Mixing
Prove that $mathbb{Z}_{2^k}^* cong mathbb{Z}_2 times mathbb{Z}_{2^{k-2}}$
$begingroup$
In order to compute the value of the Carmichael function for $2^k$ with $k > 2$, my notes indicate that one has to stablish that $mathbb{Z}_{2^k}^* cong mathbb{Z}_2 times mathbb{Z}_{2^{k-2}}$ .
However, I have not been able to find this theorem in the bibliography (essentially Shoup's number theory), nor googling nor on stack exchange.
Does anyone know proof of this fact?
Edit
Here is a python script for the powers I was asked to compute:
for i in range(3,40):
find = True
for j in range(1,280):
if(5**j % 2**i == 1 and find == True):
find = False
print(2**i," has finished with ", j)
and here the results:
8 has finished with 2
16 has finished with 4
32 has finished with 8
64 has finished with 16
128 has finished with 32
256 has finished with 64
512 has finished with 128
1024 has finished with 256
So experimentally, it seems that $ord(5) = 2^{k-2}$ for $k > 2$.
abstract-algebra group-theory number-theory
asked Jan 9 at 0:12
JavierJavier
2,01821133
$endgroup$
add a comment |
$begingroup$
In order to compute the value of the Carmichael function for $2^k$ with $k > 2$, my notes indicate that one has to stablish that $mathbb{Z}_{2^k}^* cong mathbb{Z}_2 times mathbb{Z}_{2^{k-2}}$ .
However, I have not been able to find this theorem in the bibliography (essentially Shoup's number theory), nor googling nor on stack exchange.
Does anyone know proof of this fact?
Edit
Here is a python script for the powers I was asked to compute:
for i in range(3,40):
find = True
for j in range(1,280):
if(5**j % 2**i == 1 and find == True):
find = False
print(2**i," has finished with ", j)
and here the results:
8 has finished with 2
16 has finished with 4
32 has finished with 8
64 has finished with 16
128 has finished with 32
256 has finished with 64
512 has finished with 128
1024 has finished with 256
So experimentally, it seems that $ord(5) = 2^{k-2}$ for $k > 2$.
abstract-algebra group-theory number-theory
asked Jan 9 at 0:12
JavierJavier
2,01821133
$endgroup$
add a comment |
$begingroup$
In order to compute the value of the Carmichael function for $2^k$ with $k > 2$, my notes indicate that one has to stablish that $mathbb{Z}_{2^k}^* cong mathbb{Z}_2 times mathbb{Z}_{2^{k-2}}$ .
However, I have not been able to find this theorem in the bibliography (essentially Shoup's number theory), nor googling nor on stack exchange.
Does anyone know proof of this fact?
Edit
Here is a python script for the powers I was asked to compute:
for i in range(3,40):
find = True
for j in range(1,280):
if(5**j % 2**i == 1 and find == True):
find = False
print(2**i," has finished with ", j)
and here the results:
8 has finished with 2
16 has finished with 4
32 has finished with 8
64 has finished with 16
128 has finished with 32
256 has finished with 64
512 has finished with 128
1024 has finished with 256
So experimentally, it seems that $ord(5) = 2^{k-2}$ for $k > 2$.
abstract-algebra group-theory number-theory
asked Jan 9 at 0:12
JavierJavier
2,01821133
$endgroup$
In order to compute the value of the Carmichael function for $2^k$ with $k > 2$, my notes indicate that one has to stablish that $mathbb{Z}_{2^k}^* cong mathbb{Z}_2 times mathbb{Z}_{2^{k-2}}$ .
However, I have not been able to find this theorem in the bibliography (essentially Shoup's number theory), nor googling nor on stack exchange.
Does anyone know proof of this fact?
Edit
Here is a python script for the powers I was asked to compute:
for i in range(3,40):
find = True
for j in range(1,280):
if(5**j % 2**i == 1 and find == True):
find = False
print(2**i," has finished with ", j)
and here the results:
8 has finished with 2
16 has finished with 4
32 has finished with 8
64 has finished with 16
128 has finished with 32
256 has finished with 64
512 has finished with 128
1024 has finished with 256
So experimentally, it seems that $ord(5) = 2^{k-2}$ for $k > 2$.
abstract-algebra group-theory number-theory
abstract-algebra group-theory number-theory
asked Jan 9 at 0:12
JavierJavier
2,01821133
asked Jan 9 at 0:12
JavierJavier
2,01821133
edited Jan 9 at 0:32
Javier
asked Jan 9 at 0:12
JavierJavier
2,01821133
asked Jan 9 at 0:12
JavierJavier
2,01821133
asked Jan 9 at 0:12
JavierJavier
2,01821133
2,01821133
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: compute the order of $5$ in $(mathbb{Z}/(2^k))^{times}$.
Edit: I was asked to be more specific.
I suggest that you prove that $2^k$ does not divide $5^{2^{k-3}}-1$ but divides $5^{2^{k-2}}-1$. Thus, you can deduce that $(x,y) in mathbb{Z}/(2) times mathbb{Z}/(2^{k-2}) rightarrow (-1)^x5^y in (mathbb{Z}/(2^k))^{times}$ is an isomorphism.
answered Jan 9 at 0:14
MindlackMindlack
3,49217
$endgroup$
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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active
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$begingroup$
Hint: compute the order of $5$ in $(mathbb{Z}/(2^k))^{times}$.
Edit: I was asked to be more specific.
I suggest that you prove that $2^k$ does not divide $5^{2^{k-3}}-1$ but divides $5^{2^{k-2}}-1$. Thus, you can deduce that $(x,y) in mathbb{Z}/(2) times mathbb{Z}/(2^{k-2}) rightarrow (-1)^x5^y in (mathbb{Z}/(2^k))^{times}$ is an isomorphism.
answered Jan 9 at 0:14
MindlackMindlack
3,49217
$endgroup$
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
add a comment |
$begingroup$
Hint: compute the order of $5$ in $(mathbb{Z}/(2^k))^{times}$.
Edit: I was asked to be more specific.
I suggest that you prove that $2^k$ does not divide $5^{2^{k-3}}-1$ but divides $5^{2^{k-2}}-1$. Thus, you can deduce that $(x,y) in mathbb{Z}/(2) times mathbb{Z}/(2^{k-2}) rightarrow (-1)^x5^y in (mathbb{Z}/(2^k))^{times}$ is an isomorphism.
answered Jan 9 at 0:14
MindlackMindlack
3,49217
$endgroup$
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
add a comment |
$begingroup$
Hint: compute the order of $5$ in $(mathbb{Z}/(2^k))^{times}$.
Edit: I was asked to be more specific.
I suggest that you prove that $2^k$ does not divide $5^{2^{k-3}}-1$ but divides $5^{2^{k-2}}-1$. Thus, you can deduce that $(x,y) in mathbb{Z}/(2) times mathbb{Z}/(2^{k-2}) rightarrow (-1)^x5^y in (mathbb{Z}/(2^k))^{times}$ is an isomorphism.
answered Jan 9 at 0:14
MindlackMindlack
3,49217
$endgroup$
Hint: compute the order of $5$ in $(mathbb{Z}/(2^k))^{times}$.
Edit: I was asked to be more specific.
I suggest that you prove that $2^k$ does not divide $5^{2^{k-3}}-1$ but divides $5^{2^{k-2}}-1$. Thus, you can deduce that $(x,y) in mathbb{Z}/(2) times mathbb{Z}/(2^{k-2}) rightarrow (-1)^x5^y in (mathbb{Z}/(2^k))^{times}$ is an isomorphism.
answered Jan 9 at 0:14
MindlackMindlack
3,49217
edited Jan 9 at 0:52
answered Jan 9 at 0:14
MindlackMindlack
3,49217
answered Jan 9 at 0:14
MindlackMindlack
3,49217
answered Jan 9 at 0:14
MindlackMindlack
3,49217
3,49217
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
add a comment |
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
Is this using the classification group of finite abelian groups?
$endgroup$
– Javier
Jan 9 at 0:36
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
No. Just elementary number theory and basic finite group theory.
$endgroup$
– Mindlack
Jan 9 at 0:37
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
$begingroup$
To be honest I'm a bit lost. Could you provide some extra information?
$endgroup$
– Javier
Jan 9 at 0:46
add a comment |
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