Mixing
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Finite groups isomorphic to a subgroup of $F^+$
Clash Royale CLAN TAG#URR8PPP
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Let $F$ be a field. Determine the possible finite groups $G$ that are isomorphic to a subgroup of $F^+,$ the additive group of $F$.
This problem is a bit confusing for me. $F$ is an arbitrary field, and $F^+$ is an abelian group (finite or infinite). How can I determine the subgroups of a group about which I only know that it's abelian? Does the fact that the set $F-0$ is a multiplicative group (and other field axioms) tell me anything about $F^+$?
abstract-algebra group-theory field-theory finite-groups finite-fields
asked Jul 27 at 0:49
user437309
558212
 |Â
show 1 more comment
up vote
2
down vote
favorite
Let $F$ be a field. Determine the possible finite groups $G$ that are isomorphic to a subgroup of $F^+,$ the additive group of $F$.
This problem is a bit confusing for me. $F$ is an arbitrary field, and $F^+$ is an abelian group (finite or infinite). How can I determine the subgroups of a group about which I only know that it's abelian? Does the fact that the set $F-0$ is a multiplicative group (and other field axioms) tell me anything about $F^+$?
abstract-algebra group-theory field-theory finite-groups finite-fields
asked Jul 27 at 0:49
user437309
558212
2
Do you know about the characteristic of a field?
– Alex Wertheim
Jul 27 at 1:05
Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $mathbb F_p$?
– user437309
Jul 27 at 1:12
Yep, you got it! And any finite additive subgroup will be a vector subspace over $mathbbF_p$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $mathbbF_p$.)
– Alex Wertheim
Jul 27 at 1:14
@AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $mathbb F_p$.
– user437309
Jul 27 at 1:33
If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $Delta$ be a finite additive subgroup of $F$, i.e. a finite $mathbbZ$-submodule. Since every element of $Delta$ is $p$-torsion, the $mathbbZ$-module structure on $Delta$ descends to a $mathbbZ/pmathbbZ$-module structure on $Delta$.
– Alex Wertheim
Jul 27 at 1:38
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $F$ be a field. Determine the possible finite groups $G$ that are isomorphic to a subgroup of $F^+,$ the additive group of $F$.
This problem is a bit confusing for me. $F$ is an arbitrary field, and $F^+$ is an abelian group (finite or infinite). How can I determine the subgroups of a group about which I only know that it's abelian? Does the fact that the set $F-0$ is a multiplicative group (and other field axioms) tell me anything about $F^+$?
abstract-algebra group-theory field-theory finite-groups finite-fields
asked Jul 27 at 0:49
user437309
558212
Let $F$ be a field. Determine the possible finite groups $G$ that are isomorphic to a subgroup of $F^+,$ the additive group of $F$.
This problem is a bit confusing for me. $F$ is an arbitrary field, and $F^+$ is an abelian group (finite or infinite). How can I determine the subgroups of a group about which I only know that it's abelian? Does the fact that the set $F-0$ is a multiplicative group (and other field axioms) tell me anything about $F^+$?
abstract-algebra group-theory field-theory finite-groups finite-fields
asked Jul 27 at 0:49
user437309
558212
asked Jul 27 at 0:49
user437309
558212
asked Jul 27 at 0:49
user437309
558212
asked Jul 27 at 0:49
user437309
558212
558212
2
Do you know about the characteristic of a field?
– Alex Wertheim
Jul 27 at 1:05
Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $mathbb F_p$?
– user437309
Jul 27 at 1:12
Yep, you got it! And any finite additive subgroup will be a vector subspace over $mathbbF_p$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $mathbbF_p$.)
– Alex Wertheim
Jul 27 at 1:14
@AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $mathbb F_p$.
– user437309
Jul 27 at 1:33
If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $Delta$ be a finite additive subgroup of $F$, i.e. a finite $mathbbZ$-submodule. Since every element of $Delta$ is $p$-torsion, the $mathbbZ$-module structure on $Delta$ descends to a $mathbbZ/pmathbbZ$-module structure on $Delta$.
– Alex Wertheim
Jul 27 at 1:38
 |Â
show 1 more comment
2
Do you know about the characteristic of a field?
– Alex Wertheim
Jul 27 at 1:05
Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $mathbb F_p$?
– user437309
Jul 27 at 1:12
Yep, you got it! And any finite additive subgroup will be a vector subspace over $mathbbF_p$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $mathbbF_p$.)
– Alex Wertheim
Jul 27 at 1:14
@AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $mathbb F_p$.
– user437309
Jul 27 at 1:33
If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $Delta$ be a finite additive subgroup of $F$, i.e. a finite $mathbbZ$-submodule. Since every element of $Delta$ is $p$-torsion, the $mathbbZ$-module structure on $Delta$ descends to a $mathbbZ/pmathbbZ$-module structure on $Delta$.
– Alex Wertheim
Jul 27 at 1:38
2
2
Do you know about the characteristic of a field?
– Alex Wertheim
Jul 27 at 1:05
Do you know about the characteristic of a field?
– Alex Wertheim
Jul 27 at 1:05
Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $mathbb F_p$?
– user437309
Jul 27 at 1:12
Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $mathbb F_p$?
– user437309
Jul 27 at 1:12
Yep, you got it! And any finite additive subgroup will be a vector subspace over $mathbbF_p$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $mathbbF_p$.)
– Alex Wertheim
Jul 27 at 1:14
Yep, you got it! And any finite additive subgroup will be a vector subspace over $mathbbF_p$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $mathbbF_p$.)
– Alex Wertheim
Jul 27 at 1:14
@AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $mathbb F_p$.
– user437309
Jul 27 at 1:33
@AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $mathbb F_p$.
– user437309
Jul 27 at 1:33
If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $Delta$ be a finite additive subgroup of $F$, i.e. a finite $mathbbZ$-submodule. Since every element of $Delta$ is $p$-torsion, the $mathbbZ$-module structure on $Delta$ descends to a $mathbbZ/pmathbbZ$-module structure on $Delta$.
– Alex Wertheim
Jul 27 at 1:38
If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $Delta$ be a finite additive subgroup of $F$, i.e. a finite $mathbbZ$-submodule. Since every element of $Delta$ is $p$-torsion, the $mathbbZ$-module structure on $Delta$ descends to a $mathbbZ/pmathbbZ$-module structure on $Delta$.
– Alex Wertheim
Jul 27 at 1:38
 |Â
show 1 more comment
1 Answer
1
active
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up vote
1
down vote
Hint: Let $G$ be a finite subgroup of $F^+$ and let $g in G$ have order $n$. Consider $ng=0$ as the equation $(ncdot1_F)cdot g=0$ in the field $F$.
answered Jul 27 at 1:11
lhf
155k9160365
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Let $G$ be a finite subgroup of $F^+$ and let $g in G$ have order $n$. Consider $ng=0$ as the equation $(ncdot1_F)cdot g=0$ in the field $F$.
answered Jul 27 at 1:11
lhf
155k9160365
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
add a comment |Â
up vote
1
down vote
Hint: Let $G$ be a finite subgroup of $F^+$ and let $g in G$ have order $n$. Consider $ng=0$ as the equation $(ncdot1_F)cdot g=0$ in the field $F$.
answered Jul 27 at 1:11
lhf
155k9160365
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Let $G$ be a finite subgroup of $F^+$ and let $g in G$ have order $n$. Consider $ng=0$ as the equation $(ncdot1_F)cdot g=0$ in the field $F$.
answered Jul 27 at 1:11
lhf
155k9160365
Hint: Let $G$ be a finite subgroup of $F^+$ and let $g in G$ have order $n$. Consider $ng=0$ as the equation $(ncdot1_F)cdot g=0$ in the field $F$.
answered Jul 27 at 1:11
lhf
155k9160365
edited Jul 27 at 1:24
answered Jul 27 at 1:11
lhf
155k9160365
answered Jul 27 at 1:11
lhf
155k9160365
answered Jul 27 at 1:11
lhf
155k9160365
155k9160365
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
add a comment |Â
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
In the field, $ng$ is the sum of $n$ copies of $gin F$. If you mean I need to involve characteristic here, I'm not sure how exactly.
– user437309
Jul 27 at 1:23
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
Then we could multiply both sides by $g^-1$ on the right to get $ncdot 1_F=0$, which would imply that $n$ is a multiple of a prime $p$ (which is the characteristic). That's all I can conclude.
– user437309
Jul 27 at 1:32
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
@user437309, but since $pg=0$, what then can you say about $n$ since $n$ is minimal such that $ng=0$?
– lhf
Jul 27 at 1:47
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
Assuming $pg$ means $(pcdot 1_F)cdot g$, I can say that $n=p$. So if there is a non-trivial additive subgroup in $F^+$, then all elements in it have order $p$. (That's also what Alex Wertheim wrote above.) Again, by the above comments, such a subgroup is a vector space over $mathbb F_p$ so has order $p^r$. On the other hand, each such group is a subgroup of $(mathbb F_p^r,+)$ (namely the group itself). Is that correct? I believe that's all we needed to show, isn't it?
– user437309
Jul 27 at 1:59
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2
Do you know about the characteristic of a field?
– Alex Wertheim
Jul 27 at 1:05
Absolutely. Do you mean I need to use somehow that $F$ is a vector space over $mathbb F_p$?
– user437309
Jul 27 at 1:12
Yep, you got it! And any finite additive subgroup will be a vector subspace over $mathbbF_p$ of finite dimension, so that reduces the possibilities a lot. Now all you need to check is which possibilities are realized. (Another hint: finite field extensions of $mathbbF_p$.)
– Alex Wertheim
Jul 27 at 1:14
@AlexWertheim Perhaps it's easy, but for now I cannot see why any finite additive subgroup should be a vector space over $mathbb F_p$.
– user437309
Jul 27 at 1:33
If $F$ has a nonzero finite additive subgroup, then it must have a nonzero torsion element by Lagrange. Hence, $F$ must have characteristic $p > 0$ for some prime $p$, since fields of characteristic $0$ do not have nonzero torsion elements. We conclude that every element of $F$ is (additively) $p$-torsion. Now, let $Delta$ be a finite additive subgroup of $F$, i.e. a finite $mathbbZ$-submodule. Since every element of $Delta$ is $p$-torsion, the $mathbbZ$-module structure on $Delta$ descends to a $mathbbZ/pmathbbZ$-module structure on $Delta$.
– Alex Wertheim
Jul 27 at 1:38