If $g:mathbb{Z_{10}}rightarrow U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or...
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Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?
Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.
abstract-algebra group-theory functions group-homomorphism
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closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07
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add a comment |
$begingroup$
Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?
Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.
abstract-algebra group-theory functions group-homomorphism
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closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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$1$ has order $10$ in $mathbb{Z}_{10}$.
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– Randall
Dec 10 '18 at 5:18
add a comment |
$begingroup$
Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?
Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.
abstract-algebra group-theory functions group-homomorphism
$endgroup$
Why is that if $g:mathbb{Z_{10}}$$rightarrow$$U_{20}$ is a group homomorphism, then the order of $g(1)$ is either $1$ or $2$?
Also, $g$ is a function, $mathbb{Z_{10}}$ is the group of integers modulo $10$, and $U_{20}$ is the group of units modulo $20$.
abstract-algebra group-theory functions group-homomorphism
abstract-algebra group-theory functions group-homomorphism
edited Dec 10 '18 at 5:42
Shaun
9,241113684
9,241113684
asked Dec 10 '18 at 5:15
AMN52AMN52
326
326
closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh Dec 10 '18 at 12:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shaun, Gibbs, José Carlos Santos, John B, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18
add a comment |
1
$begingroup$
$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18
1
1
$begingroup$
$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18
$begingroup$
$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18
add a comment |
2 Answers
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Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.
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In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.
$endgroup$
add a comment |
$begingroup$
Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.
$endgroup$
add a comment |
$begingroup$
Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.
$endgroup$
Hint: $operatorname{Im}(g)$ is a subgroup of $U_{20}$ and $1$ (as an element of $Bbb Z_{10}$) generates $Bbb Z_{10}$.
answered Dec 10 '18 at 5:24
ShaunShaun
9,241113684
9,241113684
add a comment |
add a comment |
$begingroup$
In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.
$endgroup$
add a comment |
$begingroup$
In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.
$endgroup$
add a comment |
$begingroup$
In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.
$endgroup$
In fact, $U(20)=mathbb Z_4oplus mathbb Z_2$. So $operatorname{Im}g=langle g(1)rangle $ has order dividing $10$ and $4$.
answered Dec 10 '18 at 6:27
Chris CusterChris Custer
13.6k3827
13.6k3827
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add a comment |
1
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$1$ has order $10$ in $mathbb{Z}_{10}$.
$endgroup$
– Randall
Dec 10 '18 at 5:18