Properties of Subgroups

I. The identity element of a subgroup is same as the identity element of the group.

Proof: Let H be a subgroup of a group G.
              Let e and e' be the identity elements of G and H respectively.
              Let a ∈ H be any element.
              ∴    ae' = a'                                                                                     [∵ e' is the identity of H]
              Also ∵ a ∈ H and H ⊆ G
              ⇒ a ∈ G
              ∴    ae = a                                                                                       [∵ e is the identity of G]
              So, we have ae = ae'
              ⇒ e = e'                                                                                            [by left cancellation law]
              Here the identity of a group and that of a subgroup is the same.

II. The inverse of any element of a subgroup is the same as the inverse of the element regarded as the element of the group.

Proof: Let e be the identity element of G and H.
              Let a ∈ H be any element.
              Since H ⊆ G
              ∴ a ∈ G.
              Let b be the inverse of a in H and c be the inverse of a in G.
              ∴ ba = e            and                 ca = e
               ⇒ ba = ca
               ⇒ b = c                                  [by right cancellation law]
               Hence the inverse of any element of a subgroup is same as the inverse of the same element regarded as an element of the group.

III. The order of any element in a subgroup is the same as the order of the element regarded as the element of the group.

Proof: Let e be the identity element of G and H.
              Let a ∈ H such that o(a) = n
              ⇒ aⁿ = e        and          a^m ≠ e  for every m < n.
              Also a ∈ H  ⇒ a ∈ G and so aⁿ = e ∈ G ⇒ o(a) = n in G.
              Hence order of any element in a subgroup is same as the order of the element regarded as the element of the group.

IV. Subgroup of an abelian group is abelian.

Proof: Let H be a subgroup of an abelian group G.
              ∴ H ⊆ G
              Let a, b ∈ H be any two elements.
              ∴ a, b ∈ G    ⇒ ab = ba                         [∵ G is abelian]
              ∴ for all a, b ∈ H      we have ab = ba
              Hence H is an abelian subgroup of G.
              The converse of above result is false.
              i.e., A subgroup may be abelian even if G is not abelian.

                        Product of two subgroups:


Let H and K be two subgroups of a group G, then the set HK defined by
HK = {hk : for all h ∈ H, k ∈ K} is called the product of the subgroups H and K.

Theorem: If H and K are two subgroups of a group G, then HK is a subgroup of G iff HK = KH.

Proof: Necessary Part: Let H and K be two subgroups of a group G such that HK is also a subgroup of G.
We shall prove that HK = KH.
Let x ∈ HK be arbitrary element.
∴ x^-1 ∈ HK, as HK is a subgroup of G.
⇒ x^-1 = hk for some h∈ H, k ∈ K
⇒ (x^-1)^-1 = (hk)^-1
⇒ x = k^-1h^-1
Since k ∈ K and K is a subgroup of G, so k^-1 ∈ K.
Similarly, h^-1 ∈ H.
∴ k^-1h^-1 ∈ KH.
⇒ x ∈ KH
∴ HK ⊆ KH.
Let now x ∈ KH be arbitrary element
∴ x = kh for some k ∈ K, h∈ H.
⇒ x^-1 = (kh)^-1 = h^-1k^-1 ∈ HK
⇒ x^-1 ∈ HK
⇒ (x^-1)^-1 ∈ HK, as HK is a subgroup of G.
⇒ x ∈ HK.
∴ KH ⊆ HK
Thus HK = KH.

Sufficient Part: Suppose that H and K are two subgroups of a group G such that
HK = KH.
We shall prove that HK is a subgroup of G.
Let x, y ∈ HK be arbitrary elements.
∴ x = h₁k₁   and   y = h₂k₂  for some h₁ , h₂ ∈ H and k₁ , k₂ ∈ K
∴ xy^-1 = (h₁k₁)(h₂k₂)^-1 = (h₁k₁)(k₂^-1h₂^-1) = h₁(k₁(k₂^-1h₂^-1))
            = h₁((k₁k₂^-1)h₂^-1)                                                         ... (1)
Now (k₁k₂^-1)h₂^-1 ∈ KH
⇒ (k₁k₂^-1)h₂^-1 ∈ HK,         since  HK = KH
⇒ (k₁k₂^-1)h₂^-1 = h₃k₃ for some h₃ ∈ H and k₃ ∈ K           ... (2)
∴ xy^-1 = h₁(h₃k₃)                                              [From (1) and (2)]
            = (h₁h₃)k₃ ∈ HK
∴ xy^-1 ∈ HK, ∀ x, y ∈ HK.
∴ HK is a subgroup of G.