Question

IfG is a group with more than one element and Ghas no proper subgroups, prove that Gis isomorphic to $Z_p$for some prime p.

Short answer

We proved that is isomorphic to $Z_p$, for some prime p.

Step-by-step solution

To show   is cyclic

We have thedefinition of a subgroupthat

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.

A subgroup H of G is called the proper subgroup of if $H\ne G.$

Let G be a group with more than one element, and H be a subgroup of G.

Suppose G has no proper subgroup.

We have to show that G is isomorphic to $Z_p$, for some prime p.

Let $x\in G$and $x\ne e$, where e is an identity element.

Let $H=<x>$ , a subgroup generated by x.

Since G has no proper subgroup, either $H=\left\{e\right\}$or $H=G$.

But $H\ne \left\{e\right\}$since $x\ne e$ .

localid="1652264558839" $\therefore H=G$

$\Rightarrow G$ is cyclic.

To show Zp is isomorphic to  for some prime  p

Let $o\left(G\right)=m$.

Also, $o(x)=m$…… role="math" localid="1652264789366" $(\because x\in G)$

Suppose m is not prime.

Then $\exists$ an integer ksuch that $k|m$ , where $1<k<m$.

Now, consider the subgroup $S=<x^k>$, subgroup generated by $x^k$.

From the hypothesis, we have either role="math" localid="1652265048003" $S=\left\{e\right\}$or $S=G$.

If $S=\left\{e\right\}$, it implies that $o\left(x\right)=k$, which is impossible as $k<m$.

On the other hand, if $S\hspace{0.17em}=G$, then $m=k$ contradicts our assumption on k .

Thus, 1 and m are the only divisors of .

This implies that m is prime.

Hence, fromTheorem 8.7, we can say that

is isomorphic to Z_p, for some prime p.