Classify $mathbb{Z}_{108}$ up to an isomorphism [closed]
$begingroup$
Note: My question is a lot more general. Can you provide me with an answer about what "classify" implies when it comes to groups and what are the steps that one should follow to carry through with this process?
Attempt:
Let $G=mathbb{Z}_{108}$
According to the Fundamental Theorem of Finite Abelian groups, if we write $G$'s order as
$$
108=2^23^3
$$
we can infer that $G$ is isomorphic to one of the following direct products of cyclic groups:
begin{align*}
1.&quad mathbb{Z}_{4} times mathbb{Z}_{27} \
2.&quad mathbb{Z}_4 times mathbb{Z}_3 timesmathbb{Z}_9 \
3.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_{27} \
4.&quad mathbb{Z}_4 times mathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
5.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 timesmathbb{Z}_9 \
6.&quad mathbb{Z}_2 timesmathbb{Z}_2 timesmathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
end{align*}
Does this process go any further or after this observation $G$ is considered as classified?
abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
$endgroup$
closed as unclear what you're asking by Derek Holt, Lord Shark the Unknown, KReiser, José Carlos Santos, Leucippus Dec 17 '18 at 8:14
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.
|
show 1 more comment
$begingroup$
Note: My question is a lot more general. Can you provide me with an answer about what "classify" implies when it comes to groups and what are the steps that one should follow to carry through with this process?
Attempt:
Let $G=mathbb{Z}_{108}$
According to the Fundamental Theorem of Finite Abelian groups, if we write $G$'s order as
$$
108=2^23^3
$$
we can infer that $G$ is isomorphic to one of the following direct products of cyclic groups:
begin{align*}
1.&quad mathbb{Z}_{4} times mathbb{Z}_{27} \
2.&quad mathbb{Z}_4 times mathbb{Z}_3 timesmathbb{Z}_9 \
3.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_{27} \
4.&quad mathbb{Z}_4 times mathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
5.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 timesmathbb{Z}_9 \
6.&quad mathbb{Z}_2 timesmathbb{Z}_2 timesmathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
end{align*}
Does this process go any further or after this observation $G$ is considered as classified?
abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
$endgroup$
closed as unclear what you're asking by Derek Holt, Lord Shark the Unknown, KReiser, José Carlos Santos, Leucippus Dec 17 '18 at 8:14
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.
$begingroup$
Are you trying to classify abelian groups of order 108?
$endgroup$
– the_fox
Dec 16 '18 at 14:49
$begingroup$
@the_fox Just $mathbb{Z}_{108}$. I'm currently studying about this theorem I mentioned and I've encountered many questions/problems similar to this one. I want to know what "classifying up to an isomorphism" means and how is it done for this example.
$endgroup$
– Jevaut
Dec 16 '18 at 14:56
$begingroup$
You can just use the Chinese remainder theorem and you see that $mathbb{Z}/108mathbb{Z} cong mathbb{Z}/4mathbb{Z} times mathbb{Z}/27mathbb{Z}$.
$endgroup$
– Mindlack
Dec 16 '18 at 14:58
$begingroup$
It doesn't make sense (to me) to ask for a classification up to isomorphism of a concrete, specific group. When a classification (up to isomorphism) of all groups with a specific property is asked for, what usually happens is that you need to provide a list of concrete groups which exhausts that list in that any group with the property is isomorphic to a group in the list. So the list you have provided is a classification of all abelian groups of order $108$.
$endgroup$
– the_fox
Dec 16 '18 at 15:06
$begingroup$
To see that your group is isomorphic to one in the list all you need is the theorem that $C_n times C_m cong C_{nm}$ when $gcd(n,m)=1$.
$endgroup$
– the_fox
Dec 16 '18 at 15:08
|
show 1 more comment
$begingroup$
Note: My question is a lot more general. Can you provide me with an answer about what "classify" implies when it comes to groups and what are the steps that one should follow to carry through with this process?
Attempt:
Let $G=mathbb{Z}_{108}$
According to the Fundamental Theorem of Finite Abelian groups, if we write $G$'s order as
$$
108=2^23^3
$$
we can infer that $G$ is isomorphic to one of the following direct products of cyclic groups:
begin{align*}
1.&quad mathbb{Z}_{4} times mathbb{Z}_{27} \
2.&quad mathbb{Z}_4 times mathbb{Z}_3 timesmathbb{Z}_9 \
3.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_{27} \
4.&quad mathbb{Z}_4 times mathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
5.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 timesmathbb{Z}_9 \
6.&quad mathbb{Z}_2 timesmathbb{Z}_2 timesmathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
end{align*}
Does this process go any further or after this observation $G$ is considered as classified?
abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
$endgroup$
Note: My question is a lot more general. Can you provide me with an answer about what "classify" implies when it comes to groups and what are the steps that one should follow to carry through with this process?
Attempt:
Let $G=mathbb{Z}_{108}$
According to the Fundamental Theorem of Finite Abelian groups, if we write $G$'s order as
$$
108=2^23^3
$$
we can infer that $G$ is isomorphic to one of the following direct products of cyclic groups:
begin{align*}
1.&quad mathbb{Z}_{4} times mathbb{Z}_{27} \
2.&quad mathbb{Z}_4 times mathbb{Z}_3 timesmathbb{Z}_9 \
3.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_{27} \
4.&quad mathbb{Z}_4 times mathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
5.&quad mathbb{Z}_2 times mathbb{Z}_2 times mathbb{Z}_3 timesmathbb{Z}_9 \
6.&quad mathbb{Z}_2 timesmathbb{Z}_2 timesmathbb{Z}_3 times mathbb{Z}_3 times mathbb{Z}_3 \
end{align*}
Does this process go any further or after this observation $G$ is considered as classified?
abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism
abstract-algebra group-theory abelian-groups cyclic-groups group-isomorphism
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
asked Dec 16 '18 at 14:43
JevautJevaut
1,166212
1,166212
closed as unclear what you're asking by Derek Holt, Lord Shark the Unknown, KReiser, José Carlos Santos, Leucippus Dec 17 '18 at 8:14
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 Derek Holt, Lord Shark the Unknown, KReiser, José Carlos Santos, Leucippus Dec 17 '18 at 8:14
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.
$begingroup$
Are you trying to classify abelian groups of order 108?
$endgroup$
– the_fox
Dec 16 '18 at 14:49
$begingroup$
@the_fox Just $mathbb{Z}_{108}$. I'm currently studying about this theorem I mentioned and I've encountered many questions/problems similar to this one. I want to know what "classifying up to an isomorphism" means and how is it done for this example.
$endgroup$
– Jevaut
Dec 16 '18 at 14:56
$begingroup$
You can just use the Chinese remainder theorem and you see that $mathbb{Z}/108mathbb{Z} cong mathbb{Z}/4mathbb{Z} times mathbb{Z}/27mathbb{Z}$.
$endgroup$
– Mindlack
Dec 16 '18 at 14:58
$begingroup$
It doesn't make sense (to me) to ask for a classification up to isomorphism of a concrete, specific group. When a classification (up to isomorphism) of all groups with a specific property is asked for, what usually happens is that you need to provide a list of concrete groups which exhausts that list in that any group with the property is isomorphic to a group in the list. So the list you have provided is a classification of all abelian groups of order $108$.
$endgroup$
– the_fox
Dec 16 '18 at 15:06
$begingroup$
To see that your group is isomorphic to one in the list all you need is the theorem that $C_n times C_m cong C_{nm}$ when $gcd(n,m)=1$.
$endgroup$
– the_fox
Dec 16 '18 at 15:08
|
show 1 more comment
$begingroup$
Are you trying to classify abelian groups of order 108?
$endgroup$
– the_fox
Dec 16 '18 at 14:49
$begingroup$
@the_fox Just $mathbb{Z}_{108}$. I'm currently studying about this theorem I mentioned and I've encountered many questions/problems similar to this one. I want to know what "classifying up to an isomorphism" means and how is it done for this example.
$endgroup$
– Jevaut
Dec 16 '18 at 14:56
$begingroup$
You can just use the Chinese remainder theorem and you see that $mathbb{Z}/108mathbb{Z} cong mathbb{Z}/4mathbb{Z} times mathbb{Z}/27mathbb{Z}$.
$endgroup$
– Mindlack
Dec 16 '18 at 14:58
$begingroup$
It doesn't make sense (to me) to ask for a classification up to isomorphism of a concrete, specific group. When a classification (up to isomorphism) of all groups with a specific property is asked for, what usually happens is that you need to provide a list of concrete groups which exhausts that list in that any group with the property is isomorphic to a group in the list. So the list you have provided is a classification of all abelian groups of order $108$.
$endgroup$
– the_fox
Dec 16 '18 at 15:06
$begingroup$
To see that your group is isomorphic to one in the list all you need is the theorem that $C_n times C_m cong C_{nm}$ when $gcd(n,m)=1$.
$endgroup$
– the_fox
Dec 16 '18 at 15:08
$begingroup$
Are you trying to classify abelian groups of order 108?
$endgroup$
– the_fox
Dec 16 '18 at 14:49
$begingroup$
Are you trying to classify abelian groups of order 108?
$endgroup$
– the_fox
Dec 16 '18 at 14:49
$begingroup$
@the_fox Just $mathbb{Z}_{108}$. I'm currently studying about this theorem I mentioned and I've encountered many questions/problems similar to this one. I want to know what "classifying up to an isomorphism" means and how is it done for this example.
$endgroup$
– Jevaut
Dec 16 '18 at 14:56
$begingroup$
@the_fox Just $mathbb{Z}_{108}$. I'm currently studying about this theorem I mentioned and I've encountered many questions/problems similar to this one. I want to know what "classifying up to an isomorphism" means and how is it done for this example.
$endgroup$
– Jevaut
Dec 16 '18 at 14:56
$begingroup$
You can just use the Chinese remainder theorem and you see that $mathbb{Z}/108mathbb{Z} cong mathbb{Z}/4mathbb{Z} times mathbb{Z}/27mathbb{Z}$.
$endgroup$
– Mindlack
Dec 16 '18 at 14:58
$begingroup$
You can just use the Chinese remainder theorem and you see that $mathbb{Z}/108mathbb{Z} cong mathbb{Z}/4mathbb{Z} times mathbb{Z}/27mathbb{Z}$.
$endgroup$
– Mindlack
Dec 16 '18 at 14:58
$begingroup$
It doesn't make sense (to me) to ask for a classification up to isomorphism of a concrete, specific group. When a classification (up to isomorphism) of all groups with a specific property is asked for, what usually happens is that you need to provide a list of concrete groups which exhausts that list in that any group with the property is isomorphic to a group in the list. So the list you have provided is a classification of all abelian groups of order $108$.
$endgroup$
– the_fox
Dec 16 '18 at 15:06
$begingroup$
It doesn't make sense (to me) to ask for a classification up to isomorphism of a concrete, specific group. When a classification (up to isomorphism) of all groups with a specific property is asked for, what usually happens is that you need to provide a list of concrete groups which exhausts that list in that any group with the property is isomorphic to a group in the list. So the list you have provided is a classification of all abelian groups of order $108$.
$endgroup$
– the_fox
Dec 16 '18 at 15:06
$begingroup$
To see that your group is isomorphic to one in the list all you need is the theorem that $C_n times C_m cong C_{nm}$ when $gcd(n,m)=1$.
$endgroup$
– the_fox
Dec 16 '18 at 15:08
$begingroup$
To see that your group is isomorphic to one in the list all you need is the theorem that $C_n times C_m cong C_{nm}$ when $gcd(n,m)=1$.
$endgroup$
– the_fox
Dec 16 '18 at 15:08
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
What you have done is correct. However, you are not yet finished! You still need to show which of these direct products actually represents $G$, since those products aren't all isomorhpic to each other. Consider the order of each element of $G$. If you find one with, say, order 27, then $G$ is isomorphic to either $1.$ or $3.$ in your list. Next, you can try to find an element with order 2. If you find one, you are done, and the result is $3.$ . If you instead find an element with order 4, then $G$ is isomorphic to $1.$ .
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
$endgroup$
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
1
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you have done is correct. However, you are not yet finished! You still need to show which of these direct products actually represents $G$, since those products aren't all isomorhpic to each other. Consider the order of each element of $G$. If you find one with, say, order 27, then $G$ is isomorphic to either $1.$ or $3.$ in your list. Next, you can try to find an element with order 2. If you find one, you are done, and the result is $3.$ . If you instead find an element with order 4, then $G$ is isomorphic to $1.$ .
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
$endgroup$
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
1
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
add a comment |
$begingroup$
What you have done is correct. However, you are not yet finished! You still need to show which of these direct products actually represents $G$, since those products aren't all isomorhpic to each other. Consider the order of each element of $G$. If you find one with, say, order 27, then $G$ is isomorphic to either $1.$ or $3.$ in your list. Next, you can try to find an element with order 2. If you find one, you are done, and the result is $3.$ . If you instead find an element with order 4, then $G$ is isomorphic to $1.$ .
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
$endgroup$
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
1
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
add a comment |
$begingroup$
What you have done is correct. However, you are not yet finished! You still need to show which of these direct products actually represents $G$, since those products aren't all isomorhpic to each other. Consider the order of each element of $G$. If you find one with, say, order 27, then $G$ is isomorphic to either $1.$ or $3.$ in your list. Next, you can try to find an element with order 2. If you find one, you are done, and the result is $3.$ . If you instead find an element with order 4, then $G$ is isomorphic to $1.$ .
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
$endgroup$
What you have done is correct. However, you are not yet finished! You still need to show which of these direct products actually represents $G$, since those products aren't all isomorhpic to each other. Consider the order of each element of $G$. If you find one with, say, order 27, then $G$ is isomorphic to either $1.$ or $3.$ in your list. Next, you can try to find an element with order 2. If you find one, you are done, and the result is $3.$ . If you instead find an element with order 4, then $G$ is isomorphic to $1.$ .
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
answered Dec 16 '18 at 14:56
UnexpectedExpectationUnexpectedExpectation
769
769
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
1
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
add a comment |
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
1
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
$begingroup$
$4$ has order $27$, but $54$ has order $2$ and $27$ has order $4$, so I'm a bit confused now. Is it 1. or 3. ?
$endgroup$
– Jevaut
Dec 16 '18 at 15:11
1
1
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
$begingroup$
54 is contained within the cyclic subgroup generated by 27.
$endgroup$
– UnexpectedExpectation
Dec 16 '18 at 15:23
add a comment |
$begingroup$
Are you trying to classify abelian groups of order 108?
$endgroup$
– the_fox
Dec 16 '18 at 14:49
$begingroup$
@the_fox Just $mathbb{Z}_{108}$. I'm currently studying about this theorem I mentioned and I've encountered many questions/problems similar to this one. I want to know what "classifying up to an isomorphism" means and how is it done for this example.
$endgroup$
– Jevaut
Dec 16 '18 at 14:56
$begingroup$
You can just use the Chinese remainder theorem and you see that $mathbb{Z}/108mathbb{Z} cong mathbb{Z}/4mathbb{Z} times mathbb{Z}/27mathbb{Z}$.
$endgroup$
– Mindlack
Dec 16 '18 at 14:58
$begingroup$
It doesn't make sense (to me) to ask for a classification up to isomorphism of a concrete, specific group. When a classification (up to isomorphism) of all groups with a specific property is asked for, what usually happens is that you need to provide a list of concrete groups which exhausts that list in that any group with the property is isomorphic to a group in the list. So the list you have provided is a classification of all abelian groups of order $108$.
$endgroup$
– the_fox
Dec 16 '18 at 15:06
$begingroup$
To see that your group is isomorphic to one in the list all you need is the theorem that $C_n times C_m cong C_{nm}$ when $gcd(n,m)=1$.
$endgroup$
– the_fox
Dec 16 '18 at 15:08