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
    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
    Jan 14, 2019 at 14:01
  • $\begingroup$ Thank you for pointing it out. I'm gonna go ahead and correct it. $\endgroup$
    – Andrew
    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
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.