The cardinality of the minimal generating set of $G = mathbbZ_2^m$ is $log_2 |G|$Why is the minimum size of a generating set for a finite group at most $log_2 n$?Generating set without elements of order 2Why is the minimum size of a generating set for a finite group at most $log_2 n$?Additive group of rationals has no minimal generating setThe minimum size of generating set of the external direct productWhat is the maximum cardinality of an independent set in a finite group?Is there an Unique Generating Set of a Group when Cardinality of the set is fixed?Why do we use the term “order” for the cardinality of a subgroup?The minimal size of generating set of quotient groupWhat is the minimum generating set of a matrix group?Minimal generating set
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The cardinality of the minimal generating set of $G = mathbbZ_2^m$ is $log_2 |G|$
Why is the minimum size of a generating set for a finite group at most $log_2 n$?Generating set without elements of order 2Why is the minimum size of a generating set for a finite group at most $log_2 n$?Additive group of rationals has no minimal generating setThe minimum size of generating set of the external direct productWhat is the maximum cardinality of an independent set in a finite group?Is there an Unique Generating Set of a Group when Cardinality of the set is fixed?Why do we use the term “order” for the cardinality of a subgroup?The minimal size of generating set of quotient groupWhat is the minimum generating set of a matrix group?Minimal generating set
$begingroup$
Referring to this thread Click Here
It is proven there that the minimum size of a generating set for a finite group at most $log_2 |G|$?
In the answer, it is noted without a proof that:
The cardinality of the minimal generating set of $G = mathbbZ_2^m$ is $log_2 |G|$
Can anyone explain why is that true?
group-theory finite-groups
edited 2 days ago
Asaf Karagila♦
306k33438769
asked 2 days ago
Um ShmumUm Shmum
1398
$endgroup$
add a comment |
$begingroup$
Referring to this thread Click Here
It is proven there that the minimum size of a generating set for a finite group at most $log_2 |G|$?
In the answer, it is noted without a proof that:
The cardinality of the minimal generating set of $G = mathbbZ_2^m$ is $log_2 |G|$
Can anyone explain why is that true?
group-theory finite-groups
edited 2 days ago
Asaf Karagila♦
306k33438769
asked 2 days ago
Um ShmumUm Shmum
1398
$endgroup$
add a comment |
$begingroup$
Referring to this thread Click Here
It is proven there that the minimum size of a generating set for a finite group at most $log_2 |G|$?
In the answer, it is noted without a proof that:
The cardinality of the minimal generating set of $G = mathbbZ_2^m$ is $log_2 |G|$
Can anyone explain why is that true?
group-theory finite-groups
edited 2 days ago
Asaf Karagila♦
306k33438769
asked 2 days ago
Um ShmumUm Shmum
1398
$endgroup$
Referring to this thread Click Here
It is proven there that the minimum size of a generating set for a finite group at most $log_2 |G|$?
In the answer, it is noted without a proof that:
The cardinality of the minimal generating set of $G = mathbbZ_2^m$ is $log_2 |G|$
Can anyone explain why is that true?
group-theory finite-groups
group-theory finite-groups
edited 2 days ago
Asaf Karagila♦
306k33438769
asked 2 days ago
Um ShmumUm Shmum
1398
edited 2 days ago
Asaf Karagila♦
306k33438769
asked 2 days ago
Um ShmumUm Shmum
1398
edited 2 days ago
Asaf Karagila♦
306k33438769
edited 2 days ago
Asaf Karagila♦
306k33438769
edited 2 days ago
Asaf Karagila♦
306k33438769
306k33438769
asked 2 days ago
Um ShmumUm Shmum
1398
asked 2 days ago
Um ShmumUm Shmum
1398
asked 2 days ago
Um ShmumUm Shmum
1398
1398
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Think of $mathbbZ_2^m$ as an $m$-dimensional vector space over $mathbbZ_2$. Then, a minimum generating set really is a basis.
answered 2 days ago
SinTan1729SinTan1729
2,672723
$endgroup$
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
1
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
|
show 3 more comments
Your Answer
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1 Answer
1
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1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
Hint: Think of $mathbbZ_2^m$ as an $m$-dimensional vector space over $mathbbZ_2$. Then, a minimum generating set really is a basis.
answered 2 days ago
SinTan1729SinTan1729
2,672723
$endgroup$
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
1
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
|
show 3 more comments
$begingroup$
Hint: Think of $mathbbZ_2^m$ as an $m$-dimensional vector space over $mathbbZ_2$. Then, a minimum generating set really is a basis.
answered 2 days ago
SinTan1729SinTan1729
2,672723
$endgroup$
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
1
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
|
show 3 more comments
$begingroup$
Hint: Think of $mathbbZ_2^m$ as an $m$-dimensional vector space over $mathbbZ_2$. Then, a minimum generating set really is a basis.
answered 2 days ago
SinTan1729SinTan1729
2,672723
$endgroup$
Hint: Think of $mathbbZ_2^m$ as an $m$-dimensional vector space over $mathbbZ_2$. Then, a minimum generating set really is a basis.
answered 2 days ago
SinTan1729SinTan1729
2,672723
edited 2 days ago
answered 2 days ago
SinTan1729SinTan1729
2,672723
answered 2 days ago
SinTan1729SinTan1729
2,672723
answered 2 days ago
SinTan1729SinTan1729
2,672723
2,672723
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
1
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
|
show 3 more comments
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
1
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
How does that help me show the result?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
@UmShmum The minimum cardinality for that minimal generator would then be $m$ and the answer you referred showed that the maximum value is $m$ too. So, it must be equal to $m$ then.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
How do you show that the dimension of the basis is greater than $m$?
$endgroup$
– Um Shmum
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
$begingroup$
@UmShmum What do you mean? Since this is an $m$ dimensional vector space, the cardinality of the basis is $m$.
$endgroup$
– SinTan1729
2 days ago
1
1
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
$begingroup$
Alternatively, since the group is abelian with exponent $2$, you could show that a subset of size $k$ generates a subgroup of order at most $2^k$. Hence to generate the whole group we must have $k ge m$.
$endgroup$
– Derek Holt
2 days ago
|
show 3 more comments
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