PROVING A SUBGROUP UNDER CERTAIN CONDITIONS
BENJAMIN BRISKEY
Abstract. In this problem, we are given certain conditions under which groups AB and BA are
derived from subgroups A and B. We will show that if BA is a subset of AB, then AB is a subgroup
of G.
Theorem 1. Let G be a group with subgroups A and B. Let AB = {x : x = ab for some a ∈ A
and some b ∈ B} and BA = {x : x = ba for some b ∈ B and some a ∈ A}. If BA ⊂ AB, then AB
is a subgroup of G.
Proof. To show that AB is a subgroup of G, we will show three criteria:
(1) e ∈ AB
(2) xy ∈ AB; for some x, y ∈ AB
(3) x−1 ∈ AB whenever x ∈ AB
Because A and B are groups, the identity element e is defined in each group. Because ee = e, then
clearly e ∈ AB.
To prove the second criteria, let x = a1 b1 ; a1 ∈ A, b1 ∈ B and y = a2 b2 ; a2 ∈ A, b2 ∈ B. When we
plug these values into xy, we get the following: xy = a1 b1 a2 b2 . We can now observe some things
about this equation.
xy = a1 b1 a2 b2
xy = a1 αb2 ; where α = b1 a2 ∈ BA
Because BA ⊂ AB, α can be written as a3 b3 for some a3 ∈ A and b3 ∈ B. We substitute this in
for α, giving us to the following equation:
xy = a1 a3 b3 b1
We can now consolidate terms; a1 a3 ∈ A by closure and b3 b1 ∈ B likewise. Therefore, we can represent a1 a3 as a4 for some a4 ∈ A and b3 b1 as b4 for some b4 ∈ B. Hence, we end the reduction as
follows:
xy = a4 b4
This shows that xy ∈ AB
By definition, some x ∈ AB can be expressed as x = ab, where a ∈ A and b ∈ B. Therefore,
x−1 = b−1 a−1 . By definition of a group, any element has an inverse in that group, in other words
b−1 ∈ B and a−1 ∈ A. Therefore b−1 a−1 ∈ BA. Again, BA ⊂ AB, so we can safely say that
x−1 ∈ AB whenever x ∈ AB.
We can now say that because the three criteria have been established, we know that AB is a
subgroup of G.
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