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One condition that delivers closure, the identity and inverses together, the subgroups every group already has, and the lattice of subgroups of a cyclic group.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to apply the one-step subgroup test and say why it replaces three separate checks, use the shorter test available for a finite subset, name the cyclic subgroups, the centre and the intersections that every group supplies, list the subgroups of a cyclic group and their sizes, and count them by counting divisors.
Every group met so far has had smaller groups sitting inside it: the rotations inside the symmetries of a square, the even integers inside the integers, the powers of one element inside anything. Those were noticed in passing. This lesson makes them the object of study and gives the cheapest way to recognise one.
$H$ is a subgroup of $G$, written $H \le G$, when $H$ is a subset of $G$ that is itself a group under the same operation. The trivial subgroup is $\{e\}$; a proper subgroup is one that is not the whole of $G$. The centre $Z(G)$ is the set of elements commuting with everything. The cyclic subgroup generated by $g$ is $\langle g \rangle$, the set of powers of $g$.
$H \le G$ means $H \subseteq G$ and $H$ is a group under $G$'s operation. Checking that directly would mean four axioms again — but two of them come free.
Associativity is inherited. The operation is the same operation, so $(ab)c = a(bc)$ holds in $H$ because it holds in $G$. This is never worth checking.
The identity of $H$ is the identity of $G$. If $f$ is an identity for $H$ then $ff = f$, and cancelling in $G$ gives $f = e$. So a subgroup cannot have an identity of its own.
That leaves closure, the identity and inverses, and one condition delivers all three.
> Subgroup test. A subset $H$ of a group $G$ is a subgroup if and only if $H$ is non-empty and $ab^{-1} \in H$ whenever $a, b \in H$.
The proof is short enough to keep in your head: take $b = a$ to get $e \in H$; take the pair $e, b$ to get $b^{-1} \in H$; take the pair $a, b^{-1}$ to get $ab \in H$.
For a finite subset there is an even shorter version: a non-empty finite subset closed under the operation is a subgroup, because the powers of any element must repeat, which produces the identity and the inverses without any extra assumption. The word finite is doing real work there — the positive integers are closed under addition and are no subgroup of $\mathbb{Z}$.
Another way: picture
Think of the Cayley table of $G$ with the elements of $H$ listed first. If $H$ is a subgroup, the top-left block of the table is complete in itself: every product of two elements of $H$ lands inside that block, so the block is the Cayley table of $H$. If $H$ is not a subgroup, some product in that corner points out of it, and that stray entry is exactly the failure of closure.
Another way: steps
To decide whether a subset is a subgroup: 1. Is it empty? If so, no — a subgroup contains the identity. 2. Does it contain the identity of $G$? If not, no, and you are finished. 3. Take general $a, b$ in the subset and ask whether $ab^{-1}$ is in it. 4. If the subset is finite, closure under the operation alone is enough. 5. Never recheck associativity.
Four constructions produce subgroups of any group, with no work at all.
Cyclic subgroups. $\langle g \rangle = \{g^{k} : k \in \mathbb{Z}\}$ is a subgroup for every $g$, because $g^{i}(g^{j})^{-1} = g^{i-j}$. Its size is the order of $g$. Every group is the union of its cyclic subgroups, since every element lies in one.
The centre. $Z(G) = \{z : zg = gz \text{ for all } g\}$. It is a subgroup, and $G$ is abelian exactly when $Z(G) = G$. For the symmetries of a square the centre is $\{e, r^{2}\}$ — only the half turn commutes with everything.
Intersections. If $H$ and $K$ are subgroups then so is $H \cap K$: the test passes in both, so it passes in the intersection. Unions are not, and the standard counterexample is $\langle 2 \rangle \cup \langle 3 \rangle$ inside $\mathbb{Z}$, which contains $2$ and $3$ and not $5$.
Centralisers. For a fixed $a$, the elements commuting with $a$ form a subgroup, and the centre is the intersection of all of them.
The subgroups of $\mathbb{Z}$ are exactly $n\mathbb{Z}$ for $n \ge 0$ — the multiples of a single number — which is the divisibility statement that reappears in unit 5 as every ideal of the integers is principal. The same argument, in a different vocabulary, and it is worth noticing the first time.
Forgetting that the subset must be non-empty. The empty set is closed under $ab^{-1}$ vacuously, and it is not a subgroup. Naming an element of $H$ is the first line of the test, not a formality.
Checking closure and stopping. For an infinite subset that is not enough: the positive integers are closed under addition and contain no inverses. The finite shortcut is only available when the subset is finite.
Rechecking associativity. It is inherited. Time spent on it is time not spent on closure, which is where subsets actually fail.
Thinking a union of subgroups is a subgroup. It almost never is. The union of two subgroups is a subgroup only when one contains the other.
Expecting a subgroup to have its own identity. It cannot: cancellation in $G$ forces the identity of $H$ to be $e$. A subset of the matrices that is a group under multiplication with a different identity matrix is a group, but it is not a subgroup.
Take $H = \{e, r, r^{2}, r^{3}\}$, the four rotations, inside the eight symmetries.
A subset to test.
It is non-empty, and $r^{i}(r^{j})^{-1} = r^{i-j}$, which is a rotation. The test passes.
One line.
So $H \le D_4$, of size $4$ inside a group of size $8$. Note $4$ divides $8$ — the first hint of Lagrange, two lessons ahead.
Size four inside size eight.
Inside the integers modulo $6$, take $H = \{0, 1, 3\}$. It contains the identity, and each element has an inverse in the group.
Two conditions look fine.
But $1 + 3 = 4$, which is not in $H$. Closure fails, so $H$ is not a subgroup.
One product leaves the set.
And it could not have been one whatever the arithmetic: $3$ does not divide $6$ evenly into a subgroup of size $3$ containing $1$, because $\langle 1 \rangle$ is the whole group.
The generated subgroup is already too big.
Non-empty: the identity matrix has determinant $1$.
The first line of the test.
For $A$ and $B$ in the set, $\det(AB^{-1}) = \det A / \det B = 1/1 = 1$.
The test, using that the determinant multiplies.
So it is a subgroup — the special linear group. The same argument shows the determinant-one matrices are the kernel of a homomorphism, which is the subject of unit 4.
Select every subset that is a subgroup of the group it sits in.
This task has no paper form; do it on a device.
Build the proof that a non-empty subset closed under $ab^{-1}$ is a subgroup.
This task has no paper form; do it on a device.
In the integers modulo $12$ under addition, say how many elements each of these subgroups has.
| How many elements | |
|---|---|
| generated by 1 | 12 |
| generated by 2 | |
| generated by 3 | |
| generated by 4 | |
| generated by 6 | |
| generated by 0 |
The two-by-two integer matrices of determinant $1$ form a subgroup of the invertible matrices. Write one of them whose entries are whole numbers and none of which is zero, with $7$ in the top right.
This task has no paper form; do it on a device.
Is $\{0, 1, 3\}$ inside the integers modulo $6$ a subgroup?
How many subgroups does the integers modulo $12$ under addition have, counting the whole group and the trivial one?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The two-by-two integer matrices of determinant $1$ form a subgroup of the invertible matrices. Write one of them whose entries are whole numbers and none of which is zero, with $7$ in the top right.
This task has no paper form; do it on a device.
You can decide whether a subset is a subgroup with one condition, and you can list the subgroups of a cyclic group. Say in your own words why associativity never has to be rechecked. Next: cyclic groups themselves, and the classification that makes them the simplest groups there are.
9. Your turn: is the set of matrices with determinant $1$ a subgroup of the invertible matrices?, step 3