Let $G=Bbb Z_30 times mathbb Z_36$.

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Let $G=Bbb Z_30times mathbb Z_36$.



Divide G to direct product of cyclic gropus from order exponent of prime. Than find all the fairs of generators of G.



I thought about $36times30=1080=2^3times3^3times5$.




group-theory finite-groups cyclic-groups




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  • 1




    Use $Bbb Z_30$ for $Bbb Z_30$.
    – Shaun
    Jul 29 at 12:30















up vote
1
down vote

favorite












Let $G=Bbb Z_30times mathbb Z_36$.



Divide G to direct product of cyclic gropus from order exponent of prime. Than find all the fairs of generators of G.



I thought about $36times30=1080=2^3times3^3times5$.




group-theory finite-groups cyclic-groups




share|cite|improve this question

















  • 1




    Use $Bbb Z_30$ for $Bbb Z_30$.
    – Shaun
    Jul 29 at 12:30













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $G=Bbb Z_30times mathbb Z_36$.



Divide G to direct product of cyclic gropus from order exponent of prime. Than find all the fairs of generators of G.



I thought about $36times30=1080=2^3times3^3times5$.




group-theory finite-groups cyclic-groups




share|cite|improve this question













Let $G=Bbb Z_30times mathbb Z_36$.



Divide G to direct product of cyclic gropus from order exponent of prime. Than find all the fairs of generators of G.



I thought about $36times30=1080=2^3times3^3times5$.





group-theory finite-groups cyclic-groups





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 12:32
























asked Jul 29 at 12:27









user579852

305




305







  • 1




    Use $Bbb Z_30$ for $Bbb Z_30$.
    – Shaun
    Jul 29 at 12:30













  • 1




    Use $Bbb Z_30$ for $Bbb Z_30$.
    – Shaun
    Jul 29 at 12:30








1




1




Use $Bbb Z_30$ for $Bbb Z_30$.
– Shaun
Jul 29 at 12:30





Use $Bbb Z_30$ for $Bbb Z_30$.
– Shaun
Jul 29 at 12:30











1 Answer
1






active

oldest

votes

















up vote
3
down vote













Hint:



$$BbbZ_mncongBbbZ_mtimesBbbZ_niffgcd(m,n)=1$$



You have managed to write $$Bbb Z_30timesBbb Z_36congBbb Z_2timesBbb Z_3timesBbb Z_5timesBbb Z_4timesBbb Z_9tag1$$ Let's try to find the number of generators in easier cases. If we look at the group $Bbb Z_3times Bbb Z_9$, we observe that no single element $(a,b)$ can generate it, since there are $3cdot 9=27$ elements, and $9(a,b)=(0,0)$ (the highest order of any element is $9$, so we can make atmost $9$ different elements with $(a,b)$). This happens because $3mid 9$.



On the other hand $Bbb Z_3timesBbb Z_8$ can be generated by a single element, for instance $(1,1)$ or $(2,3)$,... Which can easily be verified since $Bbb Z_24congBbb Z_3timesBbb Z_8$. This may suggest that $(1)$ requires two generators.



Let's denote this generators $$beginalignalpha=(a,b,c,0,0)\beta=(0,0,0,d,e)endalign$$ How many possibilities are there for $a,b,c,d,e$?



Hint:




How many generators does $Bbb Z_n$ have? $$$$ Let us suppose that $a$ is a generator of $Bbb Z_n$, then $$a,2a,3a,dots,na$$ are all distinct elements, in particular $naequiv_n0$ and $kanotequiv_n0$ for $k=1,2,dots,n-1$. This is the same as saying that the only way to have $nmid ka$ is to have $nmid k$. This is precisely the case when $gcd(a,n)=1$, that is, $a$ generates $Bbb Z_n$ when $a$ and $n$ have no common factor. The generators of $Bbb Z_10$ for instance is $1,3,7,9$.







share|cite|improve this answer























  • I know this but in this case $gcd(30,36)ne1$
    – user579852
    Jul 29 at 12:33










  • @user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
    – cansomeonehelpmeout
    Jul 29 at 12:35











  • you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
    – user579852
    Jul 29 at 12:39







  • 1




    @user579852 Yes! Now, can you expand your product further?
    – cansomeonehelpmeout
    Jul 29 at 12:42






  • 1




    both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
    – user579852
    Jul 29 at 20:54










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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
3
down vote













Hint:



$$BbbZ_mncongBbbZ_mtimesBbbZ_niffgcd(m,n)=1$$



You have managed to write $$Bbb Z_30timesBbb Z_36congBbb Z_2timesBbb Z_3timesBbb Z_5timesBbb Z_4timesBbb Z_9tag1$$ Let's try to find the number of generators in easier cases. If we look at the group $Bbb Z_3times Bbb Z_9$, we observe that no single element $(a,b)$ can generate it, since there are $3cdot 9=27$ elements, and $9(a,b)=(0,0)$ (the highest order of any element is $9$, so we can make atmost $9$ different elements with $(a,b)$). This happens because $3mid 9$.



On the other hand $Bbb Z_3timesBbb Z_8$ can be generated by a single element, for instance $(1,1)$ or $(2,3)$,... Which can easily be verified since $Bbb Z_24congBbb Z_3timesBbb Z_8$. This may suggest that $(1)$ requires two generators.



Let's denote this generators $$beginalignalpha=(a,b,c,0,0)\beta=(0,0,0,d,e)endalign$$ How many possibilities are there for $a,b,c,d,e$?



Hint:




How many generators does $Bbb Z_n$ have? $$$$ Let us suppose that $a$ is a generator of $Bbb Z_n$, then $$a,2a,3a,dots,na$$ are all distinct elements, in particular $naequiv_n0$ and $kanotequiv_n0$ for $k=1,2,dots,n-1$. This is the same as saying that the only way to have $nmid ka$ is to have $nmid k$. This is precisely the case when $gcd(a,n)=1$, that is, $a$ generates $Bbb Z_n$ when $a$ and $n$ have no common factor. The generators of $Bbb Z_10$ for instance is $1,3,7,9$.







share|cite|improve this answer























  • I know this but in this case $gcd(30,36)ne1$
    – user579852
    Jul 29 at 12:33










  • @user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
    – cansomeonehelpmeout
    Jul 29 at 12:35











  • you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
    – user579852
    Jul 29 at 12:39







  • 1




    @user579852 Yes! Now, can you expand your product further?
    – cansomeonehelpmeout
    Jul 29 at 12:42






  • 1




    both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
    – user579852
    Jul 29 at 20:54














up vote
3
down vote













Hint:



$$BbbZ_mncongBbbZ_mtimesBbbZ_niffgcd(m,n)=1$$



You have managed to write $$Bbb Z_30timesBbb Z_36congBbb Z_2timesBbb Z_3timesBbb Z_5timesBbb Z_4timesBbb Z_9tag1$$ Let's try to find the number of generators in easier cases. If we look at the group $Bbb Z_3times Bbb Z_9$, we observe that no single element $(a,b)$ can generate it, since there are $3cdot 9=27$ elements, and $9(a,b)=(0,0)$ (the highest order of any element is $9$, so we can make atmost $9$ different elements with $(a,b)$). This happens because $3mid 9$.



On the other hand $Bbb Z_3timesBbb Z_8$ can be generated by a single element, for instance $(1,1)$ or $(2,3)$,... Which can easily be verified since $Bbb Z_24congBbb Z_3timesBbb Z_8$. This may suggest that $(1)$ requires two generators.



Let's denote this generators $$beginalignalpha=(a,b,c,0,0)\beta=(0,0,0,d,e)endalign$$ How many possibilities are there for $a,b,c,d,e$?



Hint:




How many generators does $Bbb Z_n$ have? $$$$ Let us suppose that $a$ is a generator of $Bbb Z_n$, then $$a,2a,3a,dots,na$$ are all distinct elements, in particular $naequiv_n0$ and $kanotequiv_n0$ for $k=1,2,dots,n-1$. This is the same as saying that the only way to have $nmid ka$ is to have $nmid k$. This is precisely the case when $gcd(a,n)=1$, that is, $a$ generates $Bbb Z_n$ when $a$ and $n$ have no common factor. The generators of $Bbb Z_10$ for instance is $1,3,7,9$.







share|cite|improve this answer























  • I know this but in this case $gcd(30,36)ne1$
    – user579852
    Jul 29 at 12:33










  • @user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
    – cansomeonehelpmeout
    Jul 29 at 12:35











  • you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
    – user579852
    Jul 29 at 12:39







  • 1




    @user579852 Yes! Now, can you expand your product further?
    – cansomeonehelpmeout
    Jul 29 at 12:42






  • 1




    both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
    – user579852
    Jul 29 at 20:54












up vote
3
down vote










up vote
3
down vote









Hint:



$$BbbZ_mncongBbbZ_mtimesBbbZ_niffgcd(m,n)=1$$



You have managed to write $$Bbb Z_30timesBbb Z_36congBbb Z_2timesBbb Z_3timesBbb Z_5timesBbb Z_4timesBbb Z_9tag1$$ Let's try to find the number of generators in easier cases. If we look at the group $Bbb Z_3times Bbb Z_9$, we observe that no single element $(a,b)$ can generate it, since there are $3cdot 9=27$ elements, and $9(a,b)=(0,0)$ (the highest order of any element is $9$, so we can make atmost $9$ different elements with $(a,b)$). This happens because $3mid 9$.



On the other hand $Bbb Z_3timesBbb Z_8$ can be generated by a single element, for instance $(1,1)$ or $(2,3)$,... Which can easily be verified since $Bbb Z_24congBbb Z_3timesBbb Z_8$. This may suggest that $(1)$ requires two generators.



Let's denote this generators $$beginalignalpha=(a,b,c,0,0)\beta=(0,0,0,d,e)endalign$$ How many possibilities are there for $a,b,c,d,e$?



Hint:




How many generators does $Bbb Z_n$ have? $$$$ Let us suppose that $a$ is a generator of $Bbb Z_n$, then $$a,2a,3a,dots,na$$ are all distinct elements, in particular $naequiv_n0$ and $kanotequiv_n0$ for $k=1,2,dots,n-1$. This is the same as saying that the only way to have $nmid ka$ is to have $nmid k$. This is precisely the case when $gcd(a,n)=1$, that is, $a$ generates $Bbb Z_n$ when $a$ and $n$ have no common factor. The generators of $Bbb Z_10$ for instance is $1,3,7,9$.







share|cite|improve this answer















Hint:



$$BbbZ_mncongBbbZ_mtimesBbbZ_niffgcd(m,n)=1$$



You have managed to write $$Bbb Z_30timesBbb Z_36congBbb Z_2timesBbb Z_3timesBbb Z_5timesBbb Z_4timesBbb Z_9tag1$$ Let's try to find the number of generators in easier cases. If we look at the group $Bbb Z_3times Bbb Z_9$, we observe that no single element $(a,b)$ can generate it, since there are $3cdot 9=27$ elements, and $9(a,b)=(0,0)$ (the highest order of any element is $9$, so we can make atmost $9$ different elements with $(a,b)$). This happens because $3mid 9$.



On the other hand $Bbb Z_3timesBbb Z_8$ can be generated by a single element, for instance $(1,1)$ or $(2,3)$,... Which can easily be verified since $Bbb Z_24congBbb Z_3timesBbb Z_8$. This may suggest that $(1)$ requires two generators.



Let's denote this generators $$beginalignalpha=(a,b,c,0,0)\beta=(0,0,0,d,e)endalign$$ How many possibilities are there for $a,b,c,d,e$?



Hint:




How many generators does $Bbb Z_n$ have? $$$$ Let us suppose that $a$ is a generator of $Bbb Z_n$, then $$a,2a,3a,dots,na$$ are all distinct elements, in particular $naequiv_n0$ and $kanotequiv_n0$ for $k=1,2,dots,n-1$. This is the same as saying that the only way to have $nmid ka$ is to have $nmid k$. This is precisely the case when $gcd(a,n)=1$, that is, $a$ generates $Bbb Z_n$ when $a$ and $n$ have no common factor. The generators of $Bbb Z_10$ for instance is $1,3,7,9$.








share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 29 at 22:50


























answered Jul 29 at 12:30









cansomeonehelpmeout

4,7433830




4,7433830











  • I know this but in this case $gcd(30,36)ne1$
    – user579852
    Jul 29 at 12:33










  • @user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
    – cansomeonehelpmeout
    Jul 29 at 12:35











  • you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
    – user579852
    Jul 29 at 12:39







  • 1




    @user579852 Yes! Now, can you expand your product further?
    – cansomeonehelpmeout
    Jul 29 at 12:42






  • 1




    both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
    – user579852
    Jul 29 at 20:54
















  • I know this but in this case $gcd(30,36)ne1$
    – user579852
    Jul 29 at 12:33










  • @user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
    – cansomeonehelpmeout
    Jul 29 at 12:35











  • you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
    – user579852
    Jul 29 at 12:39







  • 1




    @user579852 Yes! Now, can you expand your product further?
    – cansomeonehelpmeout
    Jul 29 at 12:42






  • 1




    both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
    – user579852
    Jul 29 at 20:54















I know this but in this case $gcd(30,36)ne1$
– user579852
Jul 29 at 12:33




I know this but in this case $gcd(30,36)ne1$
– user579852
Jul 29 at 12:33












@user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
– cansomeonehelpmeout
Jul 29 at 12:35





@user579852 That's right, but what if you first look at only $mathbbZ_30$. Can you write $mathbbZ_30congmathbbZ_mtimesmathbbZ_n$?
– cansomeonehelpmeout
Jul 29 at 12:35













you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
– user579852
Jul 29 at 12:39





you mean that $Bbb Z_30 times Z_36cong Z_3 times Z_10 times Z_9 times Z_4$? but then how I find all the fairs of generators of G.
– user579852
Jul 29 at 12:39





1




1




@user579852 Yes! Now, can you expand your product further?
– cansomeonehelpmeout
Jul 29 at 12:42




@user579852 Yes! Now, can you expand your product further?
– cansomeonehelpmeout
Jul 29 at 12:42




1




1




both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
– user579852
Jul 29 at 20:54




both of them can't be genertors because $gcd(3,9)ne1$ and $gcd(6,9)ne1$, right?
– user579852
Jul 29 at 20:54












 

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