# The Group of Units

The group of units, Euler’s theorem, and Fermat’s little theorem are briefly discussed here.

## The group of units in the integers modulo n

In the previous sections, we’ve already seen that the set $\mathbb{Z}_{n}=\mathbb{Z} / n \mathbb{Z}$ consists of the elements $\{0,1,2, \ldots, n-1\}$, whereas $\mathbb{Z}_{n}$ under addition modulo $n$ (see this definition

Consequently, there’s no inverse for every element of $\mathbb{Z}_{n}$ in general. According to this definition

$(b+n \mathbb{Z})=(a \cdot b)+n \mathbb{Z}$. Therefore, for any $n=k l$ with integers $k>1$ and $l>1$, it holds that

$(k+n \mathbb{Z}) \cdot(l+n \mathbb{Z})=(k \cdot l)+n \mathbb{Z}=n+n \mathbb{Z}=0+n \mathbb{Z}=0. \quad\quad (1)$

But if $\mathbb{Z}_{n}$ under multiplication would form a group in general, there would’ve to exist an inverse element $k^{-1}$ for $k$, such that

$\left(k^{-1}+n \mathbb{Z}\right) \cdot(k+n \mathbb{Z})=1+n \mathbb{Z}.$

Hence, we multiply both sides of equation (1) by $\left(k^{-1}+n \mathbb{Z}\right)$, which yields.

$\begin{aligned} \left(k^{-1}+n \mathbb{Z}\right) \cdot(k+n \mathbb{Z}) \cdot(l+n \mathbb{Z}) &=\left(k^{-1}+n \mathbb{Z}\right) \cdot 0 \\ (1+n \mathbb{Z}) \cdot(l+n \mathbb{Z}) &=0 \\ (l+n \mathbb{Z}) &=0. \end{aligned}$

which contradicts to $1<l<n$. As a result, there’s no inverse for $k$ in general, and hence $\mathbb{Z}_{n}$ under multiplication is not necessarily a group for any $n \in \mathbb{N} .$

However, it’s possible to enforce a group structure if we constrain our attention only to the elements of $\mathbb{Z}_{n}$ that have multiplicative inverses. As a result, we get a group under multiplication $mod \space n$, that is often referred to as being the multiplicative group of integers modulo $n$ (or the group of nonzero congruence classes modulo $n$ ), which we call the **group of units** in $\mathbb{Z}_{n}$, denoted by $U_{n}$.

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