Question

If H and K are subgroups of finite group G , prove that $|H\cap K|$is a common divisor of $\left|H\right|$and $\left|K\right|$ .

Short answer

We proved that $|H\cap K|$ is a common divisor of $\left|H\right|$and $\left|K\right|$.

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 . Then $\left|H\right|||G|$ and $\left|K\right|||G|$.

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

Therefore, $o\left(\right|H\left|\right)|o(G)$and $o\left(\right|K\left|\right)|o(G)$

We have to show that $|H\cap K|$is a common divisor of $\left|H\right|$and $\left|K\right|$.

To show |H ∩ K|  is a common divisor of |H|  and  |K|

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

$\therefore$By Lagrange’s Theorem,

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

i.e., $o\left(\right|H\cap K\left|\right)|p$and $o\left(\right|H\cap K\left|\right)|q$

$\Rightarrow o\left(\right|H\cap K\left|\right)=1$

This implies that 1 is the only element (identity element) that divides $o\left(\right|H\left|\right)$and $o\left(\right|K\left|\right)$.

Hence, $|H\cap K|||H|$and $|H\cap K|||K|.$

Hence,$|H\cap K|$is a common divisor of $\left|H\right|$ and $\left|K\right|$.