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?

  • 3
    $\begingroup$ Note that in $\Bbb Z_{24}$, there are elements (such as $1$) which have order $24$. $\endgroup$
    – Arthur
    Commented Jan 14, 2019 at 13:58
  • $\begingroup$ @JyrkiLahtonen In the same vein, $\Bbb Z_{12}\times \Bbb Z_2$ is listed three times. $\endgroup$
    – Arthur
    Commented Jan 14, 2019 at 14:01
  • $\begingroup$ Thank you for pointing it out. I'm gonna go ahead and correct it. $\endgroup$
    – Andrew
    Commented Jan 14, 2019 at 14:02
  • 3
    $\begingroup$ 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 $12$. $\endgroup$
    – Yanko
    Commented Jan 14, 2019 at 14:03

1 Answer 1


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


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.


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