2
$\begingroup$

There is a question I am struggling on.

Let f : Z12 → Z12 : x ↦→ 9 x + 1 where arithmetic is done modulo 12.

(a) Show that f is neither injective, nor surjective.

(b) Now consider g, where g : Z12 → Z12 : x ↦ 7 x + 1. Show that g is invertible. Calculate g^−1(0) and g^−1(11). Find a formula for g^−1(x).

Hint: first find a number x so that 7 x is equivalent to 1 modulo 12. Now use this judiciously.

Can someone tell me if my working out is correct and how to improve the answer.

Here is what I worked out : a) f is not injective, because f(0) = 1 = f(4), but 1 and 4 are distinct mod 12.

Next, f is not surjective, because there is no x in Z12 such that f(x) = 9x + 1 = 2 (mod 12), since this is equivalent to 9x = 1 (mod 12) and 9 is not invertible mod 12, because gcd(9, 12) = 3 > 1.

(b) Given y in Z12, we want to find y in Z12 such that y = g^(-1)(x). <==> g(y) = x. <==> 7y + 1 = x (mod 12) <==> 7y = x - 1 (mod 12) <==> 7 * 7y = 7(x - 1) (mod 12) <==> y = 7x - 7 (mod 12), since 49 = 1 (mod 12) <==> y = 7x + 5 (mod 12), since -7 = 5 (mod 12).

That is, g^(-1)(x) = 7x + 5.

In particular, g^(-1)(0) = 5, and g^(-1)(11) = 82 = 10 (mod 12).

$\endgroup$
2
  • $\begingroup$ In the title, do you mean surjective $\endgroup$ Oct 8 '16 at 8:12
  • $\begingroup$ yes sorry that was a typo $\endgroup$
    – user376321
    Oct 8 '16 at 8:18
1
$\begingroup$

You could shorten you answer observing that, since $\mathbf Z/12\mathbf Z$ is a finite set, injective $\iff$ surjective$\iff$ bijective.

Furthermore, in any ring $R$, the map $x\mapsto ax+b$ is injective (resp. surjective, bijective) if and only if $x\mapsto ax$ is, and:

\begin{alignat}{2}&x\mapsto ax(+b)\;\text{is injective}&&\iff a\;\text{is a non-zero divisor},\\ &x\mapsto ax(+b)\;\text{is surjective}&&\iff a\in R^\times\;(\text{the set of units in}\;R). \end{alignat}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.