Question

P and Q are subgroups of a group G and o(P) , o(Q) are relatively prime.Prove that P intersection Q ={e}.​

Answer 1

Step-by-step explanation:

Let P \ and \ QP and Q are two subgroup of GG and O(P) \ and \ O(Q)O(P) and O(Q) are relatively prime ,i,e,gcd(O(P),O(Q))=1.gcd(O(P),O(Q))=1.

Claim: P\cap Q=\{ e\}P∩Q={e} ,where ee is the identity element of GG .

Since, intersection of two subgroup is again a subgroup.

\therefore \ P\cap Q∴ P∩Q is a subgroup of G.G.

But P\cap Q\sube P \ \text{as well as } \ P\cap Q \sube QP∩Q⊆P as well as P∩Q⊆Q , so P\cap QP∩Q is a subgroup of PP as well as QQ 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)O(P∩Q)∣O(P) and O(P∩Q)∣O(Q)

\implies O(P\cap Q)\mid gcd(O(P),O(Q)).⟹O(P∩Q)∣gcd(O(P),O(Q)).

\implies O(P\cap Q)\mid 1⟹O(P∩Q)∣1

\implies O(P\cap Q)=1⟹O(P∩Q)=1 .

Hence,P\cap Q=\{e\}P∩Q={e} .Since the subgroup of order one in a group is identity element itself.