Subgroups

In this lesson you’ll learn what a subgroup is, the efficient test for checking one, why subgroups must share the parent’s identity, and how the subgroups of a group fit together into a lattice.

A subset $H \subseteq G$ is a subgroup of $G$, written $H \le G$, if $H$ is itself a group under the same operation.

The phrase “under the same operation” is doing real work. $\{1, 2, 3, 4\}$ is a subset of $\mathbb{Z}$ and a group under multiplication mod 5, but it is not a subgroup of $(\mathbb{Z}, +)$, because the operations differ.

Checking all four axioms is wasteful, because two come free. Associativity is inherited: if it holds for every triple in $G$, it holds for every triple in $H$. And once $H$ is nonempty and closed under inverses, closure hands you the identity.

So there are only three things to check:

1. $H$ is nonempty (in practice, check $e \in H$).
2. Closed under the operation: $a, b \in H \implies ab \in H$.
3. Closed under inverses: $a \in H \implies a^{-1} \in H$.

There’s an even tighter version. For a nonempty $H$:

$$H \le G \iff ab^{-1} \in H \text{ for all } a, b \in H$$

One-step subgroup test. It works because taking $a = b$ gives $aa^{-1} = e \in H$, then $a = e$ gives $eb^{-1} = b^{-1} \in H$, and then replacing $b$ by $b^{-1}$ gives $a(b^{-1})^{-1} = ab \in H$. All three conditions fall out of the one.

For finite $H$ it’s even easier: a nonempty finite subset closed under the operation is automatically a subgroup. Repeatedly multiplying $a$ by itself in a finite set must eventually repeat, which forces a power of $a$ to be the identity, and the previous power is then $a^{-1}$.

$\{e\}$ and $G$ itself are always subgroups. They’re called the trivial and improper subgroups, and any other is a proper nontrivial subgroup. Some groups have none: $\mathbb{Z}_7$ has only $\{0\}$ and itself.

A subgroup’s identity is forced to be the parent’s identity. Suppose $f$ is the identity of $H$. Then $f f = f$ in $G$, and cancelling one $f$ (legal, since $f$ has an inverse in $G$) gives $f = e$.

The consequence is practical: a subset that omits the parent’s identity is not a subgroup, no matter how well behaved it otherwise looks. The positive integers are closed under addition and never contain $0$, so they are not a subgroup of $(\mathbb{Z}, +)$.

Pick $a \in G$. The subgroup generated by $a$ is everything you can build from it:

$$\langle a \rangle = \{ \dots, a^{-2}, a^{-1}, e, a, a^2, \dots \}$$

This is always a subgroup, and it’s the smallest one containing $a$. In additive notation it’s all integer multiples of $a$.

In $(\mathbb{Z}, +)$, $\langle 3 \rangle = \{\dots, -6, -3, 0, 3, 6, \dots\} = 3\mathbb{Z}$. In $\mathbb{Z}_{12}$, $\langle 3 \rangle = \{0, 3, 6, 9\}$, which closes up after four steps because the group is finite.

Listing every subgroup and drawing the containments gives a picture of the group’s internal structure.

$\mathbb{Z}_{12}$ has exactly six subgroups, one for each divisor of 12. That is not an accident and it is not typical of groups in general; it’s a special property of cyclic groups, proved next lesson.

Two things to notice in the diagram. Every subgroup’s order divides 12, which is Lagrange’s theorem arriving early. And the lattice mirrors the divisor lattice of 12 exactly, with $\langle a \rangle \subseteq \langle b \rangle$ precisely when $b$ divides $a$. Larger generators give smaller subgroups, which reads backwards at first and is worth pausing on.

In a Cayley table, a subgroup shows up as a block that only ever produces entries from itself.

+	0	1	2	3	4	5	6	7
0	0	1	2	3	4	5	6	7
1	1	2	3	4	5	6	7	0
2	2	3	4	5	6	7	0	1
3	3	4	5	6	7	0	1	2
4	4	5	6	7	0	1	2	3
5	5	6	7	0	1	2	3	4
6	6	7	0	1	2	3	4	5
7	7	0	1	2	3	4	5	6

Restrict attention to the shaded rows and columns and you get a complete four-element group table. Nothing in that block escapes to an odd number.

Identity: 0. Cells holding it mark inverse pairs. Shaded: 0, 2, 4, 6 (⟨2⟩) - closed, every product stays inside .

Example 1: Is the set of even integers a subgroup of $(\mathbb{Z}, +)$?

Solution. Use the one-step test on $H = 2\mathbb{Z}$.

Nonempty: $0 = 2 \cdot 0 \in H$. ✓

For $a = 2m$ and $b = 2n$ in $H$, the additive version of $ab^{-1}$ is $a - b$:

$$a - b = 2m - 2n = 2(m - n) \in H$$

Yes, $2\mathbb{Z} \le \mathbb{Z}$.

The odd integers are not, since they don’t contain $0$ - and closure fails anyway.

Example 2: Is $\mathbb{N} = \{1, 2, 3, \dots\}$ a subgroup of $(\mathbb{Z}, +)$?

Solution. Closed under addition ✓. But $0 \notin \mathbb{N}$, and no element has an inverse in $\mathbb{N}$ since $-5$ is not a positive integer.

No. This is the standard reminder that closure alone is not enough. Note the contrast with the finite case: for a finite subset closure would have sufficed. Infinitude is exactly what lets $\mathbb{N}$ be closed without ever cycling back to an identity.

Example 3: Find all subgroups of $\mathbb{Z}_8$.

Solution. Compute $\langle a \rangle$ for each element.

$$\langle 0 \rangle = \{0\}$$ $$\langle 1 \rangle = \{0,1,2,3,4,5,6,7\} = \mathbb{Z}_8$$ $$\langle 2 \rangle = \{0, 2, 4, 6\}$$ $$\langle 3 \rangle = \{0,3,6,1,4,7,2,5\} = \mathbb{Z}_8$$ $$\langle 4 \rangle = \{0, 4\}$$

$\langle 5 \rangle$, $\langle 7 \rangle$ also give all of $\mathbb{Z}_8$, and $\langle 6 \rangle = \{0,6,4,2\} = \langle 2 \rangle$.

The distinct subgroups are $\{0\}$, $\{0,4\}$, $\{0,2,4,6\}$, and $\mathbb{Z}_8$: four subgroups, of orders 1, 2, 4, 8, which are exactly the divisors of 8.

Example 4: A subset closed under the operation but not a subgroup.

In $GL_2(\mathbb{R})$, consider matrices with integer entries and determinant $\pm 1$… actually that is a subgroup. Instead take $H$ = matrices with integer entries and nonzero determinant.

Solution. Closure ✓: a product of integer matrices has integer entries, and determinants multiply so the product’s determinant is nonzero.

Inverses fail. For $A = \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$, $\det A = 2 \ne 0$, and

$$A^{-1} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix}$$

which has no integer entries, so $A^{-1} \notin H$.

Not a subgroup. Requiring $\det = \pm 1$ fixes it, because then the inverse formula divides by $\pm 1$ and stays integral. That group is $GL_2(\mathbb{Z})$ and it matters in number theory.

Example 5: Prove the intersection of two subgroups is a subgroup.

Solution. Let $H, K \le G$ and let $a, b \in H \cap K$.

$H \cap K$ is nonempty because $e$ belongs to both.

Since $a, b \in H$ and $H$ is a subgroup, $ab^{-1} \in H$. Since $a, b \in K$ likewise $ab^{-1} \in K$. So $ab^{-1} \in H \cap K$.

By the one-step test, $H \cap K \le G$. ∎

The union is a different story. In $\mathbb{Z}$, take $H = 2\mathbb{Z}$ and $K = 3\mathbb{Z}$. Then $2, 3 \in H \cup K$ but $2 + 3 = 5$ is in neither. Unions of subgroups are almost never subgroups.

Example 6: The center of a group.

Show that $Z(G) = \{z \in G : zg = gz \text{ for all } g \in G\}$ is a subgroup.

Solution. $e \in Z(G)$ since $eg = g = ge$. ✓

Take $y, z \in Z(G)$ and any $g \in G$. First, $z^{-1}$ commutes with $g$: from $zg = gz$, multiply both sides by $z^{-1}$ on the left and right to get $gz^{-1} = z^{-1}g$.

Then

$$(yz^{-1})g = y(z^{-1}g) = y(gz^{-1}) = (yg)z^{-1} = (gy)z^{-1} = g(yz^{-1})$$

So $yz^{-1} \in Z(G)$, and $Z(G) \le G$. ∎

The center measures how far a group is from abelian: $Z(G) = G$ exactly when $G$ is abelian, and a small center means most elements fail to commute. For $D_4$ the center has just two elements out of eight.

Symmetry breaking in physics. A physical system often has a large symmetry group at high energy, and as it cools the symmetry drops to a subgroup. The Higgs mechanism is exactly this: an electroweak symmetry group breaks to a subgroup, and the pattern of breaking determines which particles acquire mass.

Crystallography. A crystal’s point group is a subgroup of the full rotation group, and which subgroup it is constrains its physical properties. Only crystals whose symmetry group lacks a centre of inversion can be piezoelectric, which is why quartz works in watches and many other crystals don’t.

Error-correcting codes. A linear code is literally a subgroup of $(\mathbb{Z}_2^n, +)$: the set of valid codewords is closed under bitwise XOR and contains the all-zeros word. Decoding works by finding which coset of that subgroup a received word lies in, which is the topic of a later lesson.

Rubik’s cube subgroups. Restricting yourself to a set of moves generates a subgroup, and solving methods work by descending a chain of nested subgroups, each easier to handle than the last. Thistlethwaite’s algorithm is explicitly built as a chain of four subgroups.

Music. Within $\mathbb{Z}_{12}$, the whole-tone scale is $\langle 2 \rangle = \{0,2,4,6,8,10\}$, a subgroup of order 6. The diminished seventh chord is $\langle 3 \rangle$, of order 4. These sound “symmetric” and rootless precisely because they are subgroups, invariant under transposition by their generator.