Once you know what a group is, the very next question is: do groups hide smaller groups inside them? They do — constantly. The even integers sit inside all the integers; the pure rotations of a polygon sit inside its full set of rotations-and-flips; a single "do nothing" move sits inside every group there is. Each of these is a subgroup — a subset that is not just along for the ride, but is a fully-fledged group in its own right, using the exact same operation as its parent.

This is one of the most powerful moves in algebra. Instead of studying a big, complicated group all at once, we break it into the little groups living inside it and study those — the way you understand a machine by understanding its parts. Whether a subset counts as a subgroup is not a matter of taste: there is a crisp checklist, and this page is about learning to run it quickly. (If the word "group" itself is still hazy, revisit the examples of groups first — everything below assumes you have a group already in hand.)

What exactly is a subgroup?

Let G be a group with operation \ast and identity e. A subset H \subseteq G is a subgroup, written H \le G, when H is itself a group under the very same operation \ast inherited from G. That means H has to satisfy all the group axioms on its own:

• Identity: e \in H.
• Closure: if a, b \in H then a \ast b \in H (you can't escape H by combining its members).
• Inverses: if a \in H then a^{-1} \in H.

Associativity comes for free — it is already true everywhere in G, so it is automatically true on the smaller set H. That is the whole reason subgroups are cheap to check: you never re-verify associativity, only that the subset is closed, has the identity, and holds its inverses.

The subgroup tests — the checklist made short

Nobody wants to verify three separate conditions every time. Two theorems compress the work.

Why does "nonempty" replace "contains the identity"? Because it hands the identity to you for free: pick any a \in H (there is one, since H is nonempty); inverses give you a^{-1} \in H; closure then gives you a \ast a^{-1} = e \in H. So the identity is a consequence, not a separate thing to check.

The one-step test is the connoisseur's tool: one line does everything. Take a = b and it hands you a \ast a^{-1} = e \in H (the identity). Take a = e and it hands you e \ast b^{-1} = b^{-1} \in H (inverses). Once you know inverses live in H, replacing b by b^{-1} turns a \ast b^{-1} back into a \ast b — so you get closure too. Everything from one clean inequality-free line.

Worked example: the even integers 2\mathbb{Z} \le \mathbb{Z}

Take the group (\mathbb{Z}, +) and the subset 2\mathbb{Z} = \{\dots, -4, -2, 0, 2, 4, \dots\} of even integers. Here the operation is +, so "a \ast b^{-1}" reads as a - b (the inverse of b under addition is -b). Run the one-step test: take two evens a = 2m and b = 2n. Then

which is even, hence back in 2\mathbb{Z}. The set is nonempty (it holds 0), so the single line is passed and 2\mathbb{Z} \le \mathbb{Z}. Exactly the same argument shows k\mathbb{Z} \le \mathbb{Z} for every integer k — and, remarkably, these turn out to be the only subgroups of \mathbb{Z} there are.

Worked example: \{0, 2, 4\} \le \mathbb{Z}_6

Now a finite one. The group \mathbb{Z}_6 = \{0,1,2,3,4,5\} adds modulo 6. Is the subset of even residues H = \{0, 2, 4\} a subgroup? The cleanest way to see closure is to read it straight off the group's Cayley table — the addition table where cell (i, j) holds (i + j) \bmod 6. Reveal the steps below: the 3 \times 3 block sitting where both the row and the column belong to \{0, 2, 4\} never once produces an odd number — every entry stays inside H. That visual "the block stays inside itself" is closure.

Closure holds (the block stays put), inverses hold (0 pairs with 0, and 2 pairs with 4 since 2 + 4 = 6 \equiv 0), and the set is nonempty — so \{0, 2, 4\} \le \mathbb{Z}_6. Notice its size, 3, divides the size of the whole group, 6. That is not a coincidence, as we will see in a moment.

Worked example: rotations inside a symmetry group C_n \le D_n

The dihedral group D_n is all 2n symmetries of a regular n-gon: n rotations and n reflections. The rotations alone,

form a subgroup. Compose two rotations and you get another rotation (r^i \ast r^j = r^{(i+j) \bmod n}) — closure; the inverse of a rotation is the rotation the other way ((r^i)^{-1} = r^{n-i}) — inverses; and r^0 is the identity. So C_n \le D_n, a group of size n living inside one of size 2n. The reflections, by contrast, do not form a subgroup — but that is a trap worth its own box below.

The two subgroups every group owns

Two subgroups come free with every group G, no work required:

• the trivial subgroup \{e\} — just the identity. It is closed (e \ast e = e), it holds its inverse (e^{-1} = e), and it is nonempty. The smallest possible subgroup.
• the whole group G itself — every group is a subgroup of itself. The largest possible subgroup.

Any subgroup that is neither of these is called a proper, nontrivial subgroup — and those are where the interesting structure lives. Ordering all of a group's subgroups by inclusion draws its subgroup lattice. For \mathbb{Z}_6 it is a tidy diamond: at the bottom the trivial \{0\}; above it the two proper subgroups \{0, 3\} (size 2) and \{0, 2, 4\} (size 3); and at the top all of \mathbb{Z}_6.

Read off the sizes — 1, 2, 3, 6 — and every one of them divides 6. That pattern is no accident.