Mixing
Are all cyclic groups considered sub-groups of another group? [closed]
$begingroup$
Consider $Z_5^*$ as a group G, where the group operation is multiplication. Using 2 as a generator and element of G, a cyclic group can be generated:
- 2⋅0 (mod 5) = 0
- 2⋅1 (mod 5) = 2
- 2⋅2 (mod 5) = 4
- 2⋅3 (mod 5) = 1
- 2⋅4 (mod 5) = 3
- 2⋅5 (mod 5) = 0
Is this cyclic group and all cyclic groups that can be generated from $Z_5^*$ considered a sub-group of $Z_5^*$?
group-theory discrete-mathematics finite-groups
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
$endgroup$
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, José Carlos Santos, max_zorn, steven gregory Jan 22 at 23:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider $Z_5^*$ as a group G, where the group operation is multiplication. Using 2 as a generator and element of G, a cyclic group can be generated:
- 2⋅0 (mod 5) = 0
- 2⋅1 (mod 5) = 2
- 2⋅2 (mod 5) = 4
- 2⋅3 (mod 5) = 1
- 2⋅4 (mod 5) = 3
- 2⋅5 (mod 5) = 0
Is this cyclic group and all cyclic groups that can be generated from $Z_5^*$ considered a sub-group of $Z_5^*$?
group-theory discrete-mathematics finite-groups
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
$endgroup$
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, José Carlos Santos, max_zorn, steven gregory Jan 22 at 23:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What are you asking here? Are you asking if there's a group $G$ that, given any cyclic group $C$, there is a subgroup of $G$ isomorphic to $C$? If so, the answer is "yes", e.g. the unit circle in the complex plane, under multiplication.
$endgroup$
– Theo Bendit
Jan 22 at 4:36
$begingroup$
@TheoBendit I edited my question in an attempt to be more specific in what I'm asking.
$endgroup$
– JohnGalt
Jan 22 at 4:55
1
$begingroup$
Is $0$ an element of $Z_5^*$? If it is, then $Z_5^*$ is not a group, because $0$ has no inverse element.
$endgroup$
– Tanner Swett
Jan 22 at 5:21
$begingroup$
Every group is (isomorphic to) a subgroup of some symmetric group. See Cayley's theorem.
$endgroup$
– Shaun
Jan 22 at 11:46
add a comment |
$begingroup$
Consider $Z_5^*$ as a group G, where the group operation is multiplication. Using 2 as a generator and element of G, a cyclic group can be generated:
- 2⋅0 (mod 5) = 0
- 2⋅1 (mod 5) = 2
- 2⋅2 (mod 5) = 4
- 2⋅3 (mod 5) = 1
- 2⋅4 (mod 5) = 3
- 2⋅5 (mod 5) = 0
Is this cyclic group and all cyclic groups that can be generated from $Z_5^*$ considered a sub-group of $Z_5^*$?
group-theory discrete-mathematics finite-groups
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
$endgroup$
Consider $Z_5^*$ as a group G, where the group operation is multiplication. Using 2 as a generator and element of G, a cyclic group can be generated:
- 2⋅0 (mod 5) = 0
- 2⋅1 (mod 5) = 2
- 2⋅2 (mod 5) = 4
- 2⋅3 (mod 5) = 1
- 2⋅4 (mod 5) = 3
- 2⋅5 (mod 5) = 0
Is this cyclic group and all cyclic groups that can be generated from $Z_5^*$ considered a sub-group of $Z_5^*$?
group-theory discrete-mathematics finite-groups
group-theory discrete-mathematics finite-groups
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
edited Jan 22 at 4:54
JohnGalt
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
asked Jan 22 at 4:32
JohnGaltJohnGalt
1195
1195
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, José Carlos Santos, max_zorn, steven gregory Jan 22 at 23:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, José Carlos Santos, max_zorn, steven gregory Jan 22 at 23:05
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What are you asking here? Are you asking if there's a group $G$ that, given any cyclic group $C$, there is a subgroup of $G$ isomorphic to $C$? If so, the answer is "yes", e.g. the unit circle in the complex plane, under multiplication.
$endgroup$
– Theo Bendit
Jan 22 at 4:36
$begingroup$
@TheoBendit I edited my question in an attempt to be more specific in what I'm asking.
$endgroup$
– JohnGalt
Jan 22 at 4:55
1
$begingroup$
Is $0$ an element of $Z_5^*$? If it is, then $Z_5^*$ is not a group, because $0$ has no inverse element.
$endgroup$
– Tanner Swett
Jan 22 at 5:21
$begingroup$
Every group is (isomorphic to) a subgroup of some symmetric group. See Cayley's theorem.
$endgroup$
– Shaun
Jan 22 at 11:46
add a comment |
$begingroup$
What are you asking here? Are you asking if there's a group $G$ that, given any cyclic group $C$, there is a subgroup of $G$ isomorphic to $C$? If so, the answer is "yes", e.g. the unit circle in the complex plane, under multiplication.
$endgroup$
– Theo Bendit
Jan 22 at 4:36
$begingroup$
@TheoBendit I edited my question in an attempt to be more specific in what I'm asking.
$endgroup$
– JohnGalt
Jan 22 at 4:55
1
$begingroup$
Is $0$ an element of $Z_5^*$? If it is, then $Z_5^*$ is not a group, because $0$ has no inverse element.
$endgroup$
– Tanner Swett
Jan 22 at 5:21
$begingroup$
Every group is (isomorphic to) a subgroup of some symmetric group. See Cayley's theorem.
$endgroup$
– Shaun
Jan 22 at 11:46
$begingroup$
What are you asking here? Are you asking if there's a group $G$ that, given any cyclic group $C$, there is a subgroup of $G$ isomorphic to $C$? If so, the answer is "yes", e.g. the unit circle in the complex plane, under multiplication.
$endgroup$
– Theo Bendit
Jan 22 at 4:36
$begingroup$
What are you asking here? Are you asking if there's a group $G$ that, given any cyclic group $C$, there is a subgroup of $G$ isomorphic to $C$? If so, the answer is "yes", e.g. the unit circle in the complex plane, under multiplication.
$endgroup$
– Theo Bendit
Jan 22 at 4:36
$begingroup$
@TheoBendit I edited my question in an attempt to be more specific in what I'm asking.
$endgroup$
– JohnGalt
Jan 22 at 4:55
$begingroup$
@TheoBendit I edited my question in an attempt to be more specific in what I'm asking.
$endgroup$
– JohnGalt
Jan 22 at 4:55
1
1
$begingroup$
Is $0$ an element of $Z_5^*$? If it is, then $Z_5^*$ is not a group, because $0$ has no inverse element.
$endgroup$
– Tanner Swett
Jan 22 at 5:21
$begingroup$
Is $0$ an element of $Z_5^*$? If it is, then $Z_5^*$ is not a group, because $0$ has no inverse element.
$endgroup$
– Tanner Swett
Jan 22 at 5:21
$begingroup$
Every group is (isomorphic to) a subgroup of some symmetric group. See Cayley's theorem.
$endgroup$
– Shaun
Jan 22 at 11:46
$begingroup$
Every group is (isomorphic to) a subgroup of some symmetric group. See Cayley's theorem.
$endgroup$
– Shaun
Jan 22 at 11:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What you have written is not a cyclic group. Given that the operation in $Z^ast_5$ is multiplication, what you want is repeated multiplication by 2, i.e. taking powers, which gives you $2^0 =1$, $2^1=2$, $2^2= 4$, $2^3=8equiv 3$, $2^4=16equiv 1$.
cyclic groups generated this way are considered subgroups of $Z^ast_5$. In this instance, the subgroup is the whole group!
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
$endgroup$
add a comment |
$begingroup$
As @RyleeLyman pointed out, the list should be denoted as
$$
2^1 = 2 implies 2 equiv_{5} 2\
2^2 = 4 implies 4 equiv_{5} 4\
2^3 = 8 implies 8 equiv_{5} 3\
2^4 = 16 implies 16 equiv_{5} 1.\
$$
The reason being that you are considering the multiplicative group $mathbb Z_{5}^*$.
Is this cyclic group and all cyclic groups that can be generated from $mathbb Z_5^*$ considered a sub-group of $mathbb Z_5^*$?
$mathbb Z_5^*$ is a subgroup of $mathbb Z_5^*$, since every group $G$ is a subgroup of itself. [1]
Another way of looking at this is to use the fact that all subgroups of a cyclic group are cyclic. [2] In other words, the cyclic group that can be formed under the multiplicative group of $mathbb Z_5^*$, is $mathbb Z_5^*$ itself using the generator $2$,
$$
mathbb Z_5^* = langle 2 rangle.
$$
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you have written is not a cyclic group. Given that the operation in $Z^ast_5$ is multiplication, what you want is repeated multiplication by 2, i.e. taking powers, which gives you $2^0 =1$, $2^1=2$, $2^2= 4$, $2^3=8equiv 3$, $2^4=16equiv 1$.
cyclic groups generated this way are considered subgroups of $Z^ast_5$. In this instance, the subgroup is the whole group!
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
$endgroup$
add a comment |
$begingroup$
What you have written is not a cyclic group. Given that the operation in $Z^ast_5$ is multiplication, what you want is repeated multiplication by 2, i.e. taking powers, which gives you $2^0 =1$, $2^1=2$, $2^2= 4$, $2^3=8equiv 3$, $2^4=16equiv 1$.
cyclic groups generated this way are considered subgroups of $Z^ast_5$. In this instance, the subgroup is the whole group!
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
$endgroup$
add a comment |
$begingroup$
What you have written is not a cyclic group. Given that the operation in $Z^ast_5$ is multiplication, what you want is repeated multiplication by 2, i.e. taking powers, which gives you $2^0 =1$, $2^1=2$, $2^2= 4$, $2^3=8equiv 3$, $2^4=16equiv 1$.
cyclic groups generated this way are considered subgroups of $Z^ast_5$. In this instance, the subgroup is the whole group!
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
$endgroup$
What you have written is not a cyclic group. Given that the operation in $Z^ast_5$ is multiplication, what you want is repeated multiplication by 2, i.e. taking powers, which gives you $2^0 =1$, $2^1=2$, $2^2= 4$, $2^3=8equiv 3$, $2^4=16equiv 1$.
cyclic groups generated this way are considered subgroups of $Z^ast_5$. In this instance, the subgroup is the whole group!
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
answered Jan 22 at 5:35
Rylee LymanRylee Lyman
1529
1529
add a comment |
add a comment |
$begingroup$
As @RyleeLyman pointed out, the list should be denoted as
$$
2^1 = 2 implies 2 equiv_{5} 2\
2^2 = 4 implies 4 equiv_{5} 4\
2^3 = 8 implies 8 equiv_{5} 3\
2^4 = 16 implies 16 equiv_{5} 1.\
$$
The reason being that you are considering the multiplicative group $mathbb Z_{5}^*$.
Is this cyclic group and all cyclic groups that can be generated from $mathbb Z_5^*$ considered a sub-group of $mathbb Z_5^*$?
$mathbb Z_5^*$ is a subgroup of $mathbb Z_5^*$, since every group $G$ is a subgroup of itself. [1]
Another way of looking at this is to use the fact that all subgroups of a cyclic group are cyclic. [2] In other words, the cyclic group that can be formed under the multiplicative group of $mathbb Z_5^*$, is $mathbb Z_5^*$ itself using the generator $2$,
$$
mathbb Z_5^* = langle 2 rangle.
$$
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
$endgroup$
add a comment |
$begingroup$
As @RyleeLyman pointed out, the list should be denoted as
$$
2^1 = 2 implies 2 equiv_{5} 2\
2^2 = 4 implies 4 equiv_{5} 4\
2^3 = 8 implies 8 equiv_{5} 3\
2^4 = 16 implies 16 equiv_{5} 1.\
$$
The reason being that you are considering the multiplicative group $mathbb Z_{5}^*$.
Is this cyclic group and all cyclic groups that can be generated from $mathbb Z_5^*$ considered a sub-group of $mathbb Z_5^*$?
$mathbb Z_5^*$ is a subgroup of $mathbb Z_5^*$, since every group $G$ is a subgroup of itself. [1]
Another way of looking at this is to use the fact that all subgroups of a cyclic group are cyclic. [2] In other words, the cyclic group that can be formed under the multiplicative group of $mathbb Z_5^*$, is $mathbb Z_5^*$ itself using the generator $2$,
$$
mathbb Z_5^* = langle 2 rangle.
$$
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
$endgroup$
add a comment |
$begingroup$
As @RyleeLyman pointed out, the list should be denoted as
$$
2^1 = 2 implies 2 equiv_{5} 2\
2^2 = 4 implies 4 equiv_{5} 4\
2^3 = 8 implies 8 equiv_{5} 3\
2^4 = 16 implies 16 equiv_{5} 1.\
$$
The reason being that you are considering the multiplicative group $mathbb Z_{5}^*$.
Is this cyclic group and all cyclic groups that can be generated from $mathbb Z_5^*$ considered a sub-group of $mathbb Z_5^*$?
$mathbb Z_5^*$ is a subgroup of $mathbb Z_5^*$, since every group $G$ is a subgroup of itself. [1]
Another way of looking at this is to use the fact that all subgroups of a cyclic group are cyclic. [2] In other words, the cyclic group that can be formed under the multiplicative group of $mathbb Z_5^*$, is $mathbb Z_5^*$ itself using the generator $2$,
$$
mathbb Z_5^* = langle 2 rangle.
$$
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
$endgroup$
As @RyleeLyman pointed out, the list should be denoted as
$$
2^1 = 2 implies 2 equiv_{5} 2\
2^2 = 4 implies 4 equiv_{5} 4\
2^3 = 8 implies 8 equiv_{5} 3\
2^4 = 16 implies 16 equiv_{5} 1.\
$$
The reason being that you are considering the multiplicative group $mathbb Z_{5}^*$.
Is this cyclic group and all cyclic groups that can be generated from $mathbb Z_5^*$ considered a sub-group of $mathbb Z_5^*$?
$mathbb Z_5^*$ is a subgroup of $mathbb Z_5^*$, since every group $G$ is a subgroup of itself. [1]
Another way of looking at this is to use the fact that all subgroups of a cyclic group are cyclic. [2] In other words, the cyclic group that can be formed under the multiplicative group of $mathbb Z_5^*$, is $mathbb Z_5^*$ itself using the generator $2$,
$$
mathbb Z_5^* = langle 2 rangle.
$$
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
answered Jan 22 at 18:53
EdOverflowEdOverflow
25119
25119
add a comment |
add a comment |
$begingroup$
What are you asking here? Are you asking if there's a group $G$ that, given any cyclic group $C$, there is a subgroup of $G$ isomorphic to $C$? If so, the answer is "yes", e.g. the unit circle in the complex plane, under multiplication.
$endgroup$
– Theo Bendit
Jan 22 at 4:36
$begingroup$
@TheoBendit I edited my question in an attempt to be more specific in what I'm asking.
$endgroup$
– JohnGalt
Jan 22 at 4:55
1
$begingroup$
Is $0$ an element of $Z_5^*$? If it is, then $Z_5^*$ is not a group, because $0$ has no inverse element.
$endgroup$
– Tanner Swett
Jan 22 at 5:21
$begingroup$
Every group is (isomorphic to) a subgroup of some symmetric group. See Cayley's theorem.
$endgroup$
– Shaun
Jan 22 at 11:46