# Find all integers $n$ such that the group of units of $\mathbb{Z}/n\mathbb{Z}$ is an elementary abelian $2$-group.

Question: Find all integers $$n$$ such that the group of units of $$\mathbb{Z}/n\mathbb{Z}$$ is an elementary abelian $$2$$-group.

Thoughts: I know that $$u$$ is a unit in $$\mathbb{Z}/n\mathbb{Z}$$ if and only if $$n$$ and $$u$$ are coprime. So, in each ring, there are $$\phi(n)$$ units. But I am not sure if there is a "nice" way to see if every non-trivial element in $$U(n)$$ has order $$2$$. I was thinking of breaking it down into even and odd $$n$$, or considering how many units each ring has, but I wasn't able to notice anything useful. Any help is greatly appreciated! Thank you.

The question is asked here: For what values of $n$ is $U(\mathbb{Z}_n)$ an elementary abelian 2-group?, but I wasn't able to get to a conclusion on based on the last hint.

• If $n \neq 3$ is odd, then $2^2=4 \not \equiv 1 \pmod n$, so you only have to worry about even $n$. Jul 29, 2021 at 20:12
• You can use the Chinese Remainder Theorem to reduce it to the prime power case, and in that situation it is easy. Jul 29, 2021 at 20:31
• @ArturoMagidin Based on Robert's comment, I would only have to worry about even $n$, and based on yours, I would only have to worry about when $n=2$? I'm not sure how I would use the CRT in this case, though. Jul 29, 2021 at 20:34
• @User7238: No, he is approaching it without decomposition, I am suggesting a different approach. Jul 29, 2021 at 20:34

Using the Chinese Remainder Theorem, since the units of $$A\times B$$ are $$U(A)\times U(B)$$, and a product is elementary $$2$$-abelian if and only if both factors are elementary $$2$$-abelian, we can reduce first to the prime power case. The answer then will be any $$n$$ that can be written as a product of "suitable" prime powers.

If $$p$$ is an odd prime, then $$(\mathbb{Z}/p^a\mathbb{Z})^*$$ is cyclic of order $$p^{a-1}(p-1)$$. For this to be elementary $$2$$-abelian, we need this number to be $$2$$ (the only cyclic elementary $$2$$-abelian group). So $$a=1$$ and $$p=3$$.

For powers of $$2$$, the unit group of $$\mathbb{Z}/2^a\mathbb{Z}$$ is: trivial if $$a=1$$; cyclic of order $$2$$ if $$a=2$$; and isomorphic to $$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{a-2}\mathbb{Z}$$ if $$a\geq 3$$. So this is elementary $$2$$-abelian if and only if $$a\leq 3$$.

Thus, $$n$$ must be of the form $$n=2^a3^b$$ with $$0\leq a\leq 3$$ and $$0\leq b\leq 1$$.

By Robert Shore's comment, for odd $$n$$, we have only $$n=3$$ working.

For even $$n$$, write $$n=2^k m$$ with odd $$m$$. By CRT as suggested by Arturo Magidin, $$(\mathbb{Z}/n\mathbb{Z})^* \simeq (\mathbb{Z}/2^k\mathbb{Z})^* \times (\mathbb{Z}/m\mathbb{Z})^*.$$ Again if $$m>3$$, there is an element of order greater than $$2$$. Thus, we have $$m=1, 3$$.

If $$k\geq 2$$, then we have $$(\mathbb{Z}/2^k\mathbb{Z})^* \simeq (\mathbb{Z}/2\mathbb{Z})\times (\mathbb{Z}/2^{k-2}\mathbb{Z}).$$ So, the group has an element of order greater than $$2$$ if $$k>3$$.

Hence, we must have $$n=1,2,3,4,6,8,12,24$$.

Here, $$n=1,2$$ yield trivial groups. $$n=3,4,6$$ yield $$\mathbb{Z}/2\mathbb{Z}$$, and $$n=8,12$$ yield $$(\mathbb{Z}/2\mathbb{Z})^2$$, $$n=24$$ yield $$(\mathbb{Z}/2\mathbb{Z})^3$$.