and some $x$ gives $ax \equiv 1$. The group has $p - 1$ elements, so problem 10 gives $a^{p-1} \equiv 1$.
This is Fermat's little theorem, which falls out of Lagrange in a line. The same argument with the units mod $n$ gives Euler's theorem, the engine of RSA encryption.
Permutations
Every finite group lives inside one of these.
Show that every permutation of $\{1, \dots, n\}$ can be written as a product of disjoint cycles, and that its order is the least common multiple of the cycle lengths.
try (1 2 3)(4 5)
Follow $1 \to \sigma(1) \to \sigma(\sigma(1)) \to \cdots$ until you get back to $1$. That's one cycle. Start again from any point you haven't visited. Disjoint cycles commute.
Follow $1 \to \sigma(1) \to \sigma^2(1) \to \cdots$. It must repeat, and the first repeat is back at $1$, since $\sigma$ is one-to-one. That's one cycle. Start again from a point not yet visited, and continue until every point is used. The cycles are disjoint, and disjoint cycles commute. So $\sigma^m$ is the product of the $m$th powers of the cycles, which is the identity exactly when every cycle length divides $m$. The smallest such $m$ is the lcm.
Cycle decomposition. The element $(1\,2\,3)(4\,5)$ has order 6, the most any element of $S_5$ can have.
Every permutation is a product of transpositions (2-cycles). Show that multiplying a permutation by one transposition changes its number of cycles, counting fixed points as cycles of length 1, by exactly one. Deduce that a permutation can't be written both as a product of an even number of transpositions and as a product of an odd number.
Multiply by $(a\,b)$. If $a$ and $b$ are in the same cycle it splits in two. If they're in different cycles, the two merge.
The identity has $n$ cycles. After $k$ transpositions, the number of cycles has the same parity as $n - k$.
Check small cases: $(1\,2)(1\,2\,3) = (2\,3)$ splits one cycle on $\{1, 2, 3\}$ into two, the fixed point 1 and $(2\,3)$. $(1\,2)(1)(2\,3) = (1\,2\,3)$ merges two into one. In general, multiplying by $(a\,b)$ splits the cycle when $a, b$ share one, and merges the two cycles otherwise. So each transposition changes the number of cycles by exactly 1. The identity has $n$ cycles, so after $k$ transpositions the count has the same parity as $n - k$. A permutation written with $k$ and with $k'$ transpositions therefore has $n - k \equiv n - k' \pmod 2$.
The sign of a permutation is well defined. The even permutations form the alternating group $A_n$.
In the 15-puzzle, you slide numbered tiles into an empty square on a $4 \times 4$ board. Show that you can't start from the solved position and end with the 14 and 15 swapped and everything else in place, including the empty square.
Treat the empty square as a sixteenth tile. Each move is a transposition of 16 things.
Colour the board like a chessboard. Every move changes the colour the empty square is on. It ends where it started, so the number of moves is even.
Treat the gap as tile 16. Each move swaps the gap with a neighbouring tile, a transposition. Colour the board like a chessboard: each move changes the colour under the gap, and the gap ends where it started, so the number of moves is even. So any reachable arrangement with the gap back in its corner is an even permutation. Swapping 14 and 15 alone is a single transposition, which is odd.
Sam Loyd offered $1,000 for the solution in the 1880s, knowing it was impossible. Parity is the whole proof.
Let $\sigma$ be a permutation. Show that $\sigma(a_1\,a_2\,\dots\,a_k)\sigma^{-1} = (\sigma(a_1)\,\sigma(a_2)\,\dots\,\sigma(a_k))$. Deduce that two permutations are conjugate, meaning $\tau = \sigma\pi\sigma^{-1}$ for some $\sigma$, exactly when they have the same cycle lengths.
conjugate (1 2 3) by (1 4) in the scratchpad
Check where $\sigma(a_1)$ goes under $\sigma(a_1\dots a_k)\sigma^{-1}$. For the converse, line the two cycle decompositions up and let $\sigma$ send one to the other.
$\sigma c\sigma^{-1}$ sends $\sigma(a_i)$ to $\sigma(c(a_i)) = \sigma(a_{i+1})$. Any other point $x$ has $\sigma^{-1}(x)$ outside the cycle, so it's fixed. That makes $\sigma c\sigma^{-1} = (\sigma(a_1)\,\dots\,\sigma(a_k))$. Conjugating a product of disjoint cycles conjugates each one, so cycle lengths are preserved. Conversely, write $\pi$ and $\tau$ as disjoint cycles, fixed points included, with lengths lined up, and let $\sigma$ send each entry of $\pi$'s list to the entry in the same place in $\tau$'s. Then $\sigma\pi\sigma^{-1} = \tau$.
Conjugation is relabelling. Conjugate elements are the same move seen from a different point of view.
Maps and quotients
This is where most people get stuck.
A homomorphism is a function $\varphi : G \to H$ with $\varphi(ab) = \varphi(a)\varphi(b)$. Its kernel is $\{g : \varphi(g) = e\}$. Show that the kernel is a subgroup, that $gkg^{-1}$ is in the kernel whenever $k$ is, and that $\varphi$ is one-to-one exactly when the kernel is $\{e\}$.
First show $\varphi(e) = e$ and $\varphi(g^{-1}) = \varphi(g)^{-1}$. For one-to-one, look at $\varphi(ab^{-1})$.
$\varphi(e) = \varphi(ee) = \varphi(e)\varphi(e)$, so $\varphi(e) = e$ by cancellation. Then $\varphi(g)\varphi(g^{-1}) = \varphi(e) = e$. So the kernel contains $e$ and is closed under products and inverses. $\varphi(gkg^{-1}) = \varphi(g)\,e\,\varphi(g)^{-1} = e$. Finally, $\varphi(a) = \varphi(b)$ exactly when $\varphi(ab^{-1}) = e$, so $\varphi$ is one-to-one exactly when only $e$ maps to $e$.
A subgroup $N$ with $gNg^{-1} = N$ for every $g$ is called normal. Kernels are always normal. Problem 20 shows that's the only kind of normal subgroup there is.
Let $N$ be a subgroup of $G$. We'd like to multiply cosets by the rule $(aN)(bN) = abN$. Show that this rule is well defined, meaning the answer doesn't depend on which $a$ and $b$ you picked to name the cosets, exactly when $N$ is normal.
If $N$ is normal and $a' = an$, $b' = bm$, then $a'b' = ab(b^{-1}nb)m$.
Conversely, $eN = nN$ for $n \in N$. So $(eN)(g^{-1}N)$ must equal $(nN)(g^{-1}N)$. What does that say about $gng^{-1}$?
If $N$ is normal, $a' = an$ and $b' = bm$ with $n, m \in N$, then $a'b' = ab\,(b^{-1}nb)\,m \in abN$, because $b^{-1}nb \in N$. So the product doesn't depend on the names. Conversely, suppose it's well defined, and take $n \in N$ and $g \in G$. Since $nN = eN$, $(nN)(g^{-1}N) = (eN)(g^{-1}N)$, which says $ng^{-1}N = g^{-1}N$. So $ng^{-1} = g^{-1}n'$ for some $n' \in N$, which rearranges to $gng^{-1} = n' \in N$.
The cosets of a normal subgroup form a group, the quotient group $G/N$. "Why only normal subgroups?" has 124 votes on Math Stack Exchange, and this is the answer.
Show that a subgroup with exactly two cosets is normal.
With two cosets, the left cosets are $H$ and "everything else", and so are the right cosets. So $gH = Hg$.
With two cosets, the left cosets are $H$ and $G \setminus H$, and so are the right cosets. If $g \in H$, then $gH = H = Hg$. If not, $gH = G \setminus H = Hg$. Either way $gHg^{-1} = H$.
Index 2 means normal. For example $A_n$ is normal in $S_n$, and the rotations are normal in any dihedral group.
$A_4$ is the group of even permutations of $\{1, 2, 3, 4\}$. It has 12 elements. Show that it has no subgroup with 6 elements.
Such a subgroup $H$ would have two cosets, so it would be normal, and $A_4/H$ would have two elements. So $g^2 \in H$ for every $g$.
Every 3-cycle $c$ is a square: $c = (c^2)^2$. How many 3-cycles does $A_4$ have?
A subgroup $H$ of order 6 would have index 2, so it would be normal (problem 18), and $A_4/H$ would have two elements. In a group of two elements everything squares to the identity, so $g^2 \in H$ for every $g$. Every 3-cycle $c$ has $c^3 = e$, so $c = c^4 = (c^2)^2$ is a square. $A_4$ has eight 3-cycles, so $H$ would contain all eight. That's too many for six elements.
The converse of Lagrange's theorem is false. A number dividing $|G|$ doesn't have to be the size of a subgroup. The question has 79 votes, and $A_4$ is the smallest counterexample.
Let $\varphi : G \to H$ be a homomorphism with kernel $K$. Show that $gK \mapsto \varphi(g)$ is a well-defined isomorphism from $G/K$ to the image of $\varphi$. Deduce that every normal subgroup is the kernel of some homomorphism.
$\varphi(g) = \varphi(g')$ exactly when $g^{-1}g' \in K$, exactly when $gK = g'K$. For the last part, use $G \to G/N$.
$\varphi(g) = \varphi(g')$ exactly when $g^{-1}g' \in K$, exactly when $gK = g'K$. Reading that one way shows the map is well defined, and the other way shows it's one-to-one. $(gK)(hK) = ghK \mapsto \varphi(gh) = \varphi(g)\varphi(h)$, so it's a homomorphism, and it's onto the image by definition. For a normal $N$, the map $g \mapsto gN$ from $G$ to $G/N$ is a homomorphism (problem 17) with kernel $N$.
The first isomorphism theorem. Squash the kernel to a point, and what's left is exactly the image.
Show that every group $G$ with $n$ elements is isomorphic to a subgroup of $S_n$, the permutations of $n$ things.
Each $g$ permutes the elements of $G$ by $x \mapsto gx$. Problem 3 says it's a permutation. Check that this gives a one-to-one homomorphism.
For each $g$ let $\lambda_g(x) = gx$. By problem 3 it's a permutation of $G$. Associativity gives $\lambda_{gh} = \lambda_g \circ \lambda_h$, so $g \mapsto \lambda_g$ is a homomorphism into the permutations of $G$. If $\lambda_g$ is the identity then $g = ge = e$, so the kernel is trivial and the map is one-to-one (problem 16). Number the elements of $G$ from 1 to $n$ to land in $S_n$.
Cayley's theorem. Every group is a group of permutations. That's why the field guide can build every one of its specimens out of cycles.
Commuting
How close to abelian can a group get without being abelian?
The centre $Z(G)$ is the set of elements that commute with everything. Show that it's a normal subgroup. Then show that if $G/Z(G)$ is cyclic, $G$ is abelian.
If $G/Z(G)$ is generated by $gZ(G)$, every element of $G$ is $g^kz$ for some $k$ and some $z \in Z(G)$. Multiply two of these.
$e$ commutes with everything, and if $a$ and $b$ do then so do $ab$ and $a^{-1}$, so $Z(G)$ is a subgroup. For $z \in Z(G)$, $gzg^{-1} = z$, so it's normal. If $G/Z(G)$ is generated by $gZ(G)$, every element is $g^kz$ with $z$ central. Then $(g^kz)(g^lz') = g^{k+l}zz' = (g^lz')(g^kz)$.
With 101 votes, this is the most upvoted of all the "prove it's abelian" questions. Its real use is the contrapositive: in a non-abelian group, $G/Z(G)$ has at least 4 elements.
Pick two elements of a finite group at random. Show that if $G$ isn't abelian, the chance that they commute is at most $5/8$. Find a group where it's exactly $5/8$.
The number of commuting pairs is $\sum_x |C(x)|$, where $C(x)$ is the set of elements commuting with $x$. It's a subgroup.
$|C(x)| = |G|$ for $x$ in the centre and $|C(x)| \le |G|/2$ otherwise. By problem 22 the centre has at most $|G|/4$ elements.
$D_4$ and $Q_8$ both reach exactly $40$ commuting pairs out of $64$.
The number of commuting pairs is $\sum_x |C(x)|$. For $x$ in the centre, $|C(x)| = |G|$. Otherwise $C(x)$ is a proper subgroup, so by Lagrange $|C(x)| \le |G|/2$. If $G$ isn't abelian, $G/Z(G)$ isn't cyclic (problem 22). Groups of size 1, 2 or 3 are cyclic, so $|G/Z(G)| \ge 4$ and $|Z(G)| \le |G|/4$. So the count is at most $|Z||G| + (|G| - |Z|)\frac{|G|}{2} = \frac{|G|^2}{2} + \frac{|Z||G|}{2} \le \frac58|G|^2$. In $D_4$ the centre is $\{e, r^2\}$ and the other six elements have centralisers of size 4: $2 \cdot 8 + 6 \cdot 4 = 40$.
This is the 5/8 theorem (Gustafson, 1973). A group can't be "mostly" commutative. Either every pair commutes, or at least 3/8 of pairs don't.
Actions
Groups doing things to sets, and counting as a result.
Suppose $G$ permutes a set $X$. For $x \in X$, its orbit is everything $x$ can be moved to, and its stabiliser is the subgroup of elements that don't move it. Show that $|G| = |\text{orbit}| \cdot |\text{stabiliser}|$. Use this to count the rotations of a cube.
Two elements send $x$ to the same place exactly when they're in the same left coset of the stabiliser.
For the cube, take $x$ to be a face.
$gx = hx$ exactly when $h^{-1}g$ fixes $x$, exactly when $g$ and $h$ are in the same left coset of the stabiliser. So the points of the orbit match up one-for-one with those cosets, and Lagrange gives $|\text{orbit}| = |G|/|\text{stabiliser}|$. For the cube, rotations can bring the top face to any of 6 faces, and the rotations keeping the top face on top are the 4 turns about its axis: $24$.
This is the orbit–stabiliser theorem. The cube has $6 \times 4 = 24$ rotations: pick where the top face goes, then turn.
Show that the number of orbits of $G$ acting on $X$ is the average, over $g \in G$, of the number of points $g$ doesn't move. Use it to count the ways to paint each face of a cube black or white, if two colourings count as the same when a rotation takes one to the other.
Count the pairs $(g, x)$ with $gx = x$ in two ways: grouped by $g$, then grouped by $x$. By problem 24, each orbit contributes exactly $|G|$.
Sort the 24 rotations by type: the identity, 90° and 180° turns about a face, 120° turns about a corner, and 180° turns about an edge. How many colourings does each one leave unchanged?
Count pairs $(g, x)$ with $gx = x$. Grouped by $g$, you get $\sum_g |\text{Fix}(g)|$. Grouped by $x$, you get $\sum_x |\text{Stab}(x)| = \sum_x |G|/|\text{orbit}(x)|$, and each orbit contributes exactly $|G|$. So the number of orbits is $\frac{1}{|G|}\sum_g|\text{Fix}(g)|$. For the cube: the identity fixes all $2^6$; the six quarter turns about a face axis keep a colouring only if the four side faces match, $2^3$; the three half turns about a face axis need opposite side faces to match, $2^4$; the eight turns about a corner cycle two sets of three faces, $2^2$; the six half turns about an edge swap three pairs of faces, $2^3$.
This is Burnside's lemma, the standard tool for counting things up to symmetry, like necklaces, molecules and dice.
Let $G$ act on itself by conjugation. Show that $|G| = |Z(G)| + \sum_i [G : C(x_i)]$, where the sum runs over one element $x_i$ from each conjugacy class with more than one element. Deduce that if $|G| = p^n$ for a prime $p$ and $n \ge 1$, then $Z(G)$ has more than one element.
The orbits are the conjugacy classes, and $x$ is alone in its class exactly when $x \in Z(G)$. When $|G| = p^n$, every term $[G : C(x_i)]$ in the sum is a positive power of $p$.
Under conjugation, the orbits are the conjugacy classes and the stabiliser of $x$ is $C(x)$. So the class of $x$ has $[G : C(x)]$ elements by problem 24, and it has one element exactly when $x \in Z(G)$. Adding the class sizes gives the equation. If $|G| = p^n$, each $[G : C(x_i)]$ is bigger than 1 and divides $p^n$, so it's a multiple of $p$. So is $|G|$. Hence $|Z(G)|$ is a multiple of $p$, and since it's at least 1, it's at least $p$.
The class equation, and its first great consequence: $p$-groups have non-trivial centres.
Show that every group with $p^2$ elements, for $p$ prime, is abelian.
By problem 26, $|Z(G)|$ is $p$ or $p^2$. If it were $p$, then $G/Z(G)$ would have $p$ elements, so it would be cyclic. Now use problem 22.
$|Z(G)|$ divides $p^2$ and is bigger than 1 (problem 26), so it's $p$ or $p^2$. If it were $p$, then $G/Z(G)$ would have $p$ elements and be cyclic (problem 10), so $G$ would be abelian (problem 22), so $Z(G) = G$, which has $p^2$ elements. Contradiction. So $Z(G) = G$.
Groups of order $p^2$ are abelian. So there are exactly two of them, $C_{p^2}$ and $C_p \times C_p$. This explains the order-4 and order-9 rows of the field guide.
Let $p$ be a prime dividing $|G|$. Consider the set $X$ of all $p$-tuples $(g_1, \dots, g_p)$ of elements of $G$ with $g_1g_2\cdots g_p = e$. Show that $|X| = |G|^{p-1}$ and that rotating a tuple, $(g_1, \dots, g_p) \mapsto (g_2, \dots, g_p, g_1)$, keeps it in $X$. Use this to show $G$ has an element of order $p$.
Choose $g_1, \dots, g_{p-1}$ freely, and then $g_p$ is forced.
The rotations form a group of order $p$ acting on $X$, so every orbit has size $1$ or $p$. The orbits of size 1 are the tuples $(g, g, \dots, g)$ with $g^p = e$. Count mod $p$.
Pick $g_1, \dots, g_{p-1}$ freely, and $g_p = (g_1\cdots g_{p-1})^{-1}$ is forced: $|X| = |G|^{p-1}$, a multiple of $p$. If $g_1g_2\cdots g_p = e$ then $g_2\cdots g_pg_1 = g_1^{-1}(g_1\cdots g_p)g_1 = e$, so rotating stays in $X$. The rotations form a group of order $p$, so every orbit has size 1 or $p$. Orbits of size 1 are the tuples $(g, \dots, g)$ with $g^p = e$. Since $|X|$ and the big orbits are all multiples of $p$, so is the number of fixed tuples. $(e, \dots, e)$ is one, so there are at least $p$, and one of them has $g \neq e$. That $g$ has order $p$.
This is Cauchy's theorem, in James McKay's one-paragraph proof from 1959. Compare it with problem 8.
The finale
Why the quintic has no formula.
$A_5$ is the even permutations of five things. It has 60 elements. Show that its conjugacy classes have sizes $1, 12, 12, 15$ and $20$.
generate A₅ in the scratchpad
By cycle type, $A_5$ has the identity, 20 three-cycles, 15 double swaps like $(1\,2)(3\,4)$, and 24 five-cycles.
In $S_5$, each of those types is one conjugacy class (problem 15). In $A_5$ a class can split. Compute the centraliser of each type in $S_5$ and check whether it contains an odd permutation.
The centraliser of a 5-cycle in $S_5$ is just its own 5 powers, all even, so in $A_5$ the class has size $60/5 = 12$, not 24.
Count: $\binom53 \cdot 2 = 20$ three-cycles, $5 \cdot 3 = 15$ double swaps, and $4! = 24$ five-cycles, plus the identity, is 60. In $S_5$ each type is one class, so its centraliser has size $120/(\text{class size})$. The centraliser in $A_5$ is the even part of it: half its size if it contains an odd permutation, all of it otherwise. $(1\,2\,3)$ has centraliser of size 6 in $S_5$, containing the odd $(4\,5)$, so size 3 in $A_5$ and a class of $60/3 = 20$. $(1\,2)(3\,4)$ has centraliser of size 8, containing the odd $(1\,2)$, so size 4 and a class of 15. A 5-cycle has centraliser of size 5, just its powers, all even, so its $A_5$ class has $60/5 = 12$ elements, and the 24 five-cycles split into two classes.
These are the conjugacy classes of $A_5$. The five-cycles split into two classes, and nothing else does.
Show that $A_5$ has no normal subgroups other than $\{e\}$ and $A_5$ itself.
A normal subgroup is a union of conjugacy classes, one of which is $\{e\}$.
By Lagrange its size must divide 60. Which sums of $1$ with some of $12, 12, 15, 20$ divide 60?
A normal subgroup contains $gng^{-1}$ whenever it contains $n$, so it's a union of whole conjugacy classes, including $\{e\}$. By Lagrange its size divides 60. The possible sizes are $1$ plus any selection from $12, 12, 15, 20$: $1, 13, 16, 21, 25, 28, 33, 36, 40, 45, 48, 60$. Only $1$ and $60$ divide 60.
$A_5$ is simple, the smallest non-abelian simple group. This is the fact behind the unsolvability of the quintic: a formula using roots would force the symmetry group of a general quintic's roots, $S_5$, to break down into abelian pieces, and $A_5$ won't break. You got here from problem 1.