Question

Let H and K, each of prime order P, be subgroups of a group G . If H $\ne$K, prove that $H\cap K=<e>$

Short answer

We proved that if $H\ne K$,then we have$H\cap K=<e>$.

Step-by-step solution

To find subgroups with their orders

We have the definition of a subgroup that,

Let $(G,*)$ be a group under binary operation, say $*$. A non-empty subset H of G is said to be a subgroup of G , if $(H,*)$ is itself a group.

Let G be a group and H,K be subgroups of G .

Suppose the order of H and Kis prime.

Let $o(H)=p$ and $o\left(K\right)=q$.

We have to show that $H\cap K=<e>$.

To show H ∩ K = <e>

Now, $H\cap K$ is a subgroup of H and $H\cap K$ is a subgroup of K.

$\therefore$By Lagrange’s Theorem,

$o(H\cap K)|o(H)$and $o(H\cap K)|o(K)$

i.e., localid="1652261117470" $o(H\cap K)|o(p)$and $o(H\cap K)|o(q)$

Now, $(p,q)=1......$(since H and K have prime orders)

role="math" localid="1652261373199" $\Rightarrow o(H\cap K)=1$

$\Rightarrow H\cap K=<e>,$where $H\ne K$

Hence, if $H\ne K$, then we have $H\cap K=<e>$.