# Intersection of two subgroups of a group is again a subgroup

**Group**** : **

It is a set equipped with a binary operation that combines any two elements to form a third element in such a way that three conditions called group axioms are satisfied, namely associativity, identity, and invertibility.

**Subgroup**** : **

If a non-void subset H of a group G is itself a group under the operation of G, we say H is a subgroup of G.

**To Prove : **

Prove that the intersection of two subgroups of a group G is again a subgroup of G.

**Proof : **

Let H_{1 } and H_{2} be any two subgroups of G.

Then,

H_{1 }∩ H_{2 }≠ ∅

Since at least the identity element ‘e’ is common to both H_{1} and H_{2} .

In order to prove that H_{1 } ∩ H_{2 } is a subgroup, it is sufficient to prove that

a ∈ H_{1 }∩ H_{2}, b ∈ H_{1}∩ H_{2}⇢ a b^{-1}∈ H_{1 }∩ H_{2}

Now,

a ∈ H_{1}∩ H_{2 }⇢ a ∈ H_{1}and a ∈ H_{2}b ∈ H_{1 }∩ H_{2}⇢ b ∈ H_{1}and b ∈ H_{2}

Since H_{1} and H_{2} are subgroups.

Therefore,

a ∈ H_{1}, b ∈ H_{1}⇢ ab^{-1}∈ H_{1}

and

a ∈ H_{2}, b ∈ H_{2}⇢ ab^{-1}_{ }∈ H_{2}

Thus,

ab^{-1 }∈ H_{1}and ab^{-1}_{ }∈ H_{2}⇢ ab^{-1}∈ H_{1}∩ H_{2}

Hence, H1 ∩ H2 is a subgroup of G and that is our theorem i.e. The intersection of two subgroups of a group is again a subgroup.

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