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For a group $G$ let $i(G) = \{n \ge 1 \mid \exists g \in G, \mathrm{ord}(g)=n \}$ denote the set of integers that occur as order of an element in $G$. Which of the following is/are true?

  1. If $n,m \in i(G)$ then $\gcd(n,m) \in i(G)$.
  2. If $n,m \in i(G)$ then $\mathrm{lcm}(n,m) \in i(G)$.
  3. For any prime $p$ and natural number $n$ there exists $G$ such that $i(G) = \left\{1,p,p^2,...,p^n\right\}$.

I am adding now my answers and I just need verification:

From Lagrange's theorem, every order of a subgroup (even of an element) divides the order of the group itself. So, if $n,m \in i(G)$ and $o(G)$ is the order of $G$, we have that: $n|o(G)$ and $m|o(G)$. So, if let's say $n \ge m$, $\gcd(n,m)=m$ (the smaller number) and $\mathrm{lcm}(n,m)=n$ (the bigger number) and both are in $i(G)$ (first two questions are true). For the 3rd question, using Lagrange's theorem again we have that: $p|o(G),p^2|o(G) \implies o(G)=kp,o(G)=lp^2 \implies p=\frac{k}{l} \implies $ let us choose $k=p$ and $l=1$. That means $o(G)=p^2$, but that can be made to be not possible, if we choose $n=p^2$ and so $|i(G)|=p^2+1>p^2$, a contradiction. So the last question is false.

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    For 1) there is even a stronger statement: if $n \in i(G)$ then $d \in i(G)$ for every divisor of $,$. Indeed if $d | n$ then for $g \in G$ of order $n$, $g^{d'}$ (where $d'd = n$) has order $d$. – 2017-02-21

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Full answer:

For $\gcd(n,m)$ as noted in the comments by Marc Bogaerts, there is a stronger result. Suppose $g\in G$ with $o(g)=n$ and $d|n$ then $g^{n/d}$ has order $d$.

For $\mathrm{lcm}(n,m)$ this does not hold - e.g. $S_3$ has elements of order $2$ and $3$ but not $6$.

For the third part, $G$ could be for example the cyclic group of order $p^n$. If $G=\langle x \mid x^{p^n}=1\rangle$ then $x^{p^{n-r}}$ has order $p^r$ and any element of $G$ has order dividing $p^n$ so $i(G)=\{1,p,\dots,p^n\}$.

Why your proof is incorrect:

You say $n|o(G)$ and $m|o(G)$ then $\gcd(n,m)=\min(n,m)$ which is false. Consider for example $\Bbb{Z}_6$ in which $2$ has order $3$ and $3$ has order $2$, but $\gcd(2,3)=1$.

Your explanation for the third part doesn't make a lot of sense. For any such $p,n$ you want to find $G$, but you seem to find $G$ and then change $n$ to show that your choice of $G$ doesn't work. If you choose $o(G)=p^2$ then $i(G)$ is $\{1,p\}$ or $\left\{1,p,p^2\right\}$. If however you choose $o(G)=p^n$ then as I've shown above there is a choice of $G$ with the correct $i(G)$.

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    I edited my question: it was $n \ge 1$ not $n \gt 1$. So first question must be true: can you verify my proof for that one? – 2017-02-21
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    The first part is true, but I'm afraid your proof is incorrect. When I get back to my computer later, I'll edit my answer to give a full explanation. – 2017-02-21