Show that $n\in\mathbb{N}$ is square-free

Question

Show that $n\in\mathbb{N}$ is square-free if and only if there is a subset different from the null set, $L\subseteq\mathbb{Z}_n$, with the property that summation and multiplication from $\mathbb Z_n$ induces on $L$ a group structure.

I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?

Answer 1

This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.

$n$ is square free if and only if there exists $\varnothing \subsetneq L\subseteq \Bbb{Z}/(n)$ such that $L$ is a group under addition.

Fails because $\Bbb{Z}/(9)$ contains $(3)$.

$n$ is square free if and only if there exists $\varnothing \subsetneq L\subseteq \Bbb{Z}/(n)$ such that $L$ is a group under multiplication.

Fails because $\Bbb{Z}/(9)$ contains $\langle 4\rangle$, which has order 3.

$n$ is square free if and only if there exists $\varnothing \subsetneq L\subseteq \Bbb{Z}/(n)$ such that $L$ is a group under addition and multiplication.

We can always take $L=\{0\}$ to satisfy this, and if we require $L\ne \{0\}$, then this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $x\in L$, $x\ne 0$, we have $x\cdot 0 = 0$, so $L$ cannot be a group under multiplication.

$n$ is square free if and only if there exists $\varnothing \subsetneq L\subseteq \Bbb{Z}/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $L\setminus \{0\}$ is a group under multiplication).

Let $x\in L$, $x\ne 0$. Then $x$ must be invertible in $\Bbb{Z}/(n)$, which means that it is relatively prime to $\Bbb{Z}/(n)$. Hence $x$ generates the additive group of $\Bbb{Z}/(n)$, which implies that $L=\Bbb{Z}/(n)$. Then $\Bbb{Z}/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.