Intersection of two subgroups trivial, union is the whole group

[ToNoAvail27]

Homework Statement

Let $G$ be a group of order $n$ where $n$ is an odd squarefree prime (that is, $n=p_1p_2\cdots p_r$ where $p_i$ is an odd prime that appears only once, each $p_i$ distinct). Let $N$ be normal in $G$. If I have that $|G/N|=p_j$ for some prime in the prime factorization of $n$, and $|N|=\frac{n}{p_j}$, then are the following true?
$N \cap G/N = \{e\}$
and
$N \sqcup G/N = G$

Homework Equations

The Attempt at a Solution

For the first claim, if you take $N \cap G/N$ and obtain a non-trivial element, that element will generate $G/N$. And since that element is also in $N$, I believe that would mean $G/N < N$, but by LaGrange's theorem, since their orders are relatively prime, $G/N$ cannot be a subgroup of $N$. Thus, the intersection is trivial. I'm a bit stuck on getting started on the second condition.

 

[jbunniii]

$N \cap G/N = \{e\}$

What does $N \cap G/N$ mean? $N$ is a subset of $G$, but $G/N$ is not: it is a collection of subsets of $G$. What do you mean by the intersection of these objects?

 

[ToNoAvail27]

What $N \cap G/N$ mean? $N$ is a subset of $G$, but $G/N$ is not: it is a collection of subsets of $G$. What do you mean by the intersection of these objects?

Yes I apologize. When writing up the question, I got excited that this lemma would help me prove a bigger problem I had. Indeed, $G/N$ is a group since $N\vartriangleright G$, but of course $G/N$ itself is not a subgroup of $G$, which renders the lemma useless in my case, though, I do like the lemma. Thank you.