Question

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

Short answer

We proved that, if $H\ne K$, then we have .

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 $G$.

Suppose order of $H$and $K$is prime.

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

We have to show that .

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,

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

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

Now, $\left(p, q\right)=1$……… (since, localid="1659446755440" $H$and$K$have prime orders)

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

, where $H\ne K$

Hence, if $H\ne K$, then we have .