What are the invertible elements of multiplication in the ring $(\mathbb{Z}/12\mathbb{Z},+,\cdot)$?
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1Welcome to MSE! What have you tried? We will have an easier time helping you if we know what is giving you trouble ^_^ – HallaSurvivor Feb 17 '20 at 23:46
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There are only $12$ elements. Just check one by one if you have to. – lulu Feb 17 '20 at 23:46
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1There's only $12$ elements. But you want to find some $a*b = 1 + 12k$. – fleablood Feb 18 '20 at 00:07
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.... or in other words $ab + 12k = 1$ or more to the point when CAN'T you find $ab + 12k = 1$. – fleablood Feb 18 '20 at 00:08
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I actually meant Bezout say $ab+12 =1$ is possible when $b$ (and $a$) are relatively prime but not when they aren't. – fleablood Feb 18 '20 at 05:41
3 Answers
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The $\phi(12)=4$ elements relatively prime to $12$. Namely, $1,5,7,11$.
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The general answer is that the elements of $(\mathbf Z/n\mathbf Z)^\times$ are represented by those numbers between $1$ and $n-1$ which are coprime to $n$.
As a corollary, if $n$ is prime, $\mathbf Z/n\mathbf Z$ is a field.
Bernard
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Note: This is an unorthodox answer using claims that have not been verified by the mathematical community, but I couldn't resist taking the theory 'out for drive' and demonstrate how it doesn't 'fall off a cliff'.
Observe that when using this method we don't need Euler's totient function to get the count for the number of invertibles or identify them by checking for the numbers coprime to $21$. Rather, all the invertibles are 'lurking' in the factorization of larger numbers and are 'discovered' there by a 'combinatorial balancing' algorithm.
$\tag 1 1 \times 1 \equiv 1 \pmod{12}$
$\tag 2 11 \times 11 \equiv 1 \pmod{12}$
We will use our $\psi$-theory to find the remaining invertible elements.
$\quad 11,\quad \psi(11) = 36 \gt 12, \; \lambda(12,11) = 13$
$\quad 10,\quad \psi(10) = 30 \gt 12, \; \lambda(12,10) = 25 = 5 \times 5$
$\quad \;\,9,\quad \psi(9) \;\,= 25 \gt 12, \; \lambda(12,9) \;\,= 37$
$\quad \;\,8,\quad \psi(8) \;\,= 20 \gt 12, \; \lambda(12,8) \;\,= 49 = 7\times 7$
$\quad \;\,7,\quad \psi(7) \;\,= 16 \gt 12, \; \lambda(12,7) \;\,= 61$
$\quad \;\,6,\quad \psi(6) \;\,= 12 \le 12, \; \text{STOP}$
So the set of invertible elements, $J$, is given by
$\quad J = \{\overline1,\overline5,\overline7,\overline{11}\}$
and if $a \in J$ then $a^2 = 1$.
CopyPasteIt
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