List all of the subgroups of $(mathbb Z_11setminus [0], cdot)$Find all proper subgroups of multiplicative group $BbbZ_13$.List all subgroups of the symmetry group of $n$-gonIs there a special term for groups, the only subgroups of which are the trivial group and the group itself?What are the finite subgroups of $GL_2(mathbbZ)$?All distinct subgroups of $mathbbZ_4 times mathbbZ_4$ isomorphic to $mathbbZ_4$Find all normal subgroups of the followingorder of an element modulo safe primeGenerators of two groups with prime order $p$ already induce all the generators of the product group $G times H$Prove that $(T, cdot)$ is a group and find all of its subgroupsList the Sylow 2-subgroups and Sylow 3-subgrousp of $U_45$Find all generators of $(mathbb Z setminus 11 mathbb Z)^times$
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List all of the subgroups of $(mathbb Z_11setminus [0], cdot)$
Find all proper subgroups of multiplicative group $BbbZ_13$.List all subgroups of the symmetry group of $n$-gonIs there a special term for groups, the only subgroups of which are the trivial group and the group itself?What are the finite subgroups of $GL_2(mathbbZ)$?All distinct subgroups of $mathbbZ_4 times mathbbZ_4$ isomorphic to $mathbbZ_4$Find all normal subgroups of the followingorder of an element modulo safe primeGenerators of two groups with prime order $p$ already induce all the generators of the product group $G times H$Prove that $(T, cdot)$ is a group and find all of its subgroupsList the Sylow 2-subgroups and Sylow 3-subgrousp of $U_45$Find all generators of $(mathbb Z setminus 11 mathbb Z)^times$
$begingroup$
So obviously the order of this group is $10$. Now I need to find a generator, so I can find the subgroups.
Since $3$ and $10$ are relatively prime, $3$ would be a generator for this group, right? And it would help with getting the subgroups?
abstract-algebra group-theory
edited 17 hours ago
the_fox
2,90021537
asked 22 hours ago
Marco11Marco11
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
So obviously the order of this group is $10$. Now I need to find a generator, so I can find the subgroups.
Since $3$ and $10$ are relatively prime, $3$ would be a generator for this group, right? And it would help with getting the subgroups?
abstract-algebra group-theory
edited 17 hours ago
the_fox
2,90021537
asked 22 hours ago
Marco11Marco11
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
$3^5equiv1pmod11$. You want a primitive root modulo $11$.
$endgroup$
– Lord Shark the Unknown
22 hours ago
$begingroup$
Okay. So a generator would be 2?
$endgroup$
– Marco11
22 hours ago
1
$begingroup$
reference : (math.stackexchange.com/questions/2973114/…)
$endgroup$
– Chinnapparaj R
21 hours ago
add a comment |
$begingroup$
So obviously the order of this group is $10$. Now I need to find a generator, so I can find the subgroups.
Since $3$ and $10$ are relatively prime, $3$ would be a generator for this group, right? And it would help with getting the subgroups?
abstract-algebra group-theory
edited 17 hours ago
the_fox
2,90021537
asked 22 hours ago
Marco11Marco11
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So obviously the order of this group is $10$. Now I need to find a generator, so I can find the subgroups.
Since $3$ and $10$ are relatively prime, $3$ would be a generator for this group, right? And it would help with getting the subgroups?
abstract-algebra group-theory
abstract-algebra group-theory
edited 17 hours ago
the_fox
2,90021537
asked 22 hours ago
Marco11Marco11
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
the_fox
2,90021537
asked 22 hours ago
Marco11Marco11
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
the_fox
2,90021537
edited 17 hours ago
the_fox
2,90021537
edited 17 hours ago
the_fox
2,90021537
2,90021537
asked 22 hours ago
Marco11Marco11
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 22 hours ago
Marco11Marco11
112
asked 22 hours ago
Marco11Marco11
112
112
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Marco11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
$3^5equiv1pmod11$. You want a primitive root modulo $11$.
$endgroup$
– Lord Shark the Unknown
22 hours ago
$begingroup$
Okay. So a generator would be 2?
$endgroup$
– Marco11
22 hours ago
1
$begingroup$
reference : (math.stackexchange.com/questions/2973114/…)
$endgroup$
– Chinnapparaj R
21 hours ago
add a comment |
1
$begingroup$
$3^5equiv1pmod11$. You want a primitive root modulo $11$.
$endgroup$
– Lord Shark the Unknown
22 hours ago
$begingroup$
Okay. So a generator would be 2?
$endgroup$
– Marco11
22 hours ago
1
$begingroup$
reference : (math.stackexchange.com/questions/2973114/…)
$endgroup$
– Chinnapparaj R
21 hours ago
1
1
$begingroup$
$3^5equiv1pmod11$. You want a primitive root modulo $11$.
$endgroup$
– Lord Shark the Unknown
22 hours ago
$begingroup$
$3^5equiv1pmod11$. You want a primitive root modulo $11$.
$endgroup$
– Lord Shark the Unknown
22 hours ago
$begingroup$
Okay. So a generator would be 2?
$endgroup$
– Marco11
22 hours ago
$begingroup$
Okay. So a generator would be 2?
$endgroup$
– Marco11
22 hours ago
1
1
$begingroup$
reference : (math.stackexchange.com/questions/2973114/…)
$endgroup$
– Chinnapparaj R
21 hours ago
$begingroup$
reference : (math.stackexchange.com/questions/2973114/…)
$endgroup$
– Chinnapparaj R
21 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the comments and your insight that $2$ generates $Bbb Z_11^*$, we can say that for each $d$ such that $dmid 10$, we have that $2^frac10d$ generates a subgroup of order $d$.
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
add a comment |
$begingroup$
The element $-1=10$ surely generates a subgroup of order two, namely $1,-1=10$. Take another element, for instance $2$; then
$$
2^2=4,quad 2^5=10
$$
OK, the subgroup generated by $2$ is the whole group. Therefore $2^2=4$ generates the only subgroup of order five.
answered 16 hours ago
egregegreg
183k1486205
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Using the comments and your insight that $2$ generates $Bbb Z_11^*$, we can say that for each $d$ such that $dmid 10$, we have that $2^frac10d$ generates a subgroup of order $d$.
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
add a comment |
$begingroup$
Using the comments and your insight that $2$ generates $Bbb Z_11^*$, we can say that for each $d$ such that $dmid 10$, we have that $2^frac10d$ generates a subgroup of order $d$.
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
add a comment |
$begingroup$
Using the comments and your insight that $2$ generates $Bbb Z_11^*$, we can say that for each $d$ such that $dmid 10$, we have that $2^frac10d$ generates a subgroup of order $d$.
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
$endgroup$
Using the comments and your insight that $2$ generates $Bbb Z_11^*$, we can say that for each $d$ such that $dmid 10$, we have that $2^frac10d$ generates a subgroup of order $d$.
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
answered 20 hours ago
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
add a comment |
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
Any element of the multiplicative group is generator
$endgroup$
– PerelMan
5 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
No. There are only $varphi (10)=4$ generators. They are $2,2^3,2^7$ and $2^9$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
@PerelMan that is, $2,6,7$ and $8$ are the only generators. For instance, $3$ has order $5$; so it's not a generator.
$endgroup$
– Chris Custer
4 hours ago
add a comment |
$begingroup$
The element $-1=10$ surely generates a subgroup of order two, namely $1,-1=10$. Take another element, for instance $2$; then
$$
2^2=4,quad 2^5=10
$$
OK, the subgroup generated by $2$ is the whole group. Therefore $2^2=4$ generates the only subgroup of order five.
answered 16 hours ago
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The element $-1=10$ surely generates a subgroup of order two, namely $1,-1=10$. Take another element, for instance $2$; then
$$
2^2=4,quad 2^5=10
$$
OK, the subgroup generated by $2$ is the whole group. Therefore $2^2=4$ generates the only subgroup of order five.
answered 16 hours ago
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
The element $-1=10$ surely generates a subgroup of order two, namely $1,-1=10$. Take another element, for instance $2$; then
$$
2^2=4,quad 2^5=10
$$
OK, the subgroup generated by $2$ is the whole group. Therefore $2^2=4$ generates the only subgroup of order five.
answered 16 hours ago
egregegreg
183k1486205
$endgroup$
The element $-1=10$ surely generates a subgroup of order two, namely $1,-1=10$. Take another element, for instance $2$; then
$$
2^2=4,quad 2^5=10
$$
OK, the subgroup generated by $2$ is the whole group. Therefore $2^2=4$ generates the only subgroup of order five.
answered 16 hours ago
egregegreg
183k1486205
answered 16 hours ago
egregegreg
183k1486205
answered 16 hours ago
egregegreg
183k1486205
answered 16 hours ago
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
Marco11 is a new contributor. Be nice, and check out our Code of Conduct.
Marco11 is a new contributor. Be nice, and check out our Code of Conduct.
Marco11 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
$3^5equiv1pmod11$. You want a primitive root modulo $11$.
$endgroup$
– Lord Shark the Unknown
22 hours ago
$begingroup$
Okay. So a generator would be 2?
$endgroup$
– Marco11
22 hours ago
1
$begingroup$
reference : (math.stackexchange.com/questions/2973114/…)
$endgroup$
– Chinnapparaj R
21 hours ago