Which elements in $\mathbb{Z}_{26}$ do not have multiplicative inverses? 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 $\mathbb{Z}_{26}$.
 A: 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 $n$.
A: 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$.
A: 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.
A: 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.
