Problem involving $mathbb{Z} /63 mathbb{Z}$ and its subgroups.











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Let $A$ be a subgroup of $G=mathbb{Z}/63mathbb{Z}$, where $A$ is generated with $14:=14+63 mathbb{Z}$.



a) Determine the order of $A$ and all subgroups of $G/A$.



b) Is there a non-trivial homomorphism $varphi : A rightarrow G/A$? If yes, find one.



So, the order of $A$, $|A|=9$. I also determined all subgroups of $G/A$, but I'm not sure if I'm right:



$G/A={gA : g in G} = { g+c, cin {14,28,42,56,7,21,35,49,0} }$. But out of this, I concluded that $G/A=G$, so subgroups of $G/A$ are all cyclic subgroups of $mathbb{Z}/63 mathbb{Z}$, which are generated by all $a in G$, such that $gcd(a,63) > 1$.



Am I correct?



Also, what about b)? Any ideas?



Thanks in advance!










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    Let $A$ be a subgroup of $G=mathbb{Z}/63mathbb{Z}$, where $A$ is generated with $14:=14+63 mathbb{Z}$.



    a) Determine the order of $A$ and all subgroups of $G/A$.



    b) Is there a non-trivial homomorphism $varphi : A rightarrow G/A$? If yes, find one.



    So, the order of $A$, $|A|=9$. I also determined all subgroups of $G/A$, but I'm not sure if I'm right:



    $G/A={gA : g in G} = { g+c, cin {14,28,42,56,7,21,35,49,0} }$. But out of this, I concluded that $G/A=G$, so subgroups of $G/A$ are all cyclic subgroups of $mathbb{Z}/63 mathbb{Z}$, which are generated by all $a in G$, such that $gcd(a,63) > 1$.



    Am I correct?



    Also, what about b)? Any ideas?



    Thanks in advance!










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
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      down vote

      favorite











      Let $A$ be a subgroup of $G=mathbb{Z}/63mathbb{Z}$, where $A$ is generated with $14:=14+63 mathbb{Z}$.



      a) Determine the order of $A$ and all subgroups of $G/A$.



      b) Is there a non-trivial homomorphism $varphi : A rightarrow G/A$? If yes, find one.



      So, the order of $A$, $|A|=9$. I also determined all subgroups of $G/A$, but I'm not sure if I'm right:



      $G/A={gA : g in G} = { g+c, cin {14,28,42,56,7,21,35,49,0} }$. But out of this, I concluded that $G/A=G$, so subgroups of $G/A$ are all cyclic subgroups of $mathbb{Z}/63 mathbb{Z}$, which are generated by all $a in G$, such that $gcd(a,63) > 1$.



      Am I correct?



      Also, what about b)? Any ideas?



      Thanks in advance!










      share|cite|improve this question













      Let $A$ be a subgroup of $G=mathbb{Z}/63mathbb{Z}$, where $A$ is generated with $14:=14+63 mathbb{Z}$.



      a) Determine the order of $A$ and all subgroups of $G/A$.



      b) Is there a non-trivial homomorphism $varphi : A rightarrow G/A$? If yes, find one.



      So, the order of $A$, $|A|=9$. I also determined all subgroups of $G/A$, but I'm not sure if I'm right:



      $G/A={gA : g in G} = { g+c, cin {14,28,42,56,7,21,35,49,0} }$. But out of this, I concluded that $G/A=G$, so subgroups of $G/A$ are all cyclic subgroups of $mathbb{Z}/63 mathbb{Z}$, which are generated by all $a in G$, such that $gcd(a,63) > 1$.



      Am I correct?



      Also, what about b)? Any ideas?



      Thanks in advance!







      abstract-algebra group-theory finite-groups cyclic-groups






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      asked Nov 21 at 1:23









      mathbbandstuff

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          So, for cyclic $G$, you can look at $frac{|G|}{gcd(|G|,n)}$ to find the order of the cyclic subgroup generated by $n$. So, the order of $langle 14 rangle$ will be $frac{63}{7}=9$, which you have found already.



          Cyclic groups are abelian, so all subgroups will be normal. The cosets of $A$ (which are the elements of $G/A$) will have to partition the group $G$. So the number of cosets (called the index of A in G) is $|G|/|A|$, because the cosets all have the same size.



          So $G/A$ is of order $7$. Seven is prime, so by Lagrange's theorem the only subgroup of $G/A$ is ${A}$, the identity in $G/A$. Moreover, this means any given non identity element generates $G/A$, so $G/A$ itself is cyclic too.



          So really, the problem of finding homomorphisms $phi$ from $A$ to $G/A$ is the same as finding homomorphisms from $Z_9$ to $Z_7$. We know that the kernel needs to be a subgroup of $Z_9$ and the image needs to be a subgroup of $Z_7$. So, any non trivial homomorphism needs to have the whole of $Z_7$ as its image. We know that from the first isomorphism theorem, when we quotient out the kernel, the quotient group is isomorphic to the image. The only choices for the kernel are ${0}$ and ${3,6,0}$. So, we would have $Z_9/{0}=Z_9$ isomorphic to $Z_7$ (not possible) or $Z_9/{3,6,0}=Z_3$ isomorphic to $Z_7$ (also not possible).



          So there cannot be any homomorphisms from $A$ to $G/A$ in this case besides the trivial homomorphism.






          share|cite|improve this answer




























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            You are correct that the order of $A$ is $9$ (to see this without counting everything, you can look at the prime decompositions of $63$ and $14$, and note that $9$ is the least integer $n$ such that $63|14n$.



            Now, your $G/A$ is not correct: notice how you've swapped from varying $g$ to varying $c$ (and your elements aren't sets). Also, $G/A$ can't be $G$, because that would contradict Lagrange's Theorem. Instead, $G/A = {g + A : g in G}$. Note that $G/A$ has order $|G|/|A| = 63/9 = 7$. In particular, $G/A$ has no non-trivial proper subgroups (any subgroup must have order dividing $7$, but $7$ is prime, so the only subgroups are those of order $1$ or $7$: that is, the trivial group, and $G/A$ itself).






            share|cite|improve this answer





















            • Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
              – mathbbandstuff
              Nov 21 at 1:40










            • Trivial homomorphism only when ker contains one element only
              – John Nash
              Nov 21 at 1:44






            • 1




              @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
              – user3482749
              Nov 21 at 11:35


















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            Since $mid Amid=9$, $mid G/Amid=7$. Hence $G/A$ is $C_7$.



            Suppose we have a nontrivial homomorphism $h:Ato G/A$. By the first isomorphism theorem $A/operatorname{ker}hcongoperatorname{im}h$. The order of $operatorname{ker}h$ is $3$ or $1$, since it's a subgroup. Either way we get a contradiction: $C_7$ doesn't have a subgroup of order $3$; neither can there be an injection from a $9$ element set to a $7$ element one.






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              So, for cyclic $G$, you can look at $frac{|G|}{gcd(|G|,n)}$ to find the order of the cyclic subgroup generated by $n$. So, the order of $langle 14 rangle$ will be $frac{63}{7}=9$, which you have found already.



              Cyclic groups are abelian, so all subgroups will be normal. The cosets of $A$ (which are the elements of $G/A$) will have to partition the group $G$. So the number of cosets (called the index of A in G) is $|G|/|A|$, because the cosets all have the same size.



              So $G/A$ is of order $7$. Seven is prime, so by Lagrange's theorem the only subgroup of $G/A$ is ${A}$, the identity in $G/A$. Moreover, this means any given non identity element generates $G/A$, so $G/A$ itself is cyclic too.



              So really, the problem of finding homomorphisms $phi$ from $A$ to $G/A$ is the same as finding homomorphisms from $Z_9$ to $Z_7$. We know that the kernel needs to be a subgroup of $Z_9$ and the image needs to be a subgroup of $Z_7$. So, any non trivial homomorphism needs to have the whole of $Z_7$ as its image. We know that from the first isomorphism theorem, when we quotient out the kernel, the quotient group is isomorphic to the image. The only choices for the kernel are ${0}$ and ${3,6,0}$. So, we would have $Z_9/{0}=Z_9$ isomorphic to $Z_7$ (not possible) or $Z_9/{3,6,0}=Z_3$ isomorphic to $Z_7$ (also not possible).



              So there cannot be any homomorphisms from $A$ to $G/A$ in this case besides the trivial homomorphism.






              share|cite|improve this answer

























                up vote
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                down vote



                accepted










                So, for cyclic $G$, you can look at $frac{|G|}{gcd(|G|,n)}$ to find the order of the cyclic subgroup generated by $n$. So, the order of $langle 14 rangle$ will be $frac{63}{7}=9$, which you have found already.



                Cyclic groups are abelian, so all subgroups will be normal. The cosets of $A$ (which are the elements of $G/A$) will have to partition the group $G$. So the number of cosets (called the index of A in G) is $|G|/|A|$, because the cosets all have the same size.



                So $G/A$ is of order $7$. Seven is prime, so by Lagrange's theorem the only subgroup of $G/A$ is ${A}$, the identity in $G/A$. Moreover, this means any given non identity element generates $G/A$, so $G/A$ itself is cyclic too.



                So really, the problem of finding homomorphisms $phi$ from $A$ to $G/A$ is the same as finding homomorphisms from $Z_9$ to $Z_7$. We know that the kernel needs to be a subgroup of $Z_9$ and the image needs to be a subgroup of $Z_7$. So, any non trivial homomorphism needs to have the whole of $Z_7$ as its image. We know that from the first isomorphism theorem, when we quotient out the kernel, the quotient group is isomorphic to the image. The only choices for the kernel are ${0}$ and ${3,6,0}$. So, we would have $Z_9/{0}=Z_9$ isomorphic to $Z_7$ (not possible) or $Z_9/{3,6,0}=Z_3$ isomorphic to $Z_7$ (also not possible).



                So there cannot be any homomorphisms from $A$ to $G/A$ in this case besides the trivial homomorphism.






                share|cite|improve this answer























                  up vote
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                  accepted







                  up vote
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                  So, for cyclic $G$, you can look at $frac{|G|}{gcd(|G|,n)}$ to find the order of the cyclic subgroup generated by $n$. So, the order of $langle 14 rangle$ will be $frac{63}{7}=9$, which you have found already.



                  Cyclic groups are abelian, so all subgroups will be normal. The cosets of $A$ (which are the elements of $G/A$) will have to partition the group $G$. So the number of cosets (called the index of A in G) is $|G|/|A|$, because the cosets all have the same size.



                  So $G/A$ is of order $7$. Seven is prime, so by Lagrange's theorem the only subgroup of $G/A$ is ${A}$, the identity in $G/A$. Moreover, this means any given non identity element generates $G/A$, so $G/A$ itself is cyclic too.



                  So really, the problem of finding homomorphisms $phi$ from $A$ to $G/A$ is the same as finding homomorphisms from $Z_9$ to $Z_7$. We know that the kernel needs to be a subgroup of $Z_9$ and the image needs to be a subgroup of $Z_7$. So, any non trivial homomorphism needs to have the whole of $Z_7$ as its image. We know that from the first isomorphism theorem, when we quotient out the kernel, the quotient group is isomorphic to the image. The only choices for the kernel are ${0}$ and ${3,6,0}$. So, we would have $Z_9/{0}=Z_9$ isomorphic to $Z_7$ (not possible) or $Z_9/{3,6,0}=Z_3$ isomorphic to $Z_7$ (also not possible).



                  So there cannot be any homomorphisms from $A$ to $G/A$ in this case besides the trivial homomorphism.






                  share|cite|improve this answer












                  So, for cyclic $G$, you can look at $frac{|G|}{gcd(|G|,n)}$ to find the order of the cyclic subgroup generated by $n$. So, the order of $langle 14 rangle$ will be $frac{63}{7}=9$, which you have found already.



                  Cyclic groups are abelian, so all subgroups will be normal. The cosets of $A$ (which are the elements of $G/A$) will have to partition the group $G$. So the number of cosets (called the index of A in G) is $|G|/|A|$, because the cosets all have the same size.



                  So $G/A$ is of order $7$. Seven is prime, so by Lagrange's theorem the only subgroup of $G/A$ is ${A}$, the identity in $G/A$. Moreover, this means any given non identity element generates $G/A$, so $G/A$ itself is cyclic too.



                  So really, the problem of finding homomorphisms $phi$ from $A$ to $G/A$ is the same as finding homomorphisms from $Z_9$ to $Z_7$. We know that the kernel needs to be a subgroup of $Z_9$ and the image needs to be a subgroup of $Z_7$. So, any non trivial homomorphism needs to have the whole of $Z_7$ as its image. We know that from the first isomorphism theorem, when we quotient out the kernel, the quotient group is isomorphic to the image. The only choices for the kernel are ${0}$ and ${3,6,0}$. So, we would have $Z_9/{0}=Z_9$ isomorphic to $Z_7$ (not possible) or $Z_9/{3,6,0}=Z_3$ isomorphic to $Z_7$ (also not possible).



                  So there cannot be any homomorphisms from $A$ to $G/A$ in this case besides the trivial homomorphism.







                  share|cite|improve this answer












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                  answered Nov 21 at 2:23









                  MKeller

                  455




                  455






















                      up vote
                      1
                      down vote













                      You are correct that the order of $A$ is $9$ (to see this without counting everything, you can look at the prime decompositions of $63$ and $14$, and note that $9$ is the least integer $n$ such that $63|14n$.



                      Now, your $G/A$ is not correct: notice how you've swapped from varying $g$ to varying $c$ (and your elements aren't sets). Also, $G/A$ can't be $G$, because that would contradict Lagrange's Theorem. Instead, $G/A = {g + A : g in G}$. Note that $G/A$ has order $|G|/|A| = 63/9 = 7$. In particular, $G/A$ has no non-trivial proper subgroups (any subgroup must have order dividing $7$, but $7$ is prime, so the only subgroups are those of order $1$ or $7$: that is, the trivial group, and $G/A$ itself).






                      share|cite|improve this answer





















                      • Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
                        – mathbbandstuff
                        Nov 21 at 1:40










                      • Trivial homomorphism only when ker contains one element only
                        – John Nash
                        Nov 21 at 1:44






                      • 1




                        @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
                        – user3482749
                        Nov 21 at 11:35















                      up vote
                      1
                      down vote













                      You are correct that the order of $A$ is $9$ (to see this without counting everything, you can look at the prime decompositions of $63$ and $14$, and note that $9$ is the least integer $n$ such that $63|14n$.



                      Now, your $G/A$ is not correct: notice how you've swapped from varying $g$ to varying $c$ (and your elements aren't sets). Also, $G/A$ can't be $G$, because that would contradict Lagrange's Theorem. Instead, $G/A = {g + A : g in G}$. Note that $G/A$ has order $|G|/|A| = 63/9 = 7$. In particular, $G/A$ has no non-trivial proper subgroups (any subgroup must have order dividing $7$, but $7$ is prime, so the only subgroups are those of order $1$ or $7$: that is, the trivial group, and $G/A$ itself).






                      share|cite|improve this answer





















                      • Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
                        – mathbbandstuff
                        Nov 21 at 1:40










                      • Trivial homomorphism only when ker contains one element only
                        – John Nash
                        Nov 21 at 1:44






                      • 1




                        @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
                        – user3482749
                        Nov 21 at 11:35













                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      You are correct that the order of $A$ is $9$ (to see this without counting everything, you can look at the prime decompositions of $63$ and $14$, and note that $9$ is the least integer $n$ such that $63|14n$.



                      Now, your $G/A$ is not correct: notice how you've swapped from varying $g$ to varying $c$ (and your elements aren't sets). Also, $G/A$ can't be $G$, because that would contradict Lagrange's Theorem. Instead, $G/A = {g + A : g in G}$. Note that $G/A$ has order $|G|/|A| = 63/9 = 7$. In particular, $G/A$ has no non-trivial proper subgroups (any subgroup must have order dividing $7$, but $7$ is prime, so the only subgroups are those of order $1$ or $7$: that is, the trivial group, and $G/A$ itself).






                      share|cite|improve this answer












                      You are correct that the order of $A$ is $9$ (to see this without counting everything, you can look at the prime decompositions of $63$ and $14$, and note that $9$ is the least integer $n$ such that $63|14n$.



                      Now, your $G/A$ is not correct: notice how you've swapped from varying $g$ to varying $c$ (and your elements aren't sets). Also, $G/A$ can't be $G$, because that would contradict Lagrange's Theorem. Instead, $G/A = {g + A : g in G}$. Note that $G/A$ has order $|G|/|A| = 63/9 = 7$. In particular, $G/A$ has no non-trivial proper subgroups (any subgroup must have order dividing $7$, but $7$ is prime, so the only subgroups are those of order $1$ or $7$: that is, the trivial group, and $G/A$ itself).







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 21 at 1:37









                      user3482749

                      2,086414




                      2,086414












                      • Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
                        – mathbbandstuff
                        Nov 21 at 1:40










                      • Trivial homomorphism only when ker contains one element only
                        – John Nash
                        Nov 21 at 1:44






                      • 1




                        @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
                        – user3482749
                        Nov 21 at 11:35


















                      • Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
                        – mathbbandstuff
                        Nov 21 at 1:40










                      • Trivial homomorphism only when ker contains one element only
                        – John Nash
                        Nov 21 at 1:44






                      • 1




                        @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
                        – user3482749
                        Nov 21 at 11:35
















                      Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
                      – mathbbandstuff
                      Nov 21 at 1:40




                      Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
                      – mathbbandstuff
                      Nov 21 at 1:40












                      Trivial homomorphism only when ker contains one element only
                      – John Nash
                      Nov 21 at 1:44




                      Trivial homomorphism only when ker contains one element only
                      – John Nash
                      Nov 21 at 1:44




                      1




                      1




                      @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
                      – user3482749
                      Nov 21 at 11:35




                      @mathbbandstuff The image of a homomorphism is both a subgroup of the target (so has order dividing 7), and, by one of the isomorphism theorems, has order a factor of the order of the domain (that is: an order dividing 9). The only natural number that satisfies both of those is 1, so the only homomorphisms are those with image of order 1, and only the trivial homomorphism has this property.
                      – user3482749
                      Nov 21 at 11:35










                      up vote
                      1
                      down vote













                      Since $mid Amid=9$, $mid G/Amid=7$. Hence $G/A$ is $C_7$.



                      Suppose we have a nontrivial homomorphism $h:Ato G/A$. By the first isomorphism theorem $A/operatorname{ker}hcongoperatorname{im}h$. The order of $operatorname{ker}h$ is $3$ or $1$, since it's a subgroup. Either way we get a contradiction: $C_7$ doesn't have a subgroup of order $3$; neither can there be an injection from a $9$ element set to a $7$ element one.






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        Since $mid Amid=9$, $mid G/Amid=7$. Hence $G/A$ is $C_7$.



                        Suppose we have a nontrivial homomorphism $h:Ato G/A$. By the first isomorphism theorem $A/operatorname{ker}hcongoperatorname{im}h$. The order of $operatorname{ker}h$ is $3$ or $1$, since it's a subgroup. Either way we get a contradiction: $C_7$ doesn't have a subgroup of order $3$; neither can there be an injection from a $9$ element set to a $7$ element one.






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Since $mid Amid=9$, $mid G/Amid=7$. Hence $G/A$ is $C_7$.



                          Suppose we have a nontrivial homomorphism $h:Ato G/A$. By the first isomorphism theorem $A/operatorname{ker}hcongoperatorname{im}h$. The order of $operatorname{ker}h$ is $3$ or $1$, since it's a subgroup. Either way we get a contradiction: $C_7$ doesn't have a subgroup of order $3$; neither can there be an injection from a $9$ element set to a $7$ element one.






                          share|cite|improve this answer












                          Since $mid Amid=9$, $mid G/Amid=7$. Hence $G/A$ is $C_7$.



                          Suppose we have a nontrivial homomorphism $h:Ato G/A$. By the first isomorphism theorem $A/operatorname{ker}hcongoperatorname{im}h$. The order of $operatorname{ker}h$ is $3$ or $1$, since it's a subgroup. Either way we get a contradiction: $C_7$ doesn't have a subgroup of order $3$; neither can there be an injection from a $9$ element set to a $7$ element one.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 21 at 3:06









                          Chris Custer

                          9,7243624




                          9,7243624






























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