Question

If $\mathit{p}$ and $\mathit{q}$are primes, show that every proper subgroup of a group of order$\mathit{p}\mathit{q}$ is cyclic.

Short answer

It is shown that subgroup of a group of order$pq$ is cyclic.

Step-by-step solution

Determine Triviality

Consider the given function, $p,q$is the prime integers and $G$is the group of order $pq$.

Let’s consider the role="math" localid="1654349454454" $H$is the proper subgroup of role="math" localid="1654349451576" $G$.

By applying the Lagrange’s theorem, then the order of $H$divides $pq$then the result is the order of role="math" localid="1654349583570" $H$is role="math" localid="1654349587509" $1$or role="math" localid="1654349590675" $p$or role="math" localid="1654349594182" $q$.

When the order of$H=1$ then the result follows the triviality.

Determine pq is cyclic

Now consider the$p$is the order of$H$and$h\in H$is the non-trivial element.

Then by applying the Lagrange’s theorem, then the order of $h$ is $p$. As $p$is prime.

Therefore, $H$is cyclic group generated by $h$.

Now consider the $q$is the order of$H$ and$h\text{'}\in H$ is the non-trivial element.

Then by applying the langrage’s theorem, then the order of$h\text{'}$ is$q$ . As$q$ is prime.

Therefore,$H$ is the cyclic group generated by$h\text{'}$ .