Let "P \\ and \\ Q" are two subgroup of "G" and "O(P) \\ and \\ O(Q)" are relatively prime ,i,e,"gcd(O(P),O(Q))=1."
Claim: "P\\cap Q=\\{ e\\}" ,where "e" is the identity element of "G" .
Since, intersection of two subgroup is again a subgroup.
"\\therefore \\ P\\cap Q" is a subgroup of "G."
But "P\\cap Q\\sube P \\ \\text{as well as } \\ P\\cap Q \\sube Q" , so "P\\cap Q" is a subgroup of "P" as well as "Q" by definition of subgroup.
Therefore by Lagrange's theorem, i,e, order of every subgroup divided the order of the group .We get
"O(P\\cap Q) \\mid O(P) \\ and \\ O(P\\cap Q) \\mid O(Q)"
"\\implies O(P\\cap Q)\\mid gcd(O(P),O(Q))."
"\\implies O(P\\cap Q)\\mid 1"
"\\implies O(P\\cap Q)=1" .
Hence,"P\\cap Q=\\{e\\}" .Since the subgroup of order one in a group is identity element itself.
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