Many groups live inside bigger ones: the even integers inside $\mathbb{Z}$, the rotations inside all symmetries of a square, the matrices of determinant $1$ inside all invertible matrices, and smaller groups of units inside the groups $U(n)$. Studying these pieces is the main way we take a group apart.
Definition. A subset $H$ of a group $G$ is a subgroup, written $H \le G$, if $H$ is a group under the operation of $G$. Every group has the subgroups $\{e\}$ and $G$.
Associativity is inherited automatically, so only three things need checking: $H$ contains $e$, $H$ is closed under the operation, and $H$ is closed under inverses. These can be packed into one test.
Why it works
Proof. Pick $h \in H$. Then $e = hh^{-1} \in H$. For $b \in H$, $b^{-1} = eb^{-1} \in H$. For $a, b \in H$ we now know $b^{-1} \in H$, so $ab = a(b^{-1})^{-1} \in H$. Associativity holds in $H$ because it holds in $G$. $\square$
Finite subgroup test. If $H$ is finite and nonempty, closure alone is enough. The reason: for $a \in H$, the powers $a, a^{2}, a^{3}, \dots$ all lie in $H$, so two of them are equal, $a^{i} = a^{j}$ with $i < j$; cancelling gives $a^{j-i} = e$. Then $e \in H$ and $a^{-1} = a^{j-i-1} \in H$.
Worked example: A subgroup of $\mathbb{Z}_{12}$
- $H$ is finite, so check closure under addition mod $12$.
- Every sum of multiples of $3$ is a multiple of $3$, and reducing mod $12$ keeps it a multiple of $3$ (since $3 \mid 12$).
Answer: Yes.
Worked example: A subgroup of $U(15)$
- Squares: $4^{2} = 16 \equiv 1$, $11^{2} = 121 \equiv 1$, $14^{2} = 196 \equiv 1$.
- Mixed products: $4 \cdot 11 = 44 \equiv 14$, $4 \cdot 14 = 56 \equiv 11$, $11 \cdot 14 = 154 \equiv 4$.
- All products stay in $H$, and $H$ is finite.
Answer: Yes.
Worked example: Matrices of determinant 1
- It contains $I$, so it is nonempty.
- For $A, B$ in it, $\det(AB^{-1}) = \det A / \det B = 1/1 = 1$.
- By the one-step test, it is a subgroup.
Answer: It is a subgroup.
Common mistake
Key idea