# 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?

• Note that in $\Bbb Z_{24}$, there are elements (such as $1$) which have order $24$. – Arthur Jan 14 at 13:58
• @JyrkiLahtonen In the same vein, $\Bbb Z_{12}\times \Bbb Z_2$ is listed three times. – Arthur Jan 14 at 14:01
• Thank you for pointing it out. I'm gonna go ahead and correct it. – Jevaut Jan 14 at 14:02
• Condition (1) is false. For example $(\mathbb{Z}/2\mathbb{Z})^4$ is of order $16$ which is $2^4$ and does not divide $12$. However all of the elements are of order $12$. – Yanko Jan 14 at 14:03

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.