How to prove using Lagrange's theorem $a^{p-1}\equiv1\bmod p$.

Question

How to prove using Lagrange's theorem $a^{p-1}\equiv 1\bmod p$.

Here, exponentiation (repeated multiplication) is of concern.

So, first want to show this property in action. Below shown for two values of $p=5,7$ with operation of exponentiation, the desired property, i.e. for $4$-th power $\equiv 1\bmod 5$, and $6$-th power $\equiv 1\bmod 7$, for all elements of the group. I.e., the last column follows the rule.

Say, $$\begin{array}{|c|c|c|c|c|} \hline &a^1&a^2&a^3&a^4\\ \hline 1 & 1& 1& 1&1\\ \hline 2 & 2 & 4& 3& 1\\ \hline 3 & 3 & 4& 2& 1\\ \hline 4 & 4 & 1& 4& 1\\ \hline \end{array}$$

The same pattern occurs for $p=7$, as shown in the below table : $$\begin{array}{|c|c|c|c|c|c|c|} \hline &a^1&a^2&a^3&a^4&a^5&a^6\\ \hline 1 & 1& 1& 1&1&1&1\\ \hline 2 & 2 & 4& 1& 2&4&1\\ \hline 3 & 3 & 2& 6& 4&5&1\\ \hline 4 & 4 & 2& 1& 4&2&1\\ \hline 5 & 5 & 4& 6 & 2 &3&1\\ \hline 6 & 6 & 1& 6 & 1 &6&1\\ \hline \end{array}$$

Doubt 1: But, none of the above is a group, so can the Lagrange theorem apply?

It says: order of element divides order of group.$- \tag{1}$

There is no group (wrt, exponential operation), and no order hence comes into view.

So, cannot say that can use property (1) to derive the given property from that.

Edit : expect a simple answer, so given tag accordingly.

Edit 1: as per suggestion by @Cpc - to see instead the group properties in multiplication table, which seems is due to the reason that basically exponentiation is repeated multiplication.

So, rather than checking if $<\mathbb {Z^*}, \wedge >$ is a group, need check that $<\mathbb{Z^*}, \times > $ is a group table or not.

$$\begin{array}{|c|c|c|c|c|} \hline \times \bmod 5 &1&2&3&4\\ \hline 1 & 1& 2& 3&4\\ \hline 2 & 2 & 4& 1& 3\\ \hline 3 & 3 & 1& 4& 2\\ \hline 4 & 4 & 3& 2& 1\\ \hline \end{array}$$

$$\begin{array}{|c|c|c|c|c|c|c|} \hline \times \bmod 7 &1&2&3&4&5&6\\ \hline 1 & 1& 2& 3&4&5&6\\ \hline 2 & 2 & 4& 6& 1&3&5\\ \hline 3 & 3 & 6& 2& 5&1&4\\ \hline 4 & 4 & 1& 5& 2&6&3\\ \hline 5 & 5 & 3& 1 & 6 &4&2\\ \hline 6 & 6 & 4& 4 & 3 &2&1\\ \hline \end{array}$$

Edit 2: Also, is there any other such problem in multiplication (similar to exponentiation operation here) where instead need check addition table (similar to addition operation here) for group properties?

Answer 1

You're working in the group of invertible elements modulo $n$, denoted e.g. $(\mathbb{Z}_n^*,\times)$.

• The set is $\{i \in \mathbb{Z}_n : \gcd(i,n)=1\}$, and
• The binary operation is multiplication modulo $n$.

So yes, Lagrange's Theorem applies here. When $n$ is a prime, the order of the group is $n-1$ (noting that $\gcd(0,n) \neq 1$).

When $n$ is prime, you're rediscovering Fermat's Little Theorem, otherwise we have Euler's Theorem.

Answer 2

See you actually start with a ring, $\Bbb Z_p$. Or, actually, it's even a field. And for any field, the nonzero elements form a (multiplicative) group, also denoted $\Bbb F^*$.

The group is then$\Bbb Z_p^×\cong\Bbb Z_{p-1}$, the group of units (of the ring).

So in your examples we get $\Bbb Z_5^×\cong\Bbb Z_4$ and $\Bbb Z_7^×\cong\Bbb Z_6$.

When $p$ is prime the group of units has order $p-1$.

And by Lagrange anything raised to the power $n=|G|$ is equal to $e$.