Some Equivalent Conditions for a Finite Abelian Group to Be Cyclic

Lemma. Let $a$ be an element of largest order in a finite abelian group $G$, that is,
$$
o(a)=\max\{o(g)\mid g\in G\}.
$$
Then for every $g\in G$, we have
$$
o(g)\mid o(a),
$$
where $o(a)$ denotes the order of $a$. It follows that $o(a)$ is the least common multiple of the orders of all elements of $G$.

Proof : Suppose otherwise that there exists $g\in G$ such that
$$
o(g)\nmid o(a).
$$
Then necessarily
$$
[o(g),o(a)]>o(a),
$$
where $[o(g),o(a)]$ denotes the least common multiple of $o(g)$ and $o(a)$. Since $G$ is a finite abelian group, there exists an element in $G$ whose order is equal to
$$
[o(g),o(a)].
$$
This contradicts the definition of $a$ as an element of largest order. Therefore, for every $g\in G$,
$$
o(g)\mid o(a).
$$

Proposition 1. A finite abelian group $G$ is cyclic if and only if
$$
|G|=\max\{o(g)\mid g\in G\}.
$$

Proof : Necessity is immediate. If $G$ is cyclic and generated by $a\in G$, then
$$
|G|=o(a)=\max\{o(g)\mid g\in G\}.
$$

For sufficiency, suppose th

at
$$
o(a)=\max\{o(g)\mid g\in G\}.
$$
Let $\langle a\rangle$ be the cyclic subgroup generated by $a$. Then on the one hand,
$$
\langle a\rangle\subseteq G.
$$
On the other hand, since
$$
|G|=|\langle a\rangle|=o(a),
$$
it follows that
$$
G=\langle a\rangle.
$$
Hence $G$ is cyclic.

Proposition 2. A finite abelian group $G$ is cyclic if and only if $|G|$ is the least common multiple of the orders of all elements of $G$.

Proof : This follows immediately from the lemma and Proposition 1.

Proposition 3. A finite abelian group $G$ is cyclic if and only if $|G|$ is the smallest positive integer $n$ such that
$$
x^n=e,\qquad \forall x\in G.
$$

Proof : Necessity is immediate.

For sufficiency, argue by contradiction. Assume that $G$ is not cyclic. Let $a$ be an element of largest order in $G$, and write
$$
o(a)=m.
$$
By Proposition 1, we have
$$
m<|G|.
$$
By the lemma, $m$ is the least positive integer such that
$$
x^m=e,\qquad \forall x\in G.
$$
This contradicts the assumption that $|G|$ is the smallest positive integer with this property. Therefore $G$ must be cyclic.

Proposition 4. A finite abelian group $G$ is cyclic if and only if for every positive integer $k$, the equation
$$
x^k=e
$$
has at most $k$ solutions in $G$.

Proof : First assume that $G$ is cyclic. For any positive integer $k$, let
$$
H=\{x\in G\mid x^k=e\}.
$$
It is easy to verify that $H$ is a subgroup of $G$. Since $G$ is cyclic, $H$ is also cyclic. Hence, by Proposition 1 and the definition of $H$, we have
$$
|H|=\max\{o(h)\mid h\in H\}\le k.
$$
Therefore the equation
$$
x^k=e
$$
has at most $k$ solutions in $G$.

Conversely, let $a$ be an element of largest order in the finite abelian group $G$, and write
$$
o(a)=n.
$$
By the lemma, for every $g\in G$, we have
$$
o(g)\mid n,
$$
so every $g\in G$ is a solution of the equation
$$
x^n=e.
$$
Hence, by the assumption,
$$
|G|\le n.
$$
On the other hand, since $o(a)=n$, the elements
$$
e,a,a^2,\cdots,a^{n-1}
$$
are pairwise distinct, and
$$
\{e,a,a^2,\cdots,a^{n-1}\}\subseteq G.
$$
Thus
$$
|G|\ge n.
$$
Therefore
$$
|G|=n.
$$
By Proposition 1, it follows that $G$ is cyclic.

The cover image of this article was taken in Irkutsk Oblast, Russia.