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How to prove the product of the elements of $U_n$ , $B$is congruent to $1$ or $-1$.

$U_n$ is the set of all positive numbers less than and relatively prime to $n$.

We have to prove $B \equiv 1$ (mod n) where $B$ is the product of the elements of $U_n$.

My Attempt: We can see the statement is true for $n = 1 , 2$.

The number of elements of $U_n$ will be even for all natural number $n> 3$ as $\phi(n)$ is an even number for all natural number $n \geq 3$. So we will get odd number of elements of order $2$ for each group of order $n(\geq 3)$.

$B = b_1 b_2 ... b_k$ where $b_i \in U_n$ and $b_i$ is of order $2$ and $k$ is an odd number for all$n \geq 3$.

Now it remains to prove this product is congruent to $1$ or $-1$ mod n.

Can anyone please help me ?

anonymous
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  • But I did not understand the proof. I need a proof which used Group theory. Please open my question.@lulu – anonymous Mar 12 '21 at 16:08
  • This question has been asked over and over again on this site. I just linked to one duplicate in particular, but you can find others. – lulu Mar 12 '21 at 16:09
  • I could not find others. Can you please give other links ?@lulu – anonymous Mar 12 '21 at 16:11
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    This is known as Gauss' Generalization of Wilson's Theorem. There are links that appear in the duplicate I cited, or just look it up online. – lulu Mar 12 '21 at 16:15

1 Answers1

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Lemma Let $A$ be a multiplicatively notated finite elementary abelian group of order $2^k$, $k \in \mathbb{N}_{\gt 1}$. Then $\underset{a \in A}\prod a=1$.

Proof Regard $A$ as a $k$-dimensional vectorspace over the field of two elements $\mathbb{F}_2=\{0, 1\}$. Write 0 for the zero vector and 1 for the vector $(1,1, \cdots, 1)$ ($k$-coordinates). With each $v \in A$ we associate the vector 1 $− v$, when summing all vectors, i.e. all elements of $A$. Since there are $2^{k−1}$ such pairs, the sum equals $2^{k−1} \cdot $ 1 $\equiv$ 0 (mod $2$).

Corollary Let $A$ a multiplicatively notated finite abelian group and let $x=\underset{a \in A}\prod a$. Then $x=1$, unless the subgroup of all involutions, $I=\{a \in A: a^2=1\} \cong C_2$ in which case $x$ is the unique element of order $2$.
Proof $\underset{a \in A}\prod a=\underset{a \in I}\prod a \cdot \underset{a \notin I}\prod a$. If $a \notin I$, then $a \neq a^{-1}$, so $\underset{a \notin I}\prod a=1$, Now apply Lemma to $I$ and the proof is complete.

Note The corollary can be restated as follows: let $A$ be a finite abelian group and $P \in Syl_2(A)$, then the product $x$ of all elements of $A$ is $1$ if $P$ is trivial or non-cyclic, and if $P$ is non-trivial cyclic, $x$ equals the unique element of order $2$ of $A$.

Nicky Hekster
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