Group of order $n=pq$. Let be $G$ a group of order $n=pq$, where $p$ and $q$ are prime numbers.


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*Let be $x\in$ G and $x\neq$ e, determine possible values of order of $x$.

*Deduct that $G$ has at least one subgroup different from $\{e\}$ and $G$.


Hope you'll help me. Thanks.
 A: Let $g \in G \setminus \{1\}$. Consider $H = \langle g \rangle$, the subgroup generated by $g$. Recall (or prove) that $|H| = \operatorname{ord}(g)$. Since by Lagrange's theorem $|H|$ must divide $pq$, we have that $|H| \in \{p,q,pq\}$. Note that $|H|$ can't be $1$ precisely because $g \neq e$. Now, if $|H|$ is $p$ or $q$, this subgroup has less elements than $G$ and more than $\{e\}$. In particular, $H \neq G$ and $H \neq \{1\}$ which is what you want to prove. 
The remaining case is when $H$ has order $pq$, where necessarily $H =G$ by a cardinality argument and so $H = G \simeq \mathbb{Z}_{pq}$ via $\phi: g^n \in G \mapsto n \in \mathbb{Z}_{pq}$ If this last step is not clear enough, prove that this is an injective morphism and recall that injections from sets of the same size are bijective. Finally, the group $\mathbb{Z}_{pq}$ has a proper nontrivial subgroup generated by $\langle p \rangle$, so by taking the preimage via $\phi$, we get a proper nontrivial subgroup $\phi^{-1}(\langle p \rangle)$.
