The context for this question is that I am going to present this proof, along a couple others, to my classmates, who are in high school and have some very basic knowledge in number theory and group theory. Thanks in advance. I am trying to prove the following:
Let $n$ be an integer, whose associated multiplicative group mod n (reduced residue system) is $M_n = \{1,r_2,...,r_{\phi(n)}\}$. Then $$\prod_{x \in M_n}x \equiv \begin{cases} -1 &\text{the multiplicative group mod n is cyclic (there is a primitive root)}\\ 1 &\text{otherwise} \end{cases} \mod n$$
Also note that this can be refined as the multiplicative group mod n is known to be cyclic if and only if $n = \{1,2,4,p^k,2p^k\}$, where $k$ is an integer and $p$ is an odd prime.
My proof:
First consider when it is cyclic. Since mod n is cyclic, $M_n\cong C_{\phi (n)}$, where $C_{\phi(n)} = \{1,a,a^2,...,a^{\phi(n)-1}\}$. First try to find the elements s.t. $x^2 \equiv 1 \mod n$. This is the same as finding the elements of order 2 in $C_{\phi (n)}$.
If $a^i \in C_{\phi (n)}$ has order 2, then $a^{2i} = a^{k\phi(n)}$, but since $i\leq\phi(n) - 1$, $2i \leq 2\phi(n) - 2$. Therefore $i = \phi(n)/2, 0$. Such a number always exists unless $n = 1,2$, but those cases are trivial to verify anyway. Therefore, there are 2 solutions to $x^2 \equiv 1 \mod n$. It is trivial to verify that these two numbers are 1 and -1.
Therefore, when $M_n$ is cyclic, each term will can be paired off with its unique inverse within the group so that:
$$ \prod_{x \in M_n}x \equiv (r_1\times r_1^{-1})(r_2\times r_2^{-1})...(1)(-1) \mod n $$ $$ \equiv (1)(1)...(1)(-1) \equiv -1 \mod n $$
Now on to when $M_n$ is non-cyclic. Consider the subgroup of $M_n$ consisting of elements of order 2, $S_n$. If $x \in S_n$ then so is $n-x$ as $(n-x)^2=x^2+n^2-2nx \equiv x^2 \equiv 1\mod n$, and $x$ cannot equal $n-x$ as that would imply $2x=n$ making $(x,n) \neq 1$ and so $x \not \in M_n$. So $S_n$ consists of an even number of terms, where each $x \in S_n$ has an corresponding $-x \in S_n$.
Consider one such element $u\neq 1,-1$. It is easy verify that the 4 elements, $\{1,-1,u,-u\}$, constitute a subgroup of $S_n$, so one can use Lagrange's theorem for subgroups to show that $|S_n| = 4k$ for some integer $k$.
Partition the product as follows: $$ \prod_{x \in M_n}x \equiv \prod_{x \in S_n} x \prod_{x \not \in S_n} x \mod n $$ The terms of the second product can be each paired off with their own inverse similar to how the first case was resolved, so it can be disregarded. The first product has $4k$ terms: $$ \equiv \prod_{x \in S_n} x = (-x_1\times x_1)(-x_2\times x_2)...(-1)(1) \equiv (-1)^{4k/2} = (-1)^{2k} = 1 \mod n $$ Proven as required.
My questions are that:
- Is the proof correct, and if not how can it be mended
- Is it clear what my logic is all the time
- Is this a well known result and if so does it have a name
