Find all non-isomorphic abelian groups s.t. $|G| leq 30$ and $g^{12}=1, , forall gin G$












5












$begingroup$


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
$$
g^{12}=1,, forall gin Gtag{1}
$$
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 $gin G$ if and only if each
$p_i^{a_i}$ divides $12$




, obtaining the cases:
$$
begin{align*}
&bullet |G|=1 longrightarrow G={0} \
&bullet |G|=2 longrightarrow G=mathbb{Z_2} \
&... \
&bullet|G|=24 longrightarrow G= mathbb{Z}_2 times mathbb{Z}_{12}bigg|mathbb{Z}_4 times mathbb{Z}_6bigg|mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 \
end{align*}
$$
How could I verify the correctness and prove the sentence in bold? Is it a derivation of Cauchy's theorem?






abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Note that in $Bbb Z_{24}$, there are elements (such as $1$) which have order $24$.
    $endgroup$
    – Arthur
    Jan 14 at 13:58












  • $begingroup$
    @JyrkiLahtonen In the same vein, $Bbb Z_{12}times Bbb Z_2$ is listed three times.
    $endgroup$
    – Arthur
    Jan 14 at 14:01










  • $begingroup$
    Thank you for pointing it out. I'm gonna go ahead and correct it.
    $endgroup$
    – Jevaut
    Jan 14 at 14:02






  • 3




    $begingroup$
    Condition (1) is false. For example $(mathbb{Z}/2mathbb{Z})^4$ 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 at 14:03


















5












$begingroup$


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
$$
g^{12}=1,, forall gin Gtag{1}
$$
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 $gin G$ if and only if each
$p_i^{a_i}$ divides $12$




, obtaining the cases:
$$
begin{align*}
&bullet |G|=1 longrightarrow G={0} \
&bullet |G|=2 longrightarrow G=mathbb{Z_2} \
&... \
&bullet|G|=24 longrightarrow G= mathbb{Z}_2 times mathbb{Z}_{12}bigg|mathbb{Z}_4 times mathbb{Z}_6bigg|mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 \
end{align*}
$$
How could I verify the correctness and prove the sentence in bold? Is it a derivation of Cauchy's theorem?






abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Note that in $Bbb Z_{24}$, there are elements (such as $1$) which have order $24$.
    $endgroup$
    – Arthur
    Jan 14 at 13:58












  • $begingroup$
    @JyrkiLahtonen In the same vein, $Bbb Z_{12}times Bbb Z_2$ is listed three times.
    $endgroup$
    – Arthur
    Jan 14 at 14:01










  • $begingroup$
    Thank you for pointing it out. I'm gonna go ahead and correct it.
    $endgroup$
    – Jevaut
    Jan 14 at 14:02






  • 3




    $begingroup$
    Condition (1) is false. For example $(mathbb{Z}/2mathbb{Z})^4$ 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 at 14:03
















5












5








5


1



$begingroup$


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
$$
g^{12}=1,, forall gin Gtag{1}
$$
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 $gin G$ if and only if each
$p_i^{a_i}$ divides $12$




, obtaining the cases:
$$
begin{align*}
&bullet |G|=1 longrightarrow G={0} \
&bullet |G|=2 longrightarrow G=mathbb{Z_2} \
&... \
&bullet|G|=24 longrightarrow G= mathbb{Z}_2 times mathbb{Z}_{12}bigg|mathbb{Z}_4 times mathbb{Z}_6bigg|mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 \
end{align*}
$$
How could I verify the correctness and prove the sentence in bold? Is it a derivation of Cauchy's theorem?






abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism







share|cite|improve this question











$endgroup$




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
$$
g^{12}=1,, forall gin Gtag{1}
$$
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 $gin G$ if and only if each
$p_i^{a_i}$ divides $12$




, obtaining the cases:
$$
begin{align*}
&bullet |G|=1 longrightarrow G={0} \
&bullet |G|=2 longrightarrow G=mathbb{Z_2} \
&... \
&bullet|G|=24 longrightarrow G= mathbb{Z}_2 times mathbb{Z}_{12}bigg|mathbb{Z}_4 times mathbb{Z}_6bigg|mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 \
end{align*}
$$
How could I verify the correctness and prove the sentence in bold? Is it a derivation of Cauchy's theorem?






abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism




abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 14:08







Jevaut

















asked Jan 14 at 13:55









JevautJevaut

1,151112




1,151112








  • 3




    $begingroup$
    Note that in $Bbb Z_{24}$, there are elements (such as $1$) which have order $24$.
    $endgroup$
    – Arthur
    Jan 14 at 13:58












  • $begingroup$
    @JyrkiLahtonen In the same vein, $Bbb Z_{12}times Bbb Z_2$ is listed three times.
    $endgroup$
    – Arthur
    Jan 14 at 14:01










  • $begingroup$
    Thank you for pointing it out. I'm gonna go ahead and correct it.
    $endgroup$
    – Jevaut
    Jan 14 at 14:02






  • 3




    $begingroup$
    Condition (1) is false. For example $(mathbb{Z}/2mathbb{Z})^4$ 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 at 14:03
















  • 3




    $begingroup$
    Note that in $Bbb Z_{24}$, there are elements (such as $1$) which have order $24$.
    $endgroup$
    – Arthur
    Jan 14 at 13:58












  • $begingroup$
    @JyrkiLahtonen In the same vein, $Bbb Z_{12}times Bbb Z_2$ is listed three times.
    $endgroup$
    – Arthur
    Jan 14 at 14:01










  • $begingroup$
    Thank you for pointing it out. I'm gonna go ahead and correct it.
    $endgroup$
    – Jevaut
    Jan 14 at 14:02






  • 3




    $begingroup$
    Condition (1) is false. For example $(mathbb{Z}/2mathbb{Z})^4$ 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 at 14:03










3




3




$begingroup$
Note that in $Bbb Z_{24}$, there are elements (such as $1$) which have order $24$.
$endgroup$
– Arthur
Jan 14 at 13:58






$begingroup$
Note that in $Bbb Z_{24}$, there are elements (such as $1$) which have order $24$.
$endgroup$
– Arthur
Jan 14 at 13:58














$begingroup$
@JyrkiLahtonen In the same vein, $Bbb Z_{12}times Bbb Z_2$ is listed three times.
$endgroup$
– Arthur
Jan 14 at 14:01




$begingroup$
@JyrkiLahtonen In the same vein, $Bbb Z_{12}times Bbb Z_2$ is listed three times.
$endgroup$
– Arthur
Jan 14 at 14:01












$begingroup$
Thank you for pointing it out. I'm gonna go ahead and correct it.
$endgroup$
– Jevaut
Jan 14 at 14:02




$begingroup$
Thank you for pointing it out. I'm gonna go ahead and correct it.
$endgroup$
– Jevaut
Jan 14 at 14:02




3




3




$begingroup$
Condition (1) is false. For example $(mathbb{Z}/2mathbb{Z})^4$ 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 at 14:03






$begingroup$
Condition (1) is false. For example $(mathbb{Z}/2mathbb{Z})^4$ 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 at 14:03












1 Answer
1






active

oldest

votes


















2












$begingroup$

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



$$G=bigoplus_{i=1}^nmathbb{Z}/p_i^{a_i}mathbb{Z}$$



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 $p_inot = 2,3$ 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 $12=2^2cdot 3$. 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



$$G=bigoplus_{i=1}^{n_1} mathbb{Z}/2mathbb{Z} oplus bigoplus_{i=1}^{n_2} mathbb{Z}/4mathbb{Z} oplus bigoplus_{i=1}^{n_3} mathbb{Z}/3mathbb{Z}$$



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.






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    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



    $$G=bigoplus_{i=1}^nmathbb{Z}/p_i^{a_i}mathbb{Z}$$



    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 $p_inot = 2,3$ 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 $12=2^2cdot 3$. 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



    $$G=bigoplus_{i=1}^{n_1} mathbb{Z}/2mathbb{Z} oplus bigoplus_{i=1}^{n_2} mathbb{Z}/4mathbb{Z} oplus bigoplus_{i=1}^{n_3} mathbb{Z}/3mathbb{Z}$$



    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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



      $$G=bigoplus_{i=1}^nmathbb{Z}/p_i^{a_i}mathbb{Z}$$



      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 $p_inot = 2,3$ 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 $12=2^2cdot 3$. 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



      $$G=bigoplus_{i=1}^{n_1} mathbb{Z}/2mathbb{Z} oplus bigoplus_{i=1}^{n_2} mathbb{Z}/4mathbb{Z} oplus bigoplus_{i=1}^{n_3} mathbb{Z}/3mathbb{Z}$$



      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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



        $$G=bigoplus_{i=1}^nmathbb{Z}/p_i^{a_i}mathbb{Z}$$



        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 $p_inot = 2,3$ 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 $12=2^2cdot 3$. 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



        $$G=bigoplus_{i=1}^{n_1} mathbb{Z}/2mathbb{Z} oplus bigoplus_{i=1}^{n_2} mathbb{Z}/4mathbb{Z} oplus bigoplus_{i=1}^{n_3} mathbb{Z}/3mathbb{Z}$$



        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.






        share|cite|improve this answer









        $endgroup$



        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



        $$G=bigoplus_{i=1}^nmathbb{Z}/p_i^{a_i}mathbb{Z}$$



        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 $p_inot = 2,3$ 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 $12=2^2cdot 3$. 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



        $$G=bigoplus_{i=1}^{n_1} mathbb{Z}/2mathbb{Z} oplus bigoplus_{i=1}^{n_2} mathbb{Z}/4mathbb{Z} oplus bigoplus_{i=1}^{n_3} mathbb{Z}/3mathbb{Z}$$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 14:10









        YankoYanko

        6,5971529




        6,5971529






























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