Question

p and Q are subgroups of a group G and o(P) and o(Q) relatively prime prove that p intersection Q=e,e belonge to G

Answer 1

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.