Finding the group of invertible elements in the ring $Bbb Z_{15}$












0












$begingroup$


I need to find the group of invertible elements of $mathbb{Z}_{15}$.
So the invertible elements here are
$$
U(15)= left{1, 2, 4, 7, 8, 11, 13, 14right}.
$$



By chinese theorem $U(15)=U(5)times U(3)$.



I have found the generators, $a=2$ and $b=7$
So, $a^{2}=4$, $a^{3}=8$, $ab=14$, $a^{2}b=13$, $a^{3}b=11$. So here, $a$ takes order 3, and $b$ order 1



This is enough to say, that $U(15)$ isomorphic to $mathbb{Z}_4timesmathbb{Z}_2$ or no? Is my reasoning is correct and sufficent?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The order of $2$ is not three since $2^{3} neq 2$ in $mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $mathbb{Z}_{15}^{times}$ (what you call $U(15)$).
    $endgroup$
    – Adam Higgins
    Jan 15 at 15:41
















0












$begingroup$


I need to find the group of invertible elements of $mathbb{Z}_{15}$.
So the invertible elements here are
$$
U(15)= left{1, 2, 4, 7, 8, 11, 13, 14right}.
$$



By chinese theorem $U(15)=U(5)times U(3)$.



I have found the generators, $a=2$ and $b=7$
So, $a^{2}=4$, $a^{3}=8$, $ab=14$, $a^{2}b=13$, $a^{3}b=11$. So here, $a$ takes order 3, and $b$ order 1



This is enough to say, that $U(15)$ isomorphic to $mathbb{Z}_4timesmathbb{Z}_2$ or no? Is my reasoning is correct and sufficent?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    The order of $2$ is not three since $2^{3} neq 2$ in $mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $mathbb{Z}_{15}^{times}$ (what you call $U(15)$).
    $endgroup$
    – Adam Higgins
    Jan 15 at 15:41














0












0








0





$begingroup$


I need to find the group of invertible elements of $mathbb{Z}_{15}$.
So the invertible elements here are
$$
U(15)= left{1, 2, 4, 7, 8, 11, 13, 14right}.
$$



By chinese theorem $U(15)=U(5)times U(3)$.



I have found the generators, $a=2$ and $b=7$
So, $a^{2}=4$, $a^{3}=8$, $ab=14$, $a^{2}b=13$, $a^{3}b=11$. So here, $a$ takes order 3, and $b$ order 1



This is enough to say, that $U(15)$ isomorphic to $mathbb{Z}_4timesmathbb{Z}_2$ or no? Is my reasoning is correct and sufficent?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




I need to find the group of invertible elements of $mathbb{Z}_{15}$.
So the invertible elements here are
$$
U(15)= left{1, 2, 4, 7, 8, 11, 13, 14right}.
$$



By chinese theorem $U(15)=U(5)times U(3)$.



I have found the generators, $a=2$ and $b=7$
So, $a^{2}=4$, $a^{3}=8$, $ab=14$, $a^{2}b=13$, $a^{3}b=11$. So here, $a$ takes order 3, and $b$ order 1



This is enough to say, that $U(15)$ isomorphic to $mathbb{Z}_4timesmathbb{Z}_2$ or no? Is my reasoning is correct and sufficent?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 16:14









Bill Dubuque

210k29192640




210k29192640










asked Jan 15 at 15:32









TovarischTovarisch

297




297












  • $begingroup$
    The order of $2$ is not three since $2^{3} neq 2$ in $mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $mathbb{Z}_{15}^{times}$ (what you call $U(15)$).
    $endgroup$
    – Adam Higgins
    Jan 15 at 15:41


















  • $begingroup$
    The order of $2$ is not three since $2^{3} neq 2$ in $mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $mathbb{Z}_{15}^{times}$ (what you call $U(15)$).
    $endgroup$
    – Adam Higgins
    Jan 15 at 15:41
















$begingroup$
The order of $2$ is not three since $2^{3} neq 2$ in $mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $mathbb{Z}_{15}^{times}$ (what you call $U(15)$).
$endgroup$
– Adam Higgins
Jan 15 at 15:41




$begingroup$
The order of $2$ is not three since $2^{3} neq 2$ in $mathbb{Z}$. Similarly the order of $7$ is not one (the only element of order one is $1$). In fact the order of $2$ is $4$, and the order of $7$ is $4$. You are correct that $2,7$ generate $mathbb{Z}_{15}^{times}$ (what you call $U(15)$).
$endgroup$
– Adam Higgins
Jan 15 at 15:41










1 Answer
1






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oldest

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1












$begingroup$

You are correct that $mathbb{Z}_{15}^{times} cong mathbb{Z}_4 times mathbb{Z}_2$, but your reasoning is wrong. Firstly the trivial way to solve this is to note that $mathbb{Z}_{15}^{times}$ is a non-cyclic Abelian group of order 8. Since the only Abelian groups of order 8 are $mathbb{Z}_2^{3}$, and $mathbb{Z}_{4} times mathbb{Z}_2$ we would be done.



However on a more basic level, as you noted, $mathbb{Z}_{15}^{times}$ is an Abelian group generated by $a = 2, b = 7$. However $operatorname{ord}(a) = 4, operatorname{ord}(b) = 4$. But note also that $a^{2}b^{2} = 1$. Hence $mathbb{Z}_{15}^{times}$ has the presentation



$$
mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right>
$$



This might allow you to see that $mathbb{Z}_{15}^{times}$ is also generated by $a = 2$ and $c = ab = 14$, with $operatorname{ord}(c) = 2$. Hence



$$
mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right> cong left< a, c mid ac-ca, a^4, c^2right> cong left<a mid a^4 right> times left< c mid c^2 right>,
$$



and so we're done.




Essentially we have have noted here is that all of the units (invertible elements) in $mathbb{Z}_{15}$ can be written as $pm 2^{k}$ for $k=0,1,2,3$, and that every such element is a unit.






share|cite|improve this answer











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    $begingroup$

    You are correct that $mathbb{Z}_{15}^{times} cong mathbb{Z}_4 times mathbb{Z}_2$, but your reasoning is wrong. Firstly the trivial way to solve this is to note that $mathbb{Z}_{15}^{times}$ is a non-cyclic Abelian group of order 8. Since the only Abelian groups of order 8 are $mathbb{Z}_2^{3}$, and $mathbb{Z}_{4} times mathbb{Z}_2$ we would be done.



    However on a more basic level, as you noted, $mathbb{Z}_{15}^{times}$ is an Abelian group generated by $a = 2, b = 7$. However $operatorname{ord}(a) = 4, operatorname{ord}(b) = 4$. But note also that $a^{2}b^{2} = 1$. Hence $mathbb{Z}_{15}^{times}$ has the presentation



    $$
    mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right>
    $$



    This might allow you to see that $mathbb{Z}_{15}^{times}$ is also generated by $a = 2$ and $c = ab = 14$, with $operatorname{ord}(c) = 2$. Hence



    $$
    mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right> cong left< a, c mid ac-ca, a^4, c^2right> cong left<a mid a^4 right> times left< c mid c^2 right>,
    $$



    and so we're done.




    Essentially we have have noted here is that all of the units (invertible elements) in $mathbb{Z}_{15}$ can be written as $pm 2^{k}$ for $k=0,1,2,3$, and that every such element is a unit.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      You are correct that $mathbb{Z}_{15}^{times} cong mathbb{Z}_4 times mathbb{Z}_2$, but your reasoning is wrong. Firstly the trivial way to solve this is to note that $mathbb{Z}_{15}^{times}$ is a non-cyclic Abelian group of order 8. Since the only Abelian groups of order 8 are $mathbb{Z}_2^{3}$, and $mathbb{Z}_{4} times mathbb{Z}_2$ we would be done.



      However on a more basic level, as you noted, $mathbb{Z}_{15}^{times}$ is an Abelian group generated by $a = 2, b = 7$. However $operatorname{ord}(a) = 4, operatorname{ord}(b) = 4$. But note also that $a^{2}b^{2} = 1$. Hence $mathbb{Z}_{15}^{times}$ has the presentation



      $$
      mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right>
      $$



      This might allow you to see that $mathbb{Z}_{15}^{times}$ is also generated by $a = 2$ and $c = ab = 14$, with $operatorname{ord}(c) = 2$. Hence



      $$
      mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right> cong left< a, c mid ac-ca, a^4, c^2right> cong left<a mid a^4 right> times left< c mid c^2 right>,
      $$



      and so we're done.




      Essentially we have have noted here is that all of the units (invertible elements) in $mathbb{Z}_{15}$ can be written as $pm 2^{k}$ for $k=0,1,2,3$, and that every such element is a unit.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        You are correct that $mathbb{Z}_{15}^{times} cong mathbb{Z}_4 times mathbb{Z}_2$, but your reasoning is wrong. Firstly the trivial way to solve this is to note that $mathbb{Z}_{15}^{times}$ is a non-cyclic Abelian group of order 8. Since the only Abelian groups of order 8 are $mathbb{Z}_2^{3}$, and $mathbb{Z}_{4} times mathbb{Z}_2$ we would be done.



        However on a more basic level, as you noted, $mathbb{Z}_{15}^{times}$ is an Abelian group generated by $a = 2, b = 7$. However $operatorname{ord}(a) = 4, operatorname{ord}(b) = 4$. But note also that $a^{2}b^{2} = 1$. Hence $mathbb{Z}_{15}^{times}$ has the presentation



        $$
        mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right>
        $$



        This might allow you to see that $mathbb{Z}_{15}^{times}$ is also generated by $a = 2$ and $c = ab = 14$, with $operatorname{ord}(c) = 2$. Hence



        $$
        mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right> cong left< a, c mid ac-ca, a^4, c^2right> cong left<a mid a^4 right> times left< c mid c^2 right>,
        $$



        and so we're done.




        Essentially we have have noted here is that all of the units (invertible elements) in $mathbb{Z}_{15}$ can be written as $pm 2^{k}$ for $k=0,1,2,3$, and that every such element is a unit.






        share|cite|improve this answer











        $endgroup$



        You are correct that $mathbb{Z}_{15}^{times} cong mathbb{Z}_4 times mathbb{Z}_2$, but your reasoning is wrong. Firstly the trivial way to solve this is to note that $mathbb{Z}_{15}^{times}$ is a non-cyclic Abelian group of order 8. Since the only Abelian groups of order 8 are $mathbb{Z}_2^{3}$, and $mathbb{Z}_{4} times mathbb{Z}_2$ we would be done.



        However on a more basic level, as you noted, $mathbb{Z}_{15}^{times}$ is an Abelian group generated by $a = 2, b = 7$. However $operatorname{ord}(a) = 4, operatorname{ord}(b) = 4$. But note also that $a^{2}b^{2} = 1$. Hence $mathbb{Z}_{15}^{times}$ has the presentation



        $$
        mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right>
        $$



        This might allow you to see that $mathbb{Z}_{15}^{times}$ is also generated by $a = 2$ and $c = ab = 14$, with $operatorname{ord}(c) = 2$. Hence



        $$
        mathbb{Z}_{15}^{times} cong left< a, b mid ab-ba, a^4, b^4, a^{2}b^{2} right> cong left< a, c mid ac-ca, a^4, c^2right> cong left<a mid a^4 right> times left< c mid c^2 right>,
        $$



        and so we're done.




        Essentially we have have noted here is that all of the units (invertible elements) in $mathbb{Z}_{15}$ can be written as $pm 2^{k}$ for $k=0,1,2,3$, and that every such element is a unit.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 15 at 16:16

























        answered Jan 15 at 15:51









        Adam HigginsAdam Higgins

        605113




        605113






























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