Classification of abelian groups of order 72, and subgroups of order 4
Problem :
- Find all abelian groups of order 72, up to isomorphism.
- Which one of those groups are cyclic?
- Amont these subgroups, find all groups with a unique subgroup of order 4.
My answer : Let $G$ a abelian group such that $|G|=72=2^33^2$. We can find the elementary divisors of $G$ : $(2^3,3^2) ; (2,2^2,3^2) ; (2,2,2,3^2) ; (2^3,3,3);(2,2^2,3,3);(2,2,2,3,3)$.
It follows that $G$ is isomorphic to 6 abelian groups of order 72: $$C_{72}$$ $$C_2times C_{36},C_2times C_2times C_2$$ $$C_{24}times C_3, C_6times C_{12}, C_2times C_6times C_6$$
Now I can't seem to find which groups are cyclic. Quite intuitively I would say $C_{72}$ is cyclic because generated by $1$, $C_2times C_{36},C_2times C_2times C_2$ is generated by $(1,1,1)$ but I don't have a rigorous answer.
For the subgroup of order 4, I know that if $G$ is a cyclic group of order $m$ and if $n|m$, then there exists a unique subgroup of $G$ of order $n$. Then the group $C_{72}$ has a unique subgroup of order $4$ but what about the others?
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
add a comment |
Problem :
- Find all abelian groups of order 72, up to isomorphism.
- Which one of those groups are cyclic?
- Amont these subgroups, find all groups with a unique subgroup of order 4.
My answer : Let $G$ a abelian group such that $|G|=72=2^33^2$. We can find the elementary divisors of $G$ : $(2^3,3^2) ; (2,2^2,3^2) ; (2,2,2,3^2) ; (2^3,3,3);(2,2^2,3,3);(2,2,2,3,3)$.
It follows that $G$ is isomorphic to 6 abelian groups of order 72: $$C_{72}$$ $$C_2times C_{36},C_2times C_2times C_2$$ $$C_{24}times C_3, C_6times C_{12}, C_2times C_6times C_6$$
Now I can't seem to find which groups are cyclic. Quite intuitively I would say $C_{72}$ is cyclic because generated by $1$, $C_2times C_{36},C_2times C_2times C_2$ is generated by $(1,1,1)$ but I don't have a rigorous answer.
For the subgroup of order 4, I know that if $G$ is a cyclic group of order $m$ and if $n|m$, then there exists a unique subgroup of $G$ of order $n$. Then the group $C_{72}$ has a unique subgroup of order $4$ but what about the others?
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
1
The only cyclic group is the first one, of course: $;C_{72};$ , since there is only one cyclic group, up to isomorphism, for any order ( up to $;aleph_0,$). And this is a rigurous argument which, I presume, is studied way before classifying finite abelian groups up to isomorphism.
– DonAntonio
Jan 4 at 12:07
add a comment |
Problem :
- Find all abelian groups of order 72, up to isomorphism.
- Which one of those groups are cyclic?
- Amont these subgroups, find all groups with a unique subgroup of order 4.
My answer : Let $G$ a abelian group such that $|G|=72=2^33^2$. We can find the elementary divisors of $G$ : $(2^3,3^2) ; (2,2^2,3^2) ; (2,2,2,3^2) ; (2^3,3,3);(2,2^2,3,3);(2,2,2,3,3)$.
It follows that $G$ is isomorphic to 6 abelian groups of order 72: $$C_{72}$$ $$C_2times C_{36},C_2times C_2times C_2$$ $$C_{24}times C_3, C_6times C_{12}, C_2times C_6times C_6$$
Now I can't seem to find which groups are cyclic. Quite intuitively I would say $C_{72}$ is cyclic because generated by $1$, $C_2times C_{36},C_2times C_2times C_2$ is generated by $(1,1,1)$ but I don't have a rigorous answer.
For the subgroup of order 4, I know that if $G$ is a cyclic group of order $m$ and if $n|m$, then there exists a unique subgroup of $G$ of order $n$. Then the group $C_{72}$ has a unique subgroup of order $4$ but what about the others?
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
Problem :
- Find all abelian groups of order 72, up to isomorphism.
- Which one of those groups are cyclic?
- Amont these subgroups, find all groups with a unique subgroup of order 4.
My answer : Let $G$ a abelian group such that $|G|=72=2^33^2$. We can find the elementary divisors of $G$ : $(2^3,3^2) ; (2,2^2,3^2) ; (2,2,2,3^2) ; (2^3,3,3);(2,2^2,3,3);(2,2,2,3,3)$.
It follows that $G$ is isomorphic to 6 abelian groups of order 72: $$C_{72}$$ $$C_2times C_{36},C_2times C_2times C_2$$ $$C_{24}times C_3, C_6times C_{12}, C_2times C_6times C_6$$
Now I can't seem to find which groups are cyclic. Quite intuitively I would say $C_{72}$ is cyclic because generated by $1$, $C_2times C_{36},C_2times C_2times C_2$ is generated by $(1,1,1)$ but I don't have a rigorous answer.
For the subgroup of order 4, I know that if $G$ is a cyclic group of order $m$ and if $n|m$, then there exists a unique subgroup of $G$ of order $n$. Then the group $C_{72}$ has a unique subgroup of order $4$ but what about the others?
group-theory finite-groups abelian-groups cyclic-groups
group-theory finite-groups abelian-groups cyclic-groups
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
asked Jan 4 at 11:03
NotAbelianGroupNotAbelianGroup
588
588
1
The only cyclic group is the first one, of course: $;C_{72};$ , since there is only one cyclic group, up to isomorphism, for any order ( up to $;aleph_0,$). And this is a rigurous argument which, I presume, is studied way before classifying finite abelian groups up to isomorphism.
– DonAntonio
Jan 4 at 12:07
add a comment |
1
The only cyclic group is the first one, of course: $;C_{72};$ , since there is only one cyclic group, up to isomorphism, for any order ( up to $;aleph_0,$). And this is a rigurous argument which, I presume, is studied way before classifying finite abelian groups up to isomorphism.
– DonAntonio
Jan 4 at 12:07
1
1
The only cyclic group is the first one, of course: $;C_{72};$ , since there is only one cyclic group, up to isomorphism, for any order ( up to $;aleph_0,$). And this is a rigurous argument which, I presume, is studied way before classifying finite abelian groups up to isomorphism.
– DonAntonio
Jan 4 at 12:07
The only cyclic group is the first one, of course: $;C_{72};$ , since there is only one cyclic group, up to isomorphism, for any order ( up to $;aleph_0,$). And this is a rigurous argument which, I presume, is studied way before classifying finite abelian groups up to isomorphism.
– DonAntonio
Jan 4 at 12:07
add a comment |
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The only cyclic group is the first one, of course: $;C_{72};$ , since there is only one cyclic group, up to isomorphism, for any order ( up to $;aleph_0,$). And this is a rigurous argument which, I presume, is studied way before classifying finite abelian groups up to isomorphism.
– DonAntonio
Jan 4 at 12:07