Mixing
Problem involving $mathbb{Z} /63 mathbb{Z}$ and its subgroups.
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
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
add a comment |
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
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
add a comment |
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
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
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
abstract-algebra group-theory finite-groups cyclic-groups
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
asked Nov 21 '18 at 1:23
mathbbandstuff
256111
256111
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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.
answered Nov 21 '18 at 2:23
MKeller
455
add a comment |
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).
answered Nov 21 '18 at 1:37
user3482749
2,708414
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 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 '18 at 11:35
add a comment |
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.
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
add a comment |
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3 Answers
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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.
answered Nov 21 '18 at 2:23
MKeller
455
add a comment |
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.
answered Nov 21 '18 at 2:23
MKeller
455
add a comment |
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.
answered Nov 21 '18 at 2:23
MKeller
455
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.
answered Nov 21 '18 at 2:23
MKeller
455
answered Nov 21 '18 at 2:23
MKeller
455
answered Nov 21 '18 at 2:23
MKeller
455
answered Nov 21 '18 at 2:23
MKeller
455
455
add a comment |
add a comment |
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).
answered Nov 21 '18 at 1:37
user3482749
2,708414
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 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 '18 at 11:35
add a comment |
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).
answered Nov 21 '18 at 1:37
user3482749
2,708414
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 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 '18 at 11:35
add a comment |
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).
answered Nov 21 '18 at 1:37
user3482749
2,708414
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).
answered Nov 21 '18 at 1:37
user3482749
2,708414
answered Nov 21 '18 at 1:37
user3482749
2,708414
answered Nov 21 '18 at 1:37
user3482749
2,708414
answered Nov 21 '18 at 1:37
user3482749
2,708414
2,708414
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 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 '18 at 11:35
add a comment |
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 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 '18 at 11:35
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Oh, that's right. Thanks! That helps a lot. Also, do you have any idea about b)?
– mathbbandstuff
Nov 21 '18 at 1:40
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 at 1:44
Trivial homomorphism only when ker contains one element only
– John Nash
Nov 21 '18 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 '18 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 '18 at 11:35
add a comment |
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.
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
add a comment |
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.
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
add a comment |
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.
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
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.
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
answered Nov 21 '18 at 3:06
Chris Custer
10.9k3824
10.9k3824
add a comment |
add a comment |
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