Question

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

Short answer

We proved that,$\left|H\cap K\right|$ is a common divisor of role="math" localid="1659447475271" $\left|H\right|$and $\left|K\right|$.

Step-by-step solution

To find subgroups with their orders

We have the definition of subgroup that,

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

Let $G$be a group and $H,K$be subgroups of $G$.Then $\left|H\right| | \left|G\right|$and $\left|K\right| | \left|G\right|$.

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

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

We have to show that $\left|H\cap K\right|$is a common divisor of localid="1659447836746" $\left|H\right|$and $\left|K\right|$.

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

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

By Lagrange’s Theorem,

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

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

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

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

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

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