Let (G,*) be a group and let K be subset of G. Then K is a subgroupMathworldPlanetmathPlanetmath of G defined under the same operationMathworldPlanetmath if K is a group by itself (with respect to *), that is:

  • K is closed under the * operation.

  • There exists an identity elementMathworldPlanetmath eK such that for all kK, k*e=k=e*k.

  • Let kK then there exists an inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath k-1K such that k-1*k=e=k*k-1.

The subgroup is denoted likewise (K,*). We denote K being a subgroup of G by writing KG.

In addition the notion of a subgroup of a semigroupPlanetmathPlanetmath can be defined in the following manner. Let (S,*) be a semigroup and H be a subset of S. Then H is a subgroup of S if H is a subsemigroup of S and H is a group.


  • The set {e} whose only element is the identityPlanetmathPlanetmath is a subgroup of any group. It is called the trivial subgroup.

  • Every group is a subgroup of itself.

  • The null set {} is never a subgroup (since the definition of group states that the set must be non-empty).

There is a very useful theorem that allows proving a given subset is a subgroup.

If K is a nonempty subset of the group G. Then K is a subgroup of G if and only if s,tK implies that st-1K.

Proof: First we need to show if K is a subgroup of G then st-1K. Since s,tK then st-1K, because K is a group by itself.
Now, suppose that if for any s,tKG we have st-1K. We want to show that K is a subgroup, which we will accomplish by proving it holds the group axioms.

Since tt-1K by hypothesisMathworldPlanetmath, we conclude that the identity element is in K: eK. (Existence of identity)

Now that we know eK, for all t in K we have that et-1=t-1K so the inverses of elements in K are also in K. (Existence of inverses).

Let s,tK. Then we know that t-1K by last step. Applying hypothesis shows that


so K is closed under the operation. QED


  • Consider the group (,+). Show that(2,+) is a subgroup.

    The subgroup is closed under addition since the sum of even integers is even.

    The identity 0 of is also on 2 since 2 divides 0. For every k2 there is an -k2 which is the inverse under addition and satisfies -k+k=0=k+(-k). Therefore (2,+) is a subgroup of (,+).

    Another way to show (2,+) is a subgroup is by using the propositionPlanetmathPlanetmath stated above. If s,t2 then s,t are even numbers and s-t2 since the differencePlanetmathPlanetmath of even numbers is always an even number.

See also:

  • Wikipedia,

Title subgroup
Canonical name Subgroup
Date of creation 2013-03-22 12:02:10
Last modified on 2013-03-22 12:02:10
Owner Daume (40)
Last modified by Daume (40)
Numerical id 18
Author Daume (40)
Entry type Definition
Classification msc 20A05
Related topic Group
Related topic Ring
Related topic FreeGroup
Related topic Cycle2
Related topic Subring
Related topic GroupHomomorphism
Related topic QuotientGroup
Related topic ProperSubgroup
Related topic SubmonoidSubsemigroup
Related topic ProofThatGInGImpliesThatLangleGRangleLeG
Related topic AbelianGroup2
Related topic EssentialSubgroup
Related topic PGroup4
Defines trivial subgroup