Homomorphisms, normal subgroups, and quotient groups have each had a page of their own. This page ties them together. The isomorphism theorems say that quotient groups are not some exotic new thing: every homomorphic image of a group is secretly a quotient of it, and quotients of quotients, products of subgroups, and lattices of subgroups all behave exactly the way you'd hope. As usual, permutations and dihedral motions compose right to left, and $D_{2n}$ has $2n$ elements.
Four isomorphism theorems, and books can't even agree on the count. Some stop at three, some call the fourth the Correspondence Theorem. Nobody's sorry.
Everything else on the page is built from this one. Start with a homomorphism $\phi : G \rightarrow H$. Two subgroups come with it for free.
On Normal Subgroups we proved that kernels are always normal, and on Homomorphisms that the image $\phi(G)$ is always a subgroup of $H$.
The kernel $\ker \phi$ is a normal subgroup of $G$, so $G/\ker \phi$ is a group. The image $\phi(G)$ is a subgroup of $H$. The theorem says these two groups are the same.
If $\phi : G \rightarrow H$ is a homomorphism, then $\ker \phi \unlhd G$ and $$G / \ker \phi \cong \phi(G).$$ An isomorphism is given by $\psi(g \ker \phi) = \phi(g)$.
The proof is mostly a check that $\psi$ makes sense. Write $N = \ker \phi$ to keep things short.
Proof. $N \unlhd G$ because kernels are normal. Define $\psi : G/N \rightarrow \phi(G)$ by $\psi(gN) = \phi(g)$.
Therefore $\psi$ is an isomorphism, and $G/N \cong \phi(G)$. $\blacksquare$
The heart of the proof is the first bullet: $\phi(g) = \phi(g')$ exactly when $g$ and $g'$ are in the same coset of the kernel. So $\phi$ sorts $G$ into piles, one pile per element of the image, and the piles are the cosets of $\ker \phi$. Counting both sides gives a handy formula.
If $G$ is finite and $\phi : G \rightarrow H$ is a homomorphism, then $|G| = |\ker \phi| \cdot |\phi(G)|$.
Proof. $|G/\ker \phi| = |G|/|\ker \phi|$ (Lagrange), and $|G/\ker \phi| = |\phi(G)|$ by the theorem. $\blacksquare$
The Sneak Peek on Quotient Groups said $S_n/A_n \cong C_2$ was no accident. Here's why.
Two examples. The first explains a quotient from last page, and the second explains a coincidence from Homomorphisms.
Let $n \ge 2$ and let $\operatorname{sgn} : S_n \rightarrow \{\pm 1\}$ send even permutations to $1$ and odd permutations to $-1$. Show that $S_n/A_n \cong C_2$.
$\operatorname{sgn}$ is a homomorphism: a product of two even or two odd permutations is even, and a product of one of each is odd, which matches $1 \cdot 1 = (-1)(-1) = 1$ and $1 \cdot (-1) = -1$. It is surjective, since $\operatorname{sgn}(e) = 1$ and $\operatorname{sgn}((1 \ 2)) = -1$. Its kernel is the set of even permutations, $A_n$. By the 1st Isomorphism Theorem, $$S_n / A_n \cong \{\pm 1\} \cong C_2.$$
On Homomorphisms, every setting of the Multiplier Rows widget had (size of kernel) × (size of image) $= 12$. The Corollary above is why.
The second one revisits an Exercise from Homomorphisms, where $|\ker \phi| \cdot |\phi(G)|$ came out to $|G|$ and we promised that was no coincidence.
Let $\phi : \mathbb{Z}/12\mathbb{Z} \rightarrow \mathbb{Z}/12\mathbb{Z}$, $\phi(k) = 3k \pmod{12}$. Identify $(\mathbb{Z}/12\mathbb{Z})/\langle 4 \rangle$.
On Homomorphisms we found $\ker \phi = \{0, 4, 8\} = \langle 4 \rangle$ and $\phi(\mathbb{Z}/12\mathbb{Z}) = \{0, 3, 6, 9\} = \langle 3 \rangle$. By the 1st Isomorphism Theorem, $$(\mathbb{Z}/12\mathbb{Z}) / \langle 4 \rangle \cong \langle 3 \rangle \cong C_4.$$ And $|\ker \phi| \cdot |\phi(\mathbb{Z}/12\mathbb{Z})| = 3 \cdot 4 = 12$ is the Corollary.
See the piles for yourself. Each element of $G$ drops onto the element of $\phi(G)$ it maps to. Notice that every pile is a coset of the kernel, and every pile is the same height.
Pick a homomorphism. Press DROP to stack each element of $G$ above its image, then SQUASH to turn each stack into a single coset of the kernel.
The theorem turns "find the quotient" into "find a homomorphism with the right kernel." Try it on a quotient from last page.
Let $C_4 \times C_4 = \langle x \rangle \times \langle y \rangle$ and $C_4 = \langle z \rangle$, and define $\phi : C_4 \times C_4 \rightarrow C_4$ by $\phi((x^i, y^j)) = z^{i+j}$. Show that $\phi$ is a surjective homomorphism, find $\ker \phi$, and conclude that $(C_4 \times C_4)/\langle (x, y^3) \rangle \cong C_4$.
$\phi$ is well-defined because $z^4 = e$, so only $i + j \bmod 4$ matters. It is a homomorphism: $$\phi((x^i, y^j)(x^k, y^l)) = \phi((x^{i+k}, y^{j+l})) = z^{i+k+j+l} = z^{i+j} z^{k+l} = \phi((x^i, y^j))\,\phi((x^k, y^l)).$$ It is surjective, since $\phi((x, e)) = z$ generates $C_4$.
$\ker \phi = \{(x^i, y^j) : i + j \equiv 0 \pmod 4\} = \{(e, e), (x, y^3), (x^2, y^2), (x^3, y)\} = \langle (x, y^3) \rangle$.
By the 1st Isomorphism Theorem, $(C_4 \times C_4)/\langle (x, y^3) \rangle \cong C_4$. This is Example 5 from Quotient Groups, with no coset tables needed.
Before the other three theorems, we need a way to combine two subgroups into something bigger.
If $H, K \le G$, the product of $H$ and $K$ is $HK = \{hk : h \in H, k \in K\}$.
$HK$ is always a subset of $G$, but it doesn't have to be a subgroup. Its size, at least, is easy to pin down.
$HK$ isn't always a subgroup. It's a subset with ambitions.
If $H, K \le G$ are finite, then $$|HK| = \frac{|H||K|}{|H \cap K|}.$$
Proof. $HK$ is the union of the left cosets $hK$, $h \in H$, and each has $|K|$ elements. Two of them agree exactly when $h^{-1}h' \in K$. Since $h^{-1}h' \in H$ automatically, this says $h^{-1}h' \in H \cap K$, i.e. $h(H \cap K) = h'(H \cap K)$. So the distinct cosets $hK$ match up with the cosets of $H \cap K$ in $H$, and there are $|H|/|H \cap K|$ of those (Lagrange). Therefore $|HK| = |K| \cdot |H|/|H \cap K|$. $\blacksquare$
As for being a subgroup, one condition is enough for everything on this page.
If $H, K \le G$ and $H \subseteq N_G(K)$, then $HK \le G$. In particular, if $K \unlhd G$ then $HK \le G$ for every $H \le G$.
Proof. Recall $N_G(K) = \{g \in G : gKg^{-1} = K\}$ is a subgroup of $G$, so $h \in H \Rightarrow h^{-1} \in N_G(K)$ too. Use the subgroup criterion:
If $K \unlhd G$ then $N_G(K) = G$, so the condition holds for every $H$. $\blacksquare$
When both subgroups are normal and they only overlap in $e$, the product is as nice as it can possibly be.
On Groups of Order 4–8 we built $G \times H$ from two separate groups. Now we go the other way: find two subgroups inside one group that already behave like a direct product.
If $H, K \unlhd G$ and $H \cap K = \{e\}$, then $HK \cong H \times K$, via $(h, k) \mapsto hk$.
Proof. Step 1: elements of $H$ and $K$ commute. Let $h \in H$, $k \in K$. Then $$hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K \quad (K \unlhd G), \qquad hkh^{-1}k^{-1} = h(kh^{-1}k^{-1}) \in H \quad (H \unlhd G).$$ So $hkh^{-1}k^{-1} \in H \cap K = \{e\}$, which means $hk = kh$.
Step 2. Define $\mu : H \times K \rightarrow HK$ by $\mu(h, k) = hk$.
Therefore $\mu$ is an isomorphism. $\blacksquare$
When in addition $HK = G$, this has a name.
$G$ is the internal direct product of $H$ and $K$ if $H, K \unlhd G$, $H \cap K = \{e\}$, and $HK = G$. Then $G \cong H \times K$.
So to split a group as a direct product, you don't build anything: you find two normal subgroups that meet only at $e$ and are big enough. The Product Formula takes care of "big enough."
Use this to show that $D_{12} \cong D_6 \times C_2$.
Proof. Let $H = \langle s, r^2 \rangle$ and $K = \langle r^3 \rangle$. $$\begin{align*} H &= \{e, r^2, r^4, s, sr^2, sr^4\} \cong D_6, \\ K &= \{e, r^3\} \cong C_2. \end{align*}$$ ($H \cong D_6$ since $r^2$ has order $3$, $s$ has order $2$, and $r^2 s = sr^{-2}$.)
Since $H \cap K = \{e\} \Rightarrow |HK| = \frac{|H||K|}{|H \cap K|} = 12 \Rightarrow HK = D_{12}$.
$H$ has index $2$, so $H \unlhd G$ (the Spicy Bonus on Normal Subgroups). And $r^3$ commutes with $r$ and with $s$ (since $sr^3 = r^{-3}s = r^3s$), so $K \subseteq Z(D_{12})$ and $K \unlhd G$. Since $H, K \unlhd G \Rightarrow HK \cong H \times K$, so $D_{12} \cong D_6 \times C_2$. $\blacksquare$
Now a near miss. Same group, two different subgroups that also meet only at $e$ and also have product $D_{12}$.
$G = D_{12}$, $H = \langle r \rangle$, $K = \langle s \rangle$. Is $G \cong H \times K$?
$|H| = 6$, $|K| = 2$, $H \cap K = \{e\} \Rightarrow HK = D_{12}$.
But $H \cong C_6$ and $K \cong C_2$, so $D_{12} = HK \not\cong H \times K$: $C_6 \times C_2$ is Abelian, and $D_{12}$ is not.
Why? $H \unlhd G$, but $K \ntrianglelefteq G$. For instance $rsr^{-1} = r(sr^{-1}) = r(rs) = r^2s \notin K$.
So the theorem really needs both subgroups normal; "meet at $e$ and fill up the group" is not enough. Compare all the conditions side by side below, including a product that's too small to be a subgroup at all.
Pick a pair of subgroups. The grid lists every product $hk$; repeats are marked. The badges check the Product Formula, whether $HK$ is a subgroup, and whether $HK \cong H \times K$.
Let $G = \mathbb{Z}/6\mathbb{Z}$ (under addition, so "$HK$" is written $H + K$), $H = \{0, 3\}$, and $K = \{0, 2, 4\}$. Show that $G$ is the internal direct product of $H$ and $K$, and conclude $\mathbb{Z}/6\mathbb{Z} \cong C_2 \times C_3$.
$G$ is Abelian, so $H, K \unlhd G$. $H \cap K = \{0\}$. By the Product Formula, $|H + K| = \frac{2 \cdot 3}{1} = 6$, so $H + K = G$. Therefore $G$ is the internal direct product of $H$ and $K$, and $\mathbb{Z}/6\mathbb{Z} \cong H \times K \cong C_2 \times C_3$.
This matches $C_2 \times C_3 \cong C_6$ from Groups of Order 4–8, found there by hunting for a generator.
Now the remaining isomorphism theorems. The second one takes two subgroups, one normalizing the other, and compares two quotients you can build from them.
Suppose $A, B \le G$. If $A \subseteq N_G(B)$, then $AB \le G$, $B \unlhd AB$, $A \cap B \unlhd A$, and $$AB / B \cong A / (A \cap B).$$
The name comes from the picture. Draw the four subgroups as a lattice, and the two quotients in the theorem are the two parallel edges of a diamond.
It's called the Diamond Theorem because the picture is a diamond. Mathematicians are not great at names.
The proof is a single application of the 1st Isomorphism Theorem: map $A$ onto $AB/B$ and compute the kernel.
Proof. First, note that:
Define $\phi : A \rightarrow AB/B$ by $\phi(a) = aB$.
By the 1st Isomorphism Theorem, $A \cap B \unlhd A$ and $A / (A \cap B) \cong AB / B$. $\blacksquare$
Example 4 is a quick instance. In $D_{12}$, take $A = \langle s \rangle$ and $B = \langle r \rangle$. $B$ is normal, so $A \subseteq N_G(B)$ automatically, $AB = D_{12}$, and $A \cap B = \{e\}$. The theorem gives $D_{12}/\langle r \rangle \cong \langle s \rangle/\{e\} \cong C_2$.
In $G = \mathbb{Z}$ (under addition, so "$AB$" is written $A + B$), let $A = 4\mathbb{Z}$ and $B = 6\mathbb{Z}$. Find $A + B$ and $A \cap B$, and write down the isomorphism the Diamond Theorem gives. What familiar group are both sides isomorphic to?
$\mathbb{Z}$ is Abelian, so every subgroup normalizes every other. $A + B = \{4a + 6b : a, b \in \mathbb{Z}\} = 2\mathbb{Z}$: it contains $6 - 4 = 2$, and every $4a + 6b$ is even. $A \cap B$ is the set of common multiples of $4$ and $6$, which is $12\mathbb{Z}$. The theorem gives $$2\mathbb{Z} / 6\mathbb{Z} \cong 4\mathbb{Z} / 12\mathbb{Z}.$$ Both sides have $3$ elements ($\{0, 2, 4\} + 6\mathbb{Z}$ and $\{0, 4, 8\} + 12\mathbb{Z}$), so both are $\cong C_3$. (The $\gcd$ and $\operatorname{lcm}$ of $4$ and $6$ show up as $A + B$ and $A \cap B$.)
The third theorem answers a question from the last page: can you take a quotient of a quotient? You can, and it collapses to an ordinary quotient.
If $H, K \unlhd G$ and $H \le K$, then $H \unlhd K$, $K/H \unlhd G/H$, and $$(G/H)/(K/H) \cong G/K.$$
The $H$'s cancel like fractions. I know I'm not supposed to think that, but here it actually works.
Again the proof is the 1st Isomorphism Theorem, this time applied to a map out of $G/H$.
Proof. First, since $H \unlhd G$, it follows immediately that $H \unlhd K$. Next define $\phi : G/H \rightarrow G/K$ by $\phi(gH) = gK$.
By the 1st Isomorphism Theorem, $K/H \unlhd G/H$ and $(G/H)/(K/H) \cong G/K$. $\blacksquare$
The integers make it concrete.
Example. $G = \mathbb{Z}$, $K = 2\mathbb{Z}$, $H = 6\mathbb{Z}$. What does the theorem say?
$6\mathbb{Z} \le 2\mathbb{Z}$, and both are normal since $\mathbb{Z}$ is Abelian. $$(\mathbb{Z}/6\mathbb{Z}) / (2\mathbb{Z}/6\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}.$$
Concretely, $2\mathbb{Z}/6\mathbb{Z} = \{0, 2, 4\}$ inside $\mathbb{Z}/6\mathbb{Z}$. Modding it out leaves two cosets, "even" $\{0, 2, 4\}$ and "odd" $\{1, 3, 5\}$, which is exactly $\mathbb{Z}/2\mathbb{Z}$.
One more, in a non-Abelian group.
Let $G = Q_8$, $H = \{\pm 1\}$, and $K = \langle i \rangle = \{\pm 1, \pm i\}$. Check that the 3rd Isomorphism Theorem applies, and use it to identify $(Q_8/\{\pm 1\})/(\langle i \rangle/\{\pm 1\})$.
$H = Z(Q_8) \unlhd Q_8$, and $K$ has index $2$ in $Q_8$, so $K \unlhd Q_8$. Also $H \le K$. The theorem gives $$(Q_8/\{\pm 1\})/(\langle i \rangle/\{\pm 1\}) \cong Q_8/\langle i \rangle,$$ which has $8/4 = 2$ elements, so it is $\cong C_2$. (Directly: $Q_8/\{\pm 1\} \cong V_4$ by Example 7 on Quotient Groups, $\langle i \rangle/\{\pm 1\}$ is one of its subgroups of order $2$, and $V_4$ modulo a subgroup of order $2$ has $2$ elements.)
The last theorem is about the whole lattice of subgroups at once. Modding out by $N$ throws away everything below $N$ and keeps everything above it, perfectly intact.
The Lattice Explorer drew every subgroup of a group, with lines for "contained in." The Lattice Theorem says the top part of that drawing, everything containing $N$, is the lattice of $G/N$.
If $N \unlhd G$ then there is an inclusion preserving bijection between subgroups $A \le G$ which contain $N$, and subgroups of $G/N$. This bijection also preserves indices of respective subgroups.
The bijection is the obvious one, $A \mapsto A/N$. "Inclusion preserving" means $A \le B \iff A/N \le B/N$, and "preserves indices" means $|B : A| = |B/N : A/N|$. The proof is the Spicy Bonus at the end of the page. First, the example from the notes.
Example. $G = Q_8$, $N = \{\pm 1\}$. Match up the subgroups of $Q_8$ containing $N$ with the subgroups of $G/N \cong V_4$.
The lattice of subgroups of $G = Q_8$ containing $N$ corresponds directly to the subgroups of $G/N \cong V_4$:
Indices match too: $|Q_8 : \langle i \rangle| = 2 = |V_4 : H_1|$, and $|\langle i \rangle : N| = 2 = |H_1 : \{eN\}|$. The only subgroup of $Q_8$ missing from the picture is $\{1\}$, which doesn't contain $N$.
Tap around the two lattices below to see the matching, and try the other presets: in each one, the top of $G$'s lattice has the same shape as the lattice of $G/N$.
Pick a group and a normal subgroup $N$. Tap a subgroup on either side to light up its partner. Subgroups that don't contain $N$ are faded: they have no partner.
That's all four. Notice how often the 1st Isomorphism Theorem did the real work: the 2nd and 3rd were each one well-chosen homomorphism plus a kernel computation.
The Lattice Theorem was stated without proof. Here it is for you to prove, in the finite case.
Let $G$ be finite and $N \unlhd G$. Prove that $A \mapsto A/N$ is a bijection from $\{A : N \le A \le G\}$ to the set of subgroups of $G/N$, that $A \le B \iff A/N \le B/N$, and that $|B : A| = |B/N : A/N|$ whenever $N \le A \le B$.
Proof. Let $\pi : G \rightarrow G/N$, $\pi(g) = gN$. This is a surjective homomorphism, and for $N \le A$, $A/N = \{aN : a \in A\} = \pi(A)$ is a subgroup of $G/N$ (the image of a subgroup under a homomorphism is a subgroup).
An inverse map. For a subgroup $\bar{A} \le G/N$, let $\pi^{-1}(\bar{A}) = \{g \in G : gN \in \bar{A}\}$. It contains $N$ (since $nN = N = eN \in \bar{A}$), and it is a subgroup: it contains $e$, and if $gN, g'N \in \bar{A}$ then $(gg')N = (gN)(g'N) \in \bar{A}$ and $g^{-1}N = (gN)^{-1} \in \bar{A}$.
The two maps undo each other.
So $A \mapsto A/N$ is a bijection.
Inclusions. If $A \le B$ then clearly $A/N \subseteq B/N$. Conversely, if $A/N \subseteq B/N$ then $A = \pi^{-1}(A/N) \subseteq \pi^{-1}(B/N) = B$.
Indices. If $N \le A \le B$, then by Lagrange, $$|B/N : A/N| = \frac{|B/N|}{|A/N|} = \frac{|B|/|N|}{|A|/|N|} = \frac{|B|}{|A|} = |B : A|. \qquad \blacksquare$$