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Suppose $gcd(m,n)=1$, and let $F :Z_n→Z_n$ be defined by $F([a])=m[a]$. Prove that $F$ is an automorphism of the additive group $Z_n$. I find it is diffcult to prove $F$ is injective and surjective. Could you please to help me proof it with all the details. I type it roughly, and i am sorry and sincerely looking for a result.

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    $\begingroup$ Is it clear to you, that the equation $ma=0 \ mod(n)$ has trivial solution? In other words, that $\ker(F) = 0$ and hence the map is injective? An injective endomorphism of a group is automaticaly an isomorphism. $\endgroup$ – Lada Dudnikova Apr 22 at 8:14
  • $\begingroup$ @LadaDudnikova "An injective endomorphism of a group is automatically an isomorphism" - this is not true, see here. $\endgroup$ – Dietrich Burde Apr 22 at 10:48
  • $\begingroup$ @Lada Dudnikova Maybe for the finite group,then "An injective endomorphism of the group to itself is automatically an isomorphism" is true. However, could you please show me that how can i use $gcd(m,n)=1$, and the trivial solution to obtain the map is injective. I am confused about this corollary. $\endgroup$ – B1s Apr 22 at 11:03
  • $\begingroup$ @DietrichBurde sorry, forgot to mention finiteness of the group order here. $\endgroup$ – Lada Dudnikova Apr 22 at 11:03
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To show that the map is injective and surjective is equivalent to showing that the map has a two-sided inverse.

The extended Euclidean algorithm yields that there are numbers $m', n'$ such that $$ m m' + n n '= 1. $$ Consider the map $G : Z_{n} \to Z_{n}$ given by $G([b]) = m' [b]$. Then for all $a$ one has $$ G \circ F([a]) = G(F([a])) = G(m [a]) = m'm [a] = (1 - n' n) [a] = [a], $$ as $n [a] = [0]$. Similarly $F \circ G([b]) = [b]$ for all $b$.

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I shall prove two things

  1. The solution of $am = 0 \ (mod \ n)$ is trivial
  2. (1) means that the map is injective.

Let there a non-zero element s.t. $am = 0$ modulo $n$. The multiplication by integer in the group means consequtive addition(substraction) of the element to itself. Knowing that $gcd(m,n) = 1$ means that there is such a pair $(x',y')$ that $x'm+y'n = 1 \implies x'm = 1$ modulo $n$

This means $1 = a(x'm) = x'(am) = 0 $ contradiction.


Let the injectivity be violated. Then there is $ma = mb = c \neq 0$ modulo $n$

that means $m(a-b) = c \neq 0$ modulo $n$ which is a contradiction with the previous statement.

I admit that I operate rather dirty making commutative permutations and not justifying why in cyclic group I can do it. Do you need it to be justfied?

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