Mixing
On decomposition of finite Abelian groups
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It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $Asubset G$ and a subset $Bsubset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $A+B=a+b:ain A,;bin B$.
Problem. Is it true that for any finite abelian group $G$ and numbers $a,b$ with $abge|G|$ there are two subsets $A,Bsubset G$ of cardinality $|A|le a$ and $|B|le b$ such that $A+B=G$?
Remark. The answer is affirmative if the group $G$ is cyclic.
co.combinatorics gr.group-theory finite-groups additive-combinatorics abelian-groups
edited 7 hours ago
Martin Sleziak
2,75432028
asked 7 hours ago
Taras Banakh
14.1k12882
add a comment |Â
up vote
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It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $Asubset G$ and a subset $Bsubset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $A+B=a+b:ain A,;bin B$.
Problem. Is it true that for any finite abelian group $G$ and numbers $a,b$ with $abge|G|$ there are two subsets $A,Bsubset G$ of cardinality $|A|le a$ and $|B|le b$ such that $A+B=G$?
Remark. The answer is affirmative if the group $G$ is cyclic.
co.combinatorics gr.group-theory finite-groups additive-combinatorics abelian-groups
edited 7 hours ago
Martin Sleziak
2,75432028
asked 7 hours ago
Taras Banakh
14.1k12882
If $G$ has exponent 3 and $A=1,a$ has cardinal 2, then the injectivity of the sum map $Atimes Bto G$ implies its injectivity on $1,a,a^2times B$. This yields many further counterexamples.
– YCor
6 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $Asubset G$ and a subset $Bsubset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $A+B=a+b:ain A,;bin B$.
Problem. Is it true that for any finite abelian group $G$ and numbers $a,b$ with $abge|G|$ there are two subsets $A,Bsubset G$ of cardinality $|A|le a$ and $|B|le b$ such that $A+B=G$?
Remark. The answer is affirmative if the group $G$ is cyclic.
co.combinatorics gr.group-theory finite-groups additive-combinatorics abelian-groups
edited 7 hours ago
Martin Sleziak
2,75432028
asked 7 hours ago
Taras Banakh
14.1k12882
It is easy to see that for any finite Abelian group $G$ and any numbers $a,b$ with $|G|=ab$ there exist a subgroup $Asubset G$ and a subset $Bsubset G$ such that $|A|=a$, $|B|=b$ and $G=A+B$, where $A+B=a+b:ain A,;bin B$.
Problem. Is it true that for any finite abelian group $G$ and numbers $a,b$ with $abge|G|$ there are two subsets $A,Bsubset G$ of cardinality $|A|le a$ and $|B|le b$ such that $A+B=G$?
Remark. The answer is affirmative if the group $G$ is cyclic.
co.combinatorics gr.group-theory finite-groups additive-combinatorics abelian-groups
co.combinatorics gr.group-theory finite-groups additive-combinatorics abelian-groups
edited 7 hours ago
Martin Sleziak
2,75432028
asked 7 hours ago
Taras Banakh
14.1k12882
edited 7 hours ago
Martin Sleziak
2,75432028
asked 7 hours ago
Taras Banakh
14.1k12882
edited 7 hours ago
Martin Sleziak
2,75432028
edited 7 hours ago
Martin Sleziak
2,75432028
edited 7 hours ago
Martin Sleziak
2,75432028
2,75432028
asked 7 hours ago
Taras Banakh
14.1k12882
asked 7 hours ago
Taras Banakh
14.1k12882
asked 7 hours ago
Taras Banakh
14.1k12882
14.1k12882
If $G$ has exponent 3 and $A=1,a$ has cardinal 2, then the injectivity of the sum map $Atimes Bto G$ implies its injectivity on $1,a,a^2times B$. This yields many further counterexamples.
– YCor
6 hours ago
add a comment |Â
If $G$ has exponent 3 and $A=1,a$ has cardinal 2, then the injectivity of the sum map $Atimes Bto G$ implies its injectivity on $1,a,a^2times B$. This yields many further counterexamples.
– YCor
6 hours ago
If $G$ has exponent 3 and $A=1,a$ has cardinal 2, then the injectivity of the sum map $Atimes Bto G$ implies its injectivity on $1,a,a^2times B$. This yields many further counterexamples.
– YCor
6 hours ago
If $G$ has exponent 3 and $A=1,a$ has cardinal 2, then the injectivity of the sum map $Atimes Bto G$ implies its injectivity on $1,a,a^2times B$. This yields many further counterexamples.
– YCor
6 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
I don't think it's true for $G=mathbbF_2^3$ and $a=b=3$.
If there were such sets $A$ and $B$, they must have exactly three elements each.
By applying a translation and a group automorphism, we may as well take
$$A=(0,0,0),(1,0,0),(0,1,0).$$
Then $B$ either has at most one element with third coordinate zero, in which case $A+B$ has at most three of the four elements with third coordinate zero, or it has at most one element with third coordinate one, in which case $A+B$ has at most three of the four elements with third coordinate one.
answered 7 hours ago
Jeremy Rickard
20k15393
Yes indeed, you're right.
– YCor
6 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
I don't think it's true for $G=mathbbF_2^3$ and $a=b=3$.
If there were such sets $A$ and $B$, they must have exactly three elements each.
By applying a translation and a group automorphism, we may as well take
$$A=(0,0,0),(1,0,0),(0,1,0).$$
Then $B$ either has at most one element with third coordinate zero, in which case $A+B$ has at most three of the four elements with third coordinate zero, or it has at most one element with third coordinate one, in which case $A+B$ has at most three of the four elements with third coordinate one.
answered 7 hours ago
Jeremy Rickard
20k15393
Yes indeed, you're right.
– YCor
6 hours ago
add a comment |Â
up vote
6
down vote
accepted
I don't think it's true for $G=mathbbF_2^3$ and $a=b=3$.
If there were such sets $A$ and $B$, they must have exactly three elements each.
By applying a translation and a group automorphism, we may as well take
$$A=(0,0,0),(1,0,0),(0,1,0).$$
Then $B$ either has at most one element with third coordinate zero, in which case $A+B$ has at most three of the four elements with third coordinate zero, or it has at most one element with third coordinate one, in which case $A+B$ has at most three of the four elements with third coordinate one.
answered 7 hours ago
Jeremy Rickard
20k15393
Yes indeed, you're right.
– YCor
6 hours ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
I don't think it's true for $G=mathbbF_2^3$ and $a=b=3$.
If there were such sets $A$ and $B$, they must have exactly three elements each.
By applying a translation and a group automorphism, we may as well take
$$A=(0,0,0),(1,0,0),(0,1,0).$$
Then $B$ either has at most one element with third coordinate zero, in which case $A+B$ has at most three of the four elements with third coordinate zero, or it has at most one element with third coordinate one, in which case $A+B$ has at most three of the four elements with third coordinate one.
answered 7 hours ago
Jeremy Rickard
20k15393
I don't think it's true for $G=mathbbF_2^3$ and $a=b=3$.
If there were such sets $A$ and $B$, they must have exactly three elements each.
By applying a translation and a group automorphism, we may as well take
$$A=(0,0,0),(1,0,0),(0,1,0).$$
Then $B$ either has at most one element with third coordinate zero, in which case $A+B$ has at most three of the four elements with third coordinate zero, or it has at most one element with third coordinate one, in which case $A+B$ has at most three of the four elements with third coordinate one.
answered 7 hours ago
Jeremy Rickard
20k15393
answered 7 hours ago
Jeremy Rickard
20k15393
answered 7 hours ago
Jeremy Rickard
20k15393
answered 7 hours ago
Jeremy Rickard
20k15393
20k15393
Yes indeed, you're right.
– YCor
6 hours ago
add a comment |Â
Yes indeed, you're right.
– YCor
6 hours ago
Yes indeed, you're right.
– YCor
6 hours ago
Yes indeed, you're right.
– YCor
6 hours ago
add a comment |Â
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If $G$ has exponent 3 and $A=1,a$ has cardinal 2, then the injectivity of the sum map $Atimes Bto G$ implies its injectivity on $1,a,a^2times B$. This yields many further counterexamples.
– YCor
6 hours ago