Proof of U(n) being automorphism of $Z_n$.
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I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.
I am not able to prove the onto part.
abstract-algebra automorphism-group
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I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.
I am not able to prove the onto part.
abstract-algebra automorphism-group
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.
I am not able to prove the onto part.
abstract-algebra automorphism-group
I want to prove that $rin U(n), alpha : Z_n to Z_n, alpha (s)=mod(rs, n)$ is an automorphism. I have proved that $alpha$ is one-one, maps identity to identity and $alpha(s+t) = alpha(s) + alpha(t)$.
I am not able to prove the onto part.
abstract-algebra automorphism-group
abstract-algebra automorphism-group
asked Sep 8 at 8:29
Piyush Divyanakar
3,315222
3,315222
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1 Answer
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Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,
$$
alpha(kt) = ktr equiv k pmodn .
$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,
$$
alpha(kt) = ktr equiv k pmodn .
$$
add a comment |Â
up vote
1
down vote
accepted
Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,
$$
alpha(kt) = ktr equiv k pmodn .
$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,
$$
alpha(kt) = ktr equiv k pmodn .
$$
Recall that $r$ is a unit of $mathbbZ_n$ if and only if $r$ is coprime with $n$. Thus, by Bezout's identity, we have $s$ and $t$ so that $sn + tr = 1$. Modulo $n$, this gives $tr equiv 1 pmodn$. Therefore, for any $k in mathbbZ_n$,
$$
alpha(kt) = ktr equiv k pmodn .
$$
answered Sep 8 at 8:41
Guido A.
4,806728
4,806728
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