Why is the smallest subgroup of G containing A and B equal to G itself?

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SUMMARY

The smallest subgroup of a group G containing subgroups A and B is equal to G itself when |G| = 15, |A| = 5, and |B| = 3, with A ∩ B = {e_G}. This conclusion follows from Lagrange's Theorem, which states that the order of a subgroup must divide the order of the group. Since the orders of A and B do not share any common elements other than the identity, the combination of A and B generates the entire group G.

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In an example it says that, if [itex]|G| = 15[/itex] and [itex]G[/itex] has subgroups [itex]A,B[/itex] of [itex]G[/itex] with [itex]|A| = 5[/itex] and [itex]|B| = 3[/itex], then [itex]A \cap B[/itex] must equal [itex]\{e_G\}[/itex] and the smallest subgroup of [itex]G[/itex] containing both [itex]A[/itex] and [itex]B[/itex] is [itex]G[/itex] itself. Could anyone explain why? Thanks!
 
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Hint: Lagrange
 

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