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I'm trying to show that $U(n)^2$ is a proper subgroup of $U(n)$. Here

$$ U(n)^2 = \{x^2 \mid x \in U(n)\}$$

where $U(n)$ is the group of units modulo $n$.

My idea was to argue as follows:

Consider all elements of orders that are powers of $2$ in $U(n)$. Say these orders are $2^{k_1}, \dots, 2^{k_n}$ where $k_1 < \dots < k_n$. Then $U(n)^2$ does not contain an element of order $2^{k_n}$ (it will have elements of at most $2^{k_n - 1}$ for powers of $2$).

Is this correct?

I looked at the solution and it said that $1 = (n-1)^2$ implies it is a proper subgroup. But I don't think this works just like that: $n-1$ is minus one and there could be some $x \in U(n)$ such that $x^2 = -1$. So $U(n)^2$ could still contain $-1$.

What am I missing?

Jendrik Stelzner
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a student
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    You say $U(n)$ is the group of units modulo $n$, do you mean $U(n) \subset \mathbb{Z}/n\mathbb{Z}$? – walkar Feb 08 '16 at 02:33

2 Answers2

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First we need to realize that the statement does not hold for $n = 1,2$.

To make your proof work you need that $U(n)$ contains an element of order $2^\ell$ with $\ell > 1$, which amounts to showing that the order of $U(n)$ is even. This is not the case for $n = 1,2$ because $U(1)$ and $U(2)$ are the trival group. For $n > 2$ this is the statement that $\phi(n)$ is even for $n > 2$.

The method which the solutions hints at is much easier: You are right in arguing that $U(n)$ can contain $-1$, this happens for example for $n = 5$. The point is that $\varphi \colon U(n) \to U(n), x \mapsto x^2$ is a group homomorphism (because $U(n)$ is abelian) with image $U(n)^2$. Hence $U(n)^2 = U(n)$ if and only if $\varphi$ is surjective. Because $U(n)$ is finite this is equivalent to $\varphi$ being injective. Because $\varphi(1) = 1 = \varphi(-1)$ this is not the case for $n > 2$.

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Jendrik Stelzner
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$x\mapsto x^2$ is a group homomorphism $U(n)\to U(n)$. You want to show that it is not surjective. As the group is finite, this is equivalent to not beinng injective. And this can be seen from $1\ne -1$ but $1^2=(-1)^2$. (This works only for $n>2$)

Hagen von Eitzen
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