Why is the order of a multiplicative modulo group divisible by the order of an element in its group? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Let $a$ be a quadratic residue modulo $p$. Prove $a^(p-1)/2 equiv 1 bmod p$.Groups with order divisible by $d$ and no element of order $d$Order of element in group given order of conjugacy classIsomorphism type of a finite group with respect to multiplication modulo 65The order of its elements in the additive group $mathbbZ/9mathbbZ$.“Every element of Sym$(n)$ has order at most $n$”The order of a conjugacy class of some element is equal to the index of the centralizer of that element.Order of primitive root (mod p) in the multiplicative group of integers modulo $p^a$Show that $a equiv pm 3pmod 8$ is $a^2 equiv 9 pmod 16$.square roots in multiplicative group of integers modulo n
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Why is the order of a multiplicative modulo group divisible by the order of an element in its group?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Let $a$ be a quadratic residue modulo $p$. Prove $a^(p-1)/2 equiv 1 bmod p$.Groups with order divisible by $d$ and no element of order $d$Order of element in group given order of conjugacy classIsomorphism type of a finite group with respect to multiplication modulo 65The order of its elements in the additive group $mathbbZ/9mathbbZ$.“Every element of Sym$(n)$ has order at most $n$”The order of a conjugacy class of some element is equal to the index of the centralizer of that element.Order of primitive root (mod p) in the multiplicative group of integers modulo $p^a$Show that $a equiv pm 3pmod 8$ is $a^2 equiv 9 pmod 16$.square roots in multiplicative group of integers modulo n
$begingroup$
In this question:
Find natural numbers $n,m$ such that the residue class $[m]_n in mathbbZ_n$ has order $5$.
My solution was $m = 2$ and $n = 31$ as $2^5 = 32 equiv 1 pmod31$.
However, the solutions stated that:
If $(mathbbZ_n, times)$ has an element of order $5$, then we know that its order is divisible by $5$.
Is my answer not a counter-example to this statement?
group-theory finite-groups modular-arithmetic
edited Mar 27 at 21:01
gt6989b
36k22557
asked Mar 27 at 20:55
ShreeShree
134
$endgroup$
add a comment |
$begingroup$
In this question:
Find natural numbers $n,m$ such that the residue class $[m]_n in mathbbZ_n$ has order $5$.
My solution was $m = 2$ and $n = 31$ as $2^5 = 32 equiv 1 pmod31$.
However, the solutions stated that:
If $(mathbbZ_n, times)$ has an element of order $5$, then we know that its order is divisible by $5$.
Is my answer not a counter-example to this statement?
group-theory finite-groups modular-arithmetic
edited Mar 27 at 21:01
gt6989b
36k22557
asked Mar 27 at 20:55
ShreeShree
134
$endgroup$
1
$begingroup$
The order of $BbbZ_31^*$ is $30$, which is a multiple of five.
$endgroup$
– Jyrki Lahtonen
Mar 27 at 20:59
1
$begingroup$
The multiplicative group of integers $pmod 31$ has order $30$.
$endgroup$
– lulu
Mar 27 at 21:00
$begingroup$
Oh, I completely forgot that [31] is not an element. Thanks for the quick reply!
$endgroup$
– Shree
Mar 27 at 21:02
1
$begingroup$
By Lagrange, the order of an element divides the order of the group, i.e., $5mid 30$. Here we have $phi(31)=30$.
$endgroup$
– Dietrich Burde
Mar 27 at 21:10
add a comment |
$begingroup$
In this question:
Find natural numbers $n,m$ such that the residue class $[m]_n in mathbbZ_n$ has order $5$.
My solution was $m = 2$ and $n = 31$ as $2^5 = 32 equiv 1 pmod31$.
However, the solutions stated that:
If $(mathbbZ_n, times)$ has an element of order $5$, then we know that its order is divisible by $5$.
Is my answer not a counter-example to this statement?
group-theory finite-groups modular-arithmetic
edited Mar 27 at 21:01
gt6989b
36k22557
asked Mar 27 at 20:55
ShreeShree
134
$endgroup$
In this question:
Find natural numbers $n,m$ such that the residue class $[m]_n in mathbbZ_n$ has order $5$.
My solution was $m = 2$ and $n = 31$ as $2^5 = 32 equiv 1 pmod31$.
However, the solutions stated that:
If $(mathbbZ_n, times)$ has an element of order $5$, then we know that its order is divisible by $5$.
Is my answer not a counter-example to this statement?
group-theory finite-groups modular-arithmetic
group-theory finite-groups modular-arithmetic
edited Mar 27 at 21:01
gt6989b
36k22557
asked Mar 27 at 20:55
ShreeShree
134
edited Mar 27 at 21:01
gt6989b
36k22557
asked Mar 27 at 20:55
ShreeShree
134
edited Mar 27 at 21:01
gt6989b
36k22557
edited Mar 27 at 21:01
gt6989b
36k22557
edited Mar 27 at 21:01
gt6989b
36k22557
36k22557
asked Mar 27 at 20:55
ShreeShree
134
asked Mar 27 at 20:55
ShreeShree
134
asked Mar 27 at 20:55
ShreeShree
134
134
1
$begingroup$
The order of $BbbZ_31^*$ is $30$, which is a multiple of five.
$endgroup$
– Jyrki Lahtonen
Mar 27 at 20:59
1
$begingroup$
The multiplicative group of integers $pmod 31$ has order $30$.
$endgroup$
– lulu
Mar 27 at 21:00
$begingroup$
Oh, I completely forgot that [31] is not an element. Thanks for the quick reply!
$endgroup$
– Shree
Mar 27 at 21:02
1
$begingroup$
By Lagrange, the order of an element divides the order of the group, i.e., $5mid 30$. Here we have $phi(31)=30$.
$endgroup$
– Dietrich Burde
Mar 27 at 21:10
add a comment |
1
$begingroup$
The order of $BbbZ_31^*$ is $30$, which is a multiple of five.
$endgroup$
– Jyrki Lahtonen
Mar 27 at 20:59
1
$begingroup$
The multiplicative group of integers $pmod 31$ has order $30$.
$endgroup$
– lulu
Mar 27 at 21:00
$begingroup$
Oh, I completely forgot that [31] is not an element. Thanks for the quick reply!
$endgroup$
– Shree
Mar 27 at 21:02
1
$begingroup$
By Lagrange, the order of an element divides the order of the group, i.e., $5mid 30$. Here we have $phi(31)=30$.
$endgroup$
– Dietrich Burde
Mar 27 at 21:10
1
1
$begingroup$
The order of $BbbZ_31^*$ is $30$, which is a multiple of five.
$endgroup$
– Jyrki Lahtonen
Mar 27 at 20:59
$begingroup$
The order of $BbbZ_31^*$ is $30$, which is a multiple of five.
$endgroup$
– Jyrki Lahtonen
Mar 27 at 20:59
1
1
$begingroup$
The multiplicative group of integers $pmod 31$ has order $30$.
$endgroup$
– lulu
Mar 27 at 21:00
$begingroup$
The multiplicative group of integers $pmod 31$ has order $30$.
$endgroup$
– lulu
Mar 27 at 21:00
$begingroup$
Oh, I completely forgot that [31] is not an element. Thanks for the quick reply!
$endgroup$
– Shree
Mar 27 at 21:02
$begingroup$
Oh, I completely forgot that [31] is not an element. Thanks for the quick reply!
$endgroup$
– Shree
Mar 27 at 21:02
1
1
$begingroup$
By Lagrange, the order of an element divides the order of the group, i.e., $5mid 30$. Here we have $phi(31)=30$.
$endgroup$
– Dietrich Burde
Mar 27 at 21:10
$begingroup$
By Lagrange, the order of an element divides the order of the group, i.e., $5mid 30$. Here we have $phi(31)=30$.
$endgroup$
– Dietrich Burde
Mar 27 at 21:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The Lagrange Theorem for groups helps here: the element in question generates a subgroup. The order of the element is the order of the subgroup. The order of the subgroup divides that of the group, by Lagrange's theorem.
answered Mar 27 at 21:32
avsavs
4,295515
$endgroup$
add a comment |
$begingroup$
Here's an easier explanation. Note, $1^b=1$ for any integer b. This means once you get to 1, you can power your result to any power. Modular
arithmetic groups, obey normal arithmetic, otherwise they wouldn't be useful in disproving numerical results. Your order is an exponent, by the exponent rule $(x^y)^z=x^ycdot z$, we get the same result as multiplying your order by anything. This makes them congruent to each other, and to 1. Therefore any multiple of an order of 1 mod something, is also an exponent that produces 1 for the base used. The order of the group, is the highest exponent required to make any base element get to 1. After getting to one the congruences repeat, so the group order, must be a multiple of every element's individual orders.
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The Lagrange Theorem for groups helps here: the element in question generates a subgroup. The order of the element is the order of the subgroup. The order of the subgroup divides that of the group, by Lagrange's theorem.
answered Mar 27 at 21:32
avsavs
4,295515
$endgroup$
add a comment |
$begingroup$
The Lagrange Theorem for groups helps here: the element in question generates a subgroup. The order of the element is the order of the subgroup. The order of the subgroup divides that of the group, by Lagrange's theorem.
answered Mar 27 at 21:32
avsavs
4,295515
$endgroup$
add a comment |
$begingroup$
The Lagrange Theorem for groups helps here: the element in question generates a subgroup. The order of the element is the order of the subgroup. The order of the subgroup divides that of the group, by Lagrange's theorem.
answered Mar 27 at 21:32
avsavs
4,295515
$endgroup$
The Lagrange Theorem for groups helps here: the element in question generates a subgroup. The order of the element is the order of the subgroup. The order of the subgroup divides that of the group, by Lagrange's theorem.
answered Mar 27 at 21:32
avsavs
4,295515
answered Mar 27 at 21:32
avsavs
4,295515
answered Mar 27 at 21:32
avsavs
4,295515
answered Mar 27 at 21:32
avsavs
4,295515
4,295515
add a comment |
add a comment |
$begingroup$
Here's an easier explanation. Note, $1^b=1$ for any integer b. This means once you get to 1, you can power your result to any power. Modular
arithmetic groups, obey normal arithmetic, otherwise they wouldn't be useful in disproving numerical results. Your order is an exponent, by the exponent rule $(x^y)^z=x^ycdot z$, we get the same result as multiplying your order by anything. This makes them congruent to each other, and to 1. Therefore any multiple of an order of 1 mod something, is also an exponent that produces 1 for the base used. The order of the group, is the highest exponent required to make any base element get to 1. After getting to one the congruences repeat, so the group order, must be a multiple of every element's individual orders.
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
$endgroup$
add a comment |
$begingroup$
Here's an easier explanation. Note, $1^b=1$ for any integer b. This means once you get to 1, you can power your result to any power. Modular
arithmetic groups, obey normal arithmetic, otherwise they wouldn't be useful in disproving numerical results. Your order is an exponent, by the exponent rule $(x^y)^z=x^ycdot z$, we get the same result as multiplying your order by anything. This makes them congruent to each other, and to 1. Therefore any multiple of an order of 1 mod something, is also an exponent that produces 1 for the base used. The order of the group, is the highest exponent required to make any base element get to 1. After getting to one the congruences repeat, so the group order, must be a multiple of every element's individual orders.
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
$endgroup$
add a comment |
$begingroup$
Here's an easier explanation. Note, $1^b=1$ for any integer b. This means once you get to 1, you can power your result to any power. Modular
arithmetic groups, obey normal arithmetic, otherwise they wouldn't be useful in disproving numerical results. Your order is an exponent, by the exponent rule $(x^y)^z=x^ycdot z$, we get the same result as multiplying your order by anything. This makes them congruent to each other, and to 1. Therefore any multiple of an order of 1 mod something, is also an exponent that produces 1 for the base used. The order of the group, is the highest exponent required to make any base element get to 1. After getting to one the congruences repeat, so the group order, must be a multiple of every element's individual orders.
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
$endgroup$
Here's an easier explanation. Note, $1^b=1$ for any integer b. This means once you get to 1, you can power your result to any power. Modular
arithmetic groups, obey normal arithmetic, otherwise they wouldn't be useful in disproving numerical results. Your order is an exponent, by the exponent rule $(x^y)^z=x^ycdot z$, we get the same result as multiplying your order by anything. This makes them congruent to each other, and to 1. Therefore any multiple of an order of 1 mod something, is also an exponent that produces 1 for the base used. The order of the group, is the highest exponent required to make any base element get to 1. After getting to one the congruences repeat, so the group order, must be a multiple of every element's individual orders.
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
answered Mar 29 at 2:04
Roddy MacPheeRoddy MacPhee
929118
929118
add a comment |
add a comment |
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1
$begingroup$
The order of $BbbZ_31^*$ is $30$, which is a multiple of five.
$endgroup$
– Jyrki Lahtonen
Mar 27 at 20:59
1
$begingroup$
The multiplicative group of integers $pmod 31$ has order $30$.
$endgroup$
– lulu
Mar 27 at 21:00
$begingroup$
Oh, I completely forgot that [31] is not an element. Thanks for the quick reply!
$endgroup$
– Shree
Mar 27 at 21:02
1
$begingroup$
By Lagrange, the order of an element divides the order of the group, i.e., $5mid 30$. Here we have $phi(31)=30$.
$endgroup$
– Dietrich Burde
Mar 27 at 21:10