Abelian group of prime exponent $p$ is $mathbb{F}_p$-vector space?
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In reading about group cohomological, I came across the following in a statement of a lemma. "Let $p$ be a prime, and let $A$ be an abelian group of exponent dividing $p$..."
Is this just a roundabout way of saying that $A$ is an $mathbb{F}_p$-vector space? At least in the case where $A$ is finite or finitely generated, this seems obvious from the classification of finitely generated abelian groups. According to Exponent of a Group, I am correct, but perhaps I am missing an assumption there.
Are there abelian groups of exponent $p$ that are not $mathbb{F}_p$-vector spaces?
abelian-groups positive-characteristic
asked Nov 20 at 3:27
Joshua Ruiter
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up vote
1
down vote
favorite
In reading about group cohomological, I came across the following in a statement of a lemma. "Let $p$ be a prime, and let $A$ be an abelian group of exponent dividing $p$..."
Is this just a roundabout way of saying that $A$ is an $mathbb{F}_p$-vector space? At least in the case where $A$ is finite or finitely generated, this seems obvious from the classification of finitely generated abelian groups. According to Exponent of a Group, I am correct, but perhaps I am missing an assumption there.
Are there abelian groups of exponent $p$ that are not $mathbb{F}_p$-vector spaces?
abelian-groups positive-characteristic
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In reading about group cohomological, I came across the following in a statement of a lemma. "Let $p$ be a prime, and let $A$ be an abelian group of exponent dividing $p$..."
Is this just a roundabout way of saying that $A$ is an $mathbb{F}_p$-vector space? At least in the case where $A$ is finite or finitely generated, this seems obvious from the classification of finitely generated abelian groups. According to Exponent of a Group, I am correct, but perhaps I am missing an assumption there.
Are there abelian groups of exponent $p$ that are not $mathbb{F}_p$-vector spaces?
abelian-groups positive-characteristic
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
In reading about group cohomological, I came across the following in a statement of a lemma. "Let $p$ be a prime, and let $A$ be an abelian group of exponent dividing $p$..."
Is this just a roundabout way of saying that $A$ is an $mathbb{F}_p$-vector space? At least in the case where $A$ is finite or finitely generated, this seems obvious from the classification of finitely generated abelian groups. According to Exponent of a Group, I am correct, but perhaps I am missing an assumption there.
Are there abelian groups of exponent $p$ that are not $mathbb{F}_p$-vector spaces?
abelian-groups positive-characteristic
abelian-groups positive-characteristic
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
asked Nov 20 at 3:27
Joshua Ruiter
1,810619
1,810619
add a comment |
add a comment |
1 Answer
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Yes, this is the same thing as an $mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $mathbb{F}_p$ is the quotient ring $mathbb{Z}/(p)$, an $mathbb{F}_p$-module is the same thing as a $mathbb{Z}$-module in which every element is annihilated by $p$. But a $mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, this is the same thing as an $mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $mathbb{F}_p$ is the quotient ring $mathbb{Z}/(p)$, an $mathbb{F}_p$-module is the same thing as a $mathbb{Z}$-module in which every element is annihilated by $p$. But a $mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
add a comment |
up vote
2
down vote
accepted
Yes, this is the same thing as an $mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $mathbb{F}_p$ is the quotient ring $mathbb{Z}/(p)$, an $mathbb{F}_p$-module is the same thing as a $mathbb{Z}$-module in which every element is annihilated by $p$. But a $mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, this is the same thing as an $mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $mathbb{F}_p$ is the quotient ring $mathbb{Z}/(p)$, an $mathbb{F}_p$-module is the same thing as a $mathbb{Z}$-module in which every element is annihilated by $p$. But a $mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
Yes, this is the same thing as an $mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $mathbb{F}_p$ is the quotient ring $mathbb{Z}/(p)$, an $mathbb{F}_p$-module is the same thing as a $mathbb{Z}$-module in which every element is annihilated by $p$. But a $mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
answered Nov 20 at 3:42
Eric Wofsey
176k12202327
176k12202327
add a comment |
add a comment |
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