The concept of associate for finite abelian groups
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Let $G$ be a finite abelian group of order $n$, and let $a$ and $b$ be elements of $G$. If $a$ generates the same subgroup as $b$, must there be an integer $i$ prime to $n$ such that $ia=b$?
I can prove this when $G$ is cyclic, but not in general.
abstract-algebra number-theory
asked Dec 10 '18 at 8:04
falangfalang
309110
$endgroup$
|
show 3 more comments
$begingroup$
Let $G$ be a finite abelian group of order $n$, and let $a$ and $b$ be elements of $G$. If $a$ generates the same subgroup as $b$, must there be an integer $i$ prime to $n$ such that $ia=b$?
I can prove this when $G$ is cyclic, but not in general.
abstract-algebra number-theory
asked Dec 10 '18 at 8:04
falangfalang
309110
$endgroup$
$begingroup$
Where did you encounter this problem?
$endgroup$
– Shaun
Dec 10 '18 at 8:14
$begingroup$
Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group?
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:22
1
$begingroup$
@Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem.
$endgroup$
– falang
Dec 10 '18 at 8:26
$begingroup$
@Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$.
$endgroup$
– falang
Dec 10 '18 at 8:29
1
$begingroup$
@Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups.
$endgroup$
– falang
Dec 10 '18 at 8:39
|
show 3 more comments
$begingroup$
Let $G$ be a finite abelian group of order $n$, and let $a$ and $b$ be elements of $G$. If $a$ generates the same subgroup as $b$, must there be an integer $i$ prime to $n$ such that $ia=b$?
I can prove this when $G$ is cyclic, but not in general.
abstract-algebra number-theory
asked Dec 10 '18 at 8:04
falangfalang
309110
$endgroup$
Let $G$ be a finite abelian group of order $n$, and let $a$ and $b$ be elements of $G$. If $a$ generates the same subgroup as $b$, must there be an integer $i$ prime to $n$ such that $ia=b$?
I can prove this when $G$ is cyclic, but not in general.
abstract-algebra number-theory
abstract-algebra number-theory
asked Dec 10 '18 at 8:04
falangfalang
309110
asked Dec 10 '18 at 8:04
falangfalang
309110
asked Dec 10 '18 at 8:04
falangfalang
309110
asked Dec 10 '18 at 8:04
falangfalang
309110
asked Dec 10 '18 at 8:04
falangfalang
309110
309110
$begingroup$
Where did you encounter this problem?
$endgroup$
– Shaun
Dec 10 '18 at 8:14
$begingroup$
Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group?
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:22
1
$begingroup$
@Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem.
$endgroup$
– falang
Dec 10 '18 at 8:26
$begingroup$
@Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$.
$endgroup$
– falang
Dec 10 '18 at 8:29
1
$begingroup$
@Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups.
$endgroup$
– falang
Dec 10 '18 at 8:39
|
show 3 more comments
$begingroup$
Where did you encounter this problem?
$endgroup$
– Shaun
Dec 10 '18 at 8:14
$begingroup$
Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group?
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:22
1
$begingroup$
@Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem.
$endgroup$
– falang
Dec 10 '18 at 8:26
$begingroup$
@Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$.
$endgroup$
– falang
Dec 10 '18 at 8:29
1
$begingroup$
@Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups.
$endgroup$
– falang
Dec 10 '18 at 8:39
$begingroup$
Where did you encounter this problem?
$endgroup$
– Shaun
Dec 10 '18 at 8:14
$begingroup$
Where did you encounter this problem?
$endgroup$
– Shaun
Dec 10 '18 at 8:14
$begingroup$
Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group?
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:22
$begingroup$
Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group?
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:22
1
1
$begingroup$
@Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem.
$endgroup$
– falang
Dec 10 '18 at 8:26
$begingroup$
@Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem.
$endgroup$
– falang
Dec 10 '18 at 8:26
$begingroup$
@Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$.
$endgroup$
– falang
Dec 10 '18 at 8:29
$begingroup$
@Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$.
$endgroup$
– falang
Dec 10 '18 at 8:29
1
1
$begingroup$
@Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups.
$endgroup$
– falang
Dec 10 '18 at 8:39
$begingroup$
@Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups.
$endgroup$
– falang
Dec 10 '18 at 8:39
|
show 3 more comments
1 Answer
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$begingroup$
I should have thought more before posting... The answer is yes and the proof goes like this: Suppose first that $G$ is a $p$-group. By hypothesis there are integers $i$ and $j$ such that $ia=b$ and $jb=a$. We must show that it is possible to choose $i$ and $j$ prime to $p$. But the latter two equations imply that $ijb=b$, hence $(ij-1)b=0$. We may assume that $bne0$, otherwise there is nothing to prove. But then the order of $b$ divides $ij-1$. Since the order of $b$ is divisible by $p$, it follows that $i$ and $j$ are prime to $p$, which gives the required conclusion.
Finally, an arbitrary $G$ can be decomposed into a product of $p$-groups for distinct primes $p$. The result we want then follows readily from the Chinese Remainder Theorem.
answered Dec 10 '18 at 16:47
SJRSJR
51537
$endgroup$
add a comment |
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$begingroup$
I should have thought more before posting... The answer is yes and the proof goes like this: Suppose first that $G$ is a $p$-group. By hypothesis there are integers $i$ and $j$ such that $ia=b$ and $jb=a$. We must show that it is possible to choose $i$ and $j$ prime to $p$. But the latter two equations imply that $ijb=b$, hence $(ij-1)b=0$. We may assume that $bne0$, otherwise there is nothing to prove. But then the order of $b$ divides $ij-1$. Since the order of $b$ is divisible by $p$, it follows that $i$ and $j$ are prime to $p$, which gives the required conclusion.
Finally, an arbitrary $G$ can be decomposed into a product of $p$-groups for distinct primes $p$. The result we want then follows readily from the Chinese Remainder Theorem.
answered Dec 10 '18 at 16:47
SJRSJR
51537
$endgroup$
add a comment |
$begingroup$
I should have thought more before posting... The answer is yes and the proof goes like this: Suppose first that $G$ is a $p$-group. By hypothesis there are integers $i$ and $j$ such that $ia=b$ and $jb=a$. We must show that it is possible to choose $i$ and $j$ prime to $p$. But the latter two equations imply that $ijb=b$, hence $(ij-1)b=0$. We may assume that $bne0$, otherwise there is nothing to prove. But then the order of $b$ divides $ij-1$. Since the order of $b$ is divisible by $p$, it follows that $i$ and $j$ are prime to $p$, which gives the required conclusion.
Finally, an arbitrary $G$ can be decomposed into a product of $p$-groups for distinct primes $p$. The result we want then follows readily from the Chinese Remainder Theorem.
answered Dec 10 '18 at 16:47
SJRSJR
51537
$endgroup$
add a comment |
$begingroup$
I should have thought more before posting... The answer is yes and the proof goes like this: Suppose first that $G$ is a $p$-group. By hypothesis there are integers $i$ and $j$ such that $ia=b$ and $jb=a$. We must show that it is possible to choose $i$ and $j$ prime to $p$. But the latter two equations imply that $ijb=b$, hence $(ij-1)b=0$. We may assume that $bne0$, otherwise there is nothing to prove. But then the order of $b$ divides $ij-1$. Since the order of $b$ is divisible by $p$, it follows that $i$ and $j$ are prime to $p$, which gives the required conclusion.
Finally, an arbitrary $G$ can be decomposed into a product of $p$-groups for distinct primes $p$. The result we want then follows readily from the Chinese Remainder Theorem.
answered Dec 10 '18 at 16:47
SJRSJR
51537
$endgroup$
I should have thought more before posting... The answer is yes and the proof goes like this: Suppose first that $G$ is a $p$-group. By hypothesis there are integers $i$ and $j$ such that $ia=b$ and $jb=a$. We must show that it is possible to choose $i$ and $j$ prime to $p$. But the latter two equations imply that $ijb=b$, hence $(ij-1)b=0$. We may assume that $bne0$, otherwise there is nothing to prove. But then the order of $b$ divides $ij-1$. Since the order of $b$ is divisible by $p$, it follows that $i$ and $j$ are prime to $p$, which gives the required conclusion.
Finally, an arbitrary $G$ can be decomposed into a product of $p$-groups for distinct primes $p$. The result we want then follows readily from the Chinese Remainder Theorem.
answered Dec 10 '18 at 16:47
SJRSJR
51537
answered Dec 10 '18 at 16:47
SJRSJR
51537
answered Dec 10 '18 at 16:47
SJRSJR
51537
answered Dec 10 '18 at 16:47
SJRSJR
51537
51537
add a comment |
add a comment |
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$begingroup$
Where did you encounter this problem?
$endgroup$
– Shaun
Dec 10 '18 at 8:14
$begingroup$
Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group?
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 8:22
1
$begingroup$
@Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem.
$endgroup$
– falang
Dec 10 '18 at 8:26
$begingroup$
@Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$.
$endgroup$
– falang
Dec 10 '18 at 8:29
1
$begingroup$
@Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups.
$endgroup$
– falang
Dec 10 '18 at 8:39