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How would I find which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses? I assume I would find them by finding which elements in $\mathbb{Z}_{26}$ have a $\gcd(a, m) \neq 1$. Is this the correct approach? I have gotten that 2, 4, 6, 8, 10, 12, 13, 14, 16, 18, 20, 22, and 24 don't have multiplicative inverses in $\mathbb{Z}_{26}$.

Bernard
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Brendan Sullivan
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    Don't forget $0$. – Hayden Feb 25 '21 at 21:58
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    This is the correct approach, but do you understand why? You answer is correct but you forget $\gcd(0, 26)=26 \ne 1$ either ($0 = 26\times 0$ so $26$ divides $0$ and $26$ divides $26$ so $26$ is a common divisor of $0$ and $26$) so $0$ does not have an inverse. – fleablood Feb 25 '21 at 22:04
  • Yes, $a$ is invertible $!\bmod m\iff \gcd(a,m)=1,,$ see the dupe. – Bill Dubuque Apr 01 '21 at 20:58

4 Answers4

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Exactly and only those values not coprime to $26$: even values and $13$.

Perhaps more important is why there must be a multiplicative inverse for all the co-prime values. This hinges on knowing that there is only one multiplicative identity (that is, $1$) and so that $a\cdot x = x \implies a = 1$. Then we can build a running product of any coprime number $n$ (taking successive powers $n^2, n^3, \ldots$), which also can only be coprime values, and eventually we will reach $n^k=1$ - so the step before that, $n^{k-1}$, gave us the multiplicative inverse of $n$.

Joffan
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hint

$$26=2\times 13$$

$$\phi(26)=26(1-\frac 12)(1-\frac{1}{13})=12$$

So, there are $ 26-12=14 $ elements with no inverse.

$\phi(n) $ is the number of inversible elements in $ \Bbb Z_n$.

hamam_Abdallah
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That is exactly the right approach.

If $\gcd(a,m) =1$ then there will exist integers $x,y$ so that $ax + ym = 1$ and $ax = 1 + ym$ so $ax \equiv 1$ so $x$ is the inverse.

If $a$ has a multiplicative inverse $w$ then $aw \equiv 1 \pmod m$ so there is an integer $v$ so that $aw = 1 + mv$. So $aw -mv = 1$ but that's only possible if $\gcd(a,m) =1$.

So $a$ has an inverse $\pmod m$ if and only if $\gcd(a,m)=1$.

So you answer (if you include $0$) is correct.

fleablood
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The invertible elements form the group of units $\Bbb Z_{26}^×$, which has order $\varphi(26)=12$.

It's complement, $\Bbb Z_{26}\setminus\Bbb Z_{26}^×$ then consists of $14$ elements; namely, $0,2,4,6,8,10,12,13,14,16,18,20,22,24$. Those are the elements which are not coprime with $26$.

In addition they are all divisors of zero.