Mixing
Remembering how to find the elements of a certain order in a group
$begingroup$
I am working on a math problem and am stuck on remembering the correct way to solve it.
I basically need to figure out how much elements of order 8 are in Z (of 23432). (Where Z is a cyclic group under addition mod n).
I am not asking for the answer, I am looking for advice on how to remember how to find elements of an order for a group.
modular-arithmetic cyclic-groups
asked Feb 1 at 3:46
user638072user638072
61
$endgroup$
add a comment |
$begingroup$
I am working on a math problem and am stuck on remembering the correct way to solve it.
I basically need to figure out how much elements of order 8 are in Z (of 23432). (Where Z is a cyclic group under addition mod n).
I am not asking for the answer, I am looking for advice on how to remember how to find elements of an order for a group.
modular-arithmetic cyclic-groups
asked Feb 1 at 3:46
user638072user638072
61
$endgroup$
$begingroup$
This answer might get you on the right track.
$endgroup$
– The Chaz 2.0
Feb 1 at 4:05
add a comment |
$begingroup$
I am working on a math problem and am stuck on remembering the correct way to solve it.
I basically need to figure out how much elements of order 8 are in Z (of 23432). (Where Z is a cyclic group under addition mod n).
I am not asking for the answer, I am looking for advice on how to remember how to find elements of an order for a group.
modular-arithmetic cyclic-groups
asked Feb 1 at 3:46
user638072user638072
61
$endgroup$
I am working on a math problem and am stuck on remembering the correct way to solve it.
I basically need to figure out how much elements of order 8 are in Z (of 23432). (Where Z is a cyclic group under addition mod n).
I am not asking for the answer, I am looking for advice on how to remember how to find elements of an order for a group.
modular-arithmetic cyclic-groups
modular-arithmetic cyclic-groups
asked Feb 1 at 3:46
user638072user638072
61
asked Feb 1 at 3:46
user638072user638072
61
asked Feb 1 at 3:46
user638072user638072
61
asked Feb 1 at 3:46
user638072user638072
61
asked Feb 1 at 3:46
user638072user638072
61
61
$begingroup$
This answer might get you on the right track.
$endgroup$
– The Chaz 2.0
Feb 1 at 4:05
add a comment |
$begingroup$
This answer might get you on the right track.
$endgroup$
– The Chaz 2.0
Feb 1 at 4:05
$begingroup$
This answer might get you on the right track.
$endgroup$
– The Chaz 2.0
Feb 1 at 4:05
$begingroup$
This answer might get you on the right track.
$endgroup$
– The Chaz 2.0
Feb 1 at 4:05
add a comment |
1 Answer
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$begingroup$
Hint: $mid x^kmid=frac n{operatorname{gcd}(n,k)}$, where $n=mid xmid$, in any cyclic group.
Perhaps try a smaller group first. Say $Bbb Z_9$. How many elements of order $3$? The equivalence class of $3$. Also $6$. So there are two such elements. Use the hint to see these are the only two.
How many elements of order $4$? The answer is none. That's because $9$ and $4$ are relatively prime.
There is a theorem (Lagrange's theorem), that says the order of an element of a finite group has to divide the order of the group.
So, one of the first things to check is if $8mid 23432$. I think it does. Thus, there's at least one. Now use the hint to find it and any others.
This'll give you something to gnash your teeth on. Good luck!
Ps. As you will have seen in the linked answer above, the answer winds up being $varphi (n)$, where $varphi $ is Euler's totient function (if there are any solutions). Thus in the example I gave, $varphi (3)=2$. Similarly, the answer to your question will be $varphi (8)$.
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
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1 Answer
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$begingroup$
Hint: $mid x^kmid=frac n{operatorname{gcd}(n,k)}$, where $n=mid xmid$, in any cyclic group.
Perhaps try a smaller group first. Say $Bbb Z_9$. How many elements of order $3$? The equivalence class of $3$. Also $6$. So there are two such elements. Use the hint to see these are the only two.
How many elements of order $4$? The answer is none. That's because $9$ and $4$ are relatively prime.
There is a theorem (Lagrange's theorem), that says the order of an element of a finite group has to divide the order of the group.
So, one of the first things to check is if $8mid 23432$. I think it does. Thus, there's at least one. Now use the hint to find it and any others.
This'll give you something to gnash your teeth on. Good luck!
Ps. As you will have seen in the linked answer above, the answer winds up being $varphi (n)$, where $varphi $ is Euler's totient function (if there are any solutions). Thus in the example I gave, $varphi (3)=2$. Similarly, the answer to your question will be $varphi (8)$.
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
Hint: $mid x^kmid=frac n{operatorname{gcd}(n,k)}$, where $n=mid xmid$, in any cyclic group.
Perhaps try a smaller group first. Say $Bbb Z_9$. How many elements of order $3$? The equivalence class of $3$. Also $6$. So there are two such elements. Use the hint to see these are the only two.
How many elements of order $4$? The answer is none. That's because $9$ and $4$ are relatively prime.
There is a theorem (Lagrange's theorem), that says the order of an element of a finite group has to divide the order of the group.
So, one of the first things to check is if $8mid 23432$. I think it does. Thus, there's at least one. Now use the hint to find it and any others.
This'll give you something to gnash your teeth on. Good luck!
Ps. As you will have seen in the linked answer above, the answer winds up being $varphi (n)$, where $varphi $ is Euler's totient function (if there are any solutions). Thus in the example I gave, $varphi (3)=2$. Similarly, the answer to your question will be $varphi (8)$.
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
Hint: $mid x^kmid=frac n{operatorname{gcd}(n,k)}$, where $n=mid xmid$, in any cyclic group.
Perhaps try a smaller group first. Say $Bbb Z_9$. How many elements of order $3$? The equivalence class of $3$. Also $6$. So there are two such elements. Use the hint to see these are the only two.
How many elements of order $4$? The answer is none. That's because $9$ and $4$ are relatively prime.
There is a theorem (Lagrange's theorem), that says the order of an element of a finite group has to divide the order of the group.
So, one of the first things to check is if $8mid 23432$. I think it does. Thus, there's at least one. Now use the hint to find it and any others.
This'll give you something to gnash your teeth on. Good luck!
Ps. As you will have seen in the linked answer above, the answer winds up being $varphi (n)$, where $varphi $ is Euler's totient function (if there are any solutions). Thus in the example I gave, $varphi (3)=2$. Similarly, the answer to your question will be $varphi (8)$.
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
$endgroup$
Hint: $mid x^kmid=frac n{operatorname{gcd}(n,k)}$, where $n=mid xmid$, in any cyclic group.
Perhaps try a smaller group first. Say $Bbb Z_9$. How many elements of order $3$? The equivalence class of $3$. Also $6$. So there are two such elements. Use the hint to see these are the only two.
How many elements of order $4$? The answer is none. That's because $9$ and $4$ are relatively prime.
There is a theorem (Lagrange's theorem), that says the order of an element of a finite group has to divide the order of the group.
So, one of the first things to check is if $8mid 23432$. I think it does. Thus, there's at least one. Now use the hint to find it and any others.
This'll give you something to gnash your teeth on. Good luck!
Ps. As you will have seen in the linked answer above, the answer winds up being $varphi (n)$, where $varphi $ is Euler's totient function (if there are any solutions). Thus in the example I gave, $varphi (3)=2$. Similarly, the answer to your question will be $varphi (8)$.
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
edited Feb 1 at 5:40
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 4:23
Chris CusterChris Custer
14.3k3827
14.3k3827
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$begingroup$
This answer might get you on the right track.
$endgroup$
– The Chaz 2.0
Feb 1 at 4:05