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Apr 8 at 10:07 comment added Thrash Update: If there is more than one such element, we can consider all such elements together with $e$, which form a subgroup that (I think) is isomorphic to a $k$-fold copy of the cyclic group of order $2$ (where $k>1$), and it is not difficult to see that the product of all these elements yields $e$ because in each component, the element of order $2$ occurs an even number of times.
Apr 7 at 22:10 comment added Thrash What if there is more than one element of order 2? Then we just now that the product of all elements is the product of the elements of order 2 (which could be different from $e$).
Jun 20, 2022 at 5:26 comment added KConrad For prime $p$, $-1 \bmod p^\alpha$ has order $2$ unless $p^\alpha = 2$. So by the Chinese rem. thm, there is more than one unit mod $n$ of order $2$ if $n$ has more than one odd prime factor or if $n$ has one odd prime factor and is divisible by $4$. There are three elements of order $2$ in the units mod $2^\alpha$ if $\alpha \geq 3$. The only moduli $n > 1$ left for which there could be a unique unit of order $2$ are $2$, $4$, $p^\alpha$, and $2p^\alpha$ for odd primes $p$. For these moduli except $2$, only one unit has order $2$. We don’t need to know the moduli whose units are cyclic.
Jun 20, 2022 at 1:55 history edited Ofir Gorodetsky CC BY-SA 4.0
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Jun 20, 2022 at 0:37 history edited Ofir Gorodetsky CC BY-SA 4.0
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Jun 19, 2022 at 17:15 history answered Ofir Gorodetsky CC BY-SA 4.0