Multiplication modulo 10 in Cayley's Table 
In the Fourth row and second column, we have $3\cdot7 = 21$
But $9$ is the "limit", and note that there are $4$ elements in $U(10)$
Using modular arithmetic, $21= (4*5)+1$
Thus, $21 = 1 \pmod 4$
There ought to be $1$ instead of $7$ but the table implies otherwise. I'm pretty lost.
 A: For any $N$, the symbol $U(N)$ denotes the group of the integers invertible $\bmod N$ (or what amounts to say the integers having GCD $1$ with $N$) under multiplication.
Its order, i.e. the number of its elements, is $\varphi(N)$ where $\varphi$ denotes Euler's totient function.
As $U(N)$ is a group, general basic results of group theory imply that if ${\rm GCD}(a,N)=1$ then
$$
a^{\varphi(N)}\equiv1\bmod N.
$$
When $N=p$ is a prime number one has $\varphi(p)=p-1$, so this fact generalizes Fermat's Little Theorem.
Moreover, again when $N=p$ is prime, it can be proved that $U(p)$ is cyclic, i.e. there exists a class $\bar a\in U(p)$ (called a generator) such that any other class $\bar b$ is of the form $\bar b=\bar a^k$ for some $k\in\Bbb Z$. In fact, there exist always $\varphi(p-1)$ such generators in $U(p)$.
The ciclicity of $U(N)$ holds also for some composite $N$, but it is false in general: for instance $U(8)$ is not cyclic as $1^2\equiv3^2\equiv5^2\equiv7^2\equiv1\bmod8$ (thus, it is not possible to obtain $\bar 5$ taking powers of $\bar 3$ and so on)
