9.1 Subgroups

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Definition 9.1:
A subgroup of a group G is a subset H \subseteq G that is itself a group with the same operation as G. We sometimes write H \leq G.
Remark:
It is important that the group operation is the same. For example \mathbb{R}\setminus\{0\} is a subset of \mathbb{R}, but we do not regard (\mathbb{R}\setminus\{0\},\times) as a subgroup of (\mathbb{R},+) since the group operations are different.
Example:
For any group G, the trivial subgroup \{e\} and G itself are subgroups of G.
Definition 9.2:
We call a subgroup not equal to \{e\} a non-trivial subgroup, and a subgroup not equal to G a proper subgroup of G.
Example:
We have (\mathbb{Q},+) is a proper subgroup of (\mathbb{R},+). We have (\mathbb{Z},+) is a proper subgroup of both (\mathbb{Q},+) and (\mathbb{R},+).
Example:

The group (2\mathbb{Z},+) is a subgroup of (\mathbb{Z},+).

If n is a positive integer, then the group (n\mathbb{Z},+) is a subgroup of (\mathbb{Z},+).
Example:
The group of rotations of a regular n-sided polygon (i.e., \{e,a,a^2,\dots,a^{n-1}\}) is a subgroup of the dihedral group D_{2n}.

Here’s a simple description of what needs to be checked for a subset to be a subgroup.

Theorem 9.3:

Let G be a multiplicatively written group and let H\subseteq G be a subset of G, Then H is a subgroup of G if and only if the following conditions are satisfied.

  • (Closure) If x,y\in H then xy\in H.

  • (Identity) e\in H.

  • (Inverses) If x\in H then x^{-1}\in H.

Proof.

We don’t need to check associativity, since x(yz)=(xy)z is true for all elements of G, so is certainly true for all elements of H. So the conditions imply H is a group with the same operation as G.

If H is a group, then (Closure) must hold. By uniqueness of identity and inverses, the identity of H must be the same as that of G, and the inverse of x\in H is the same in H as in G, so (Identity) and (Inverses) must hold.

Proposition 9.4:

If H,K are two subgroups of a group G, then H\cap K is also a subgroup of G.

Proof.

We check the three properties in the theorem.

  • If x,y\in H\cap K then xy\in H by closure of H, and xy\in K by closure of K, and so xy\in H\cap K.

  • e\in H and e\in K, so e\in H\cap K.

  • If x\in H\cap K, then x^{-1}\in H since H has inverses, and x^{-1}\in K since K has inverses. So x^{-1}\in H\cap K.