Application of Lagrange's Theorem to $Z/2018Z$
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I want to find the order of the subgroup of $mathbb{Z}/2018mathbb{Z}$ generated by 8.
I know that since the order of $mathbb{Z}/2018mathbb{Z}$ is finite I can use Lagrange's theorem which says that the order of the sungroup generated by 8 must divide the order of $mathbb{Z}/2018mathbb{Z}$ which is 2018.
Now I am not sure where to go from here besides brute force checking which divisor of 2018 is the order of the subgroup generated by 8
Is there some other theorem or fact I could use to make my life easier?
abstract-algebra
asked Nov 14 at 15:22
deco
185
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up vote
0
down vote
favorite
I want to find the order of the subgroup of $mathbb{Z}/2018mathbb{Z}$ generated by 8.
I know that since the order of $mathbb{Z}/2018mathbb{Z}$ is finite I can use Lagrange's theorem which says that the order of the sungroup generated by 8 must divide the order of $mathbb{Z}/2018mathbb{Z}$ which is 2018.
Now I am not sure where to go from here besides brute force checking which divisor of 2018 is the order of the subgroup generated by 8
Is there some other theorem or fact I could use to make my life easier?
abstract-algebra
asked Nov 14 at 15:22
deco
185
2
This question might help.
– Théophile
Nov 14 at 15:28
1
Can you factorize $2018$?
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 15:31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to find the order of the subgroup of $mathbb{Z}/2018mathbb{Z}$ generated by 8.
I know that since the order of $mathbb{Z}/2018mathbb{Z}$ is finite I can use Lagrange's theorem which says that the order of the sungroup generated by 8 must divide the order of $mathbb{Z}/2018mathbb{Z}$ which is 2018.
Now I am not sure where to go from here besides brute force checking which divisor of 2018 is the order of the subgroup generated by 8
Is there some other theorem or fact I could use to make my life easier?
abstract-algebra
asked Nov 14 at 15:22
deco
185
I want to find the order of the subgroup of $mathbb{Z}/2018mathbb{Z}$ generated by 8.
I know that since the order of $mathbb{Z}/2018mathbb{Z}$ is finite I can use Lagrange's theorem which says that the order of the sungroup generated by 8 must divide the order of $mathbb{Z}/2018mathbb{Z}$ which is 2018.
Now I am not sure where to go from here besides brute force checking which divisor of 2018 is the order of the subgroup generated by 8
Is there some other theorem or fact I could use to make my life easier?
abstract-algebra
abstract-algebra
asked Nov 14 at 15:22
deco
185
asked Nov 14 at 15:22
deco
185
asked Nov 14 at 15:22
deco
185
asked Nov 14 at 15:22
deco
185
asked Nov 14 at 15:22
deco
185
185
2
This question might help.
– Théophile
Nov 14 at 15:28
1
Can you factorize $2018$?
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 15:31
add a comment |
2
This question might help.
– Théophile
Nov 14 at 15:28
1
Can you factorize $2018$?
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 15:31
2
2
This question might help.
– Théophile
Nov 14 at 15:28
This question might help.
– Théophile
Nov 14 at 15:28
1
1
Can you factorize $2018$?
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 15:31
Can you factorize $2018$?
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 15:31
add a comment |
2 Answers
2
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oldest
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up vote
0
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Hint: In a cyclic group of order $n$ generated by $g$, the subgroup generated by $g^k$ has order $n/gcd(n,k)$.
answered Nov 14 at 15:28
lhf
161k9165384
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up vote
0
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Hint $, 2018mid 8niff 1009mid 4niff 1009mid n$
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: In a cyclic group of order $n$ generated by $g$, the subgroup generated by $g^k$ has order $n/gcd(n,k)$.
answered Nov 14 at 15:28
lhf
161k9165384
add a comment |
up vote
0
down vote
Hint: In a cyclic group of order $n$ generated by $g$, the subgroup generated by $g^k$ has order $n/gcd(n,k)$.
answered Nov 14 at 15:28
lhf
161k9165384
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: In a cyclic group of order $n$ generated by $g$, the subgroup generated by $g^k$ has order $n/gcd(n,k)$.
answered Nov 14 at 15:28
lhf
161k9165384
Hint: In a cyclic group of order $n$ generated by $g$, the subgroup generated by $g^k$ has order $n/gcd(n,k)$.
answered Nov 14 at 15:28
lhf
161k9165384
answered Nov 14 at 15:28
lhf
161k9165384
answered Nov 14 at 15:28
lhf
161k9165384
answered Nov 14 at 15:28
lhf
161k9165384
161k9165384
add a comment |
add a comment |
up vote
0
down vote
Hint $, 2018mid 8niff 1009mid 4niff 1009mid n$
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
add a comment |
up vote
0
down vote
Hint $, 2018mid 8niff 1009mid 4niff 1009mid n$
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint $, 2018mid 8niff 1009mid 4niff 1009mid n$
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
Hint $, 2018mid 8niff 1009mid 4niff 1009mid n$
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
answered Nov 14 at 15:50
Bill Dubuque
206k29189621
206k29189621
add a comment |
add a comment |
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2
This question might help.
– Théophile
Nov 14 at 15:28
1
Can you factorize $2018$?
– GNUSupporter 8964民主女神 地下教會
Nov 14 at 15:31