Find all non-isomorphic abelian groups s.t. $|G| \leq 30$ and $g^{12}=1, \, \forall g\in G$ Assume the prime factorization of the order of $G$:
$$
|G|=p_1^{a_1}p_2^{a_2}\dots p_r^{a_r}
$$
The condition
$$
g^{12}=1,\, \forall g\in G\tag{1}
$$
in other words means that we must find those groups whose elements satisfy:
$$
\text{ord} (g) \big|12, \, \forall g \in G
$$
Intuitively, I can see that 

Condition $(1)$ holds for every $g\in G$ if and only if each
  $p_i^{a_i}$ divides $12$

, obtaining the cases:
$$
\begin{align*}
&\bullet |G|=1 \longrightarrow G=\{0\} \\
&\bullet |G|=2 \longrightarrow G=\mathbb{Z_2} \\
&... \\
&\bullet|G|=24 \longrightarrow G= \mathbb{Z}_2 \times \mathbb{Z}_{12}\bigg|\mathbb{Z}_4 \times \mathbb{Z}_6\bigg|\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \\
\end{align*}
$$
How could I verify the correctness and prove the sentence in bold? Is it a derivation of Cauchy's theorem?
 A: You should start a little different and end a little different. Let me suggest how to do it:
First you begin with a finite abelian group $G$ and so it takes the form
$$G=\bigoplus_{i=1}^n\mathbb{Z}/p_i^{a_i}\mathbb{Z}$$
For some primes $p_1,...,p_n$ (not necessarily distinct) and natural numbers $a_1,...,a_n$.

Lemma: All $p_i$ are either $2$ or $3$. If $p_i=2$ then $a_i$ is either $1$ or $2$ and if $p_i=3$ then $a_i=1$.

Proof: If $p_i\not = 2,3$ then the generator of $\mathbb{Z}/p_i^{a_i}\mathbb{Z}$ is of order $p_i^{a_i}$ but it's also of order $12=2^2\cdot 3$. Hence of order $\gcd(p_i^{a_i},12)=1$ and so it's necessarily trivial (thus $a_i=0$). I leave the rest as an exercise.
Using the Lemma we can write
$$G=\bigoplus_{i=1}^{n_1} \mathbb{Z}/2\mathbb{Z} \oplus \bigoplus_{i=1}^{n_2} \mathbb{Z}/4\mathbb{Z} \oplus \bigoplus_{i=1}^{n_3} \mathbb{Z}/3\mathbb{Z}$$
Moreover $|G|\leq 30$ and so $2^{n_1}\cdot 4^{n_2}\cdot 3^{n_3}\leq 30$.
Now you have to run over all possible choices of $n_1,n_2,n_3$ for which the above holds and you finish.
