If $N$ is a normal subgroup of $G$ and if $x^2 \in N$ for all $x\in G$, prove that every nonidentity element of $G/N$ has order $2$. I've been trying to solve this proof for a while and couldn't figure out how to prove that $(gN)^2=eN$. I've tried to set up an equation where $gN \cdot g^{-1}N=eN$ but couldn't get to the end result.
An alternative, but perhaps more intuitive formulation of Tsemo Aristide's answer:
A non-identity element of $G/N$ is a coset $gN$ for $g\notin N$. The binary operation on $G/N$ is given by $(gN)(hN)=(gh)N$, and thus $(gN)^2=(gN)(gN)=g^2N$, but by assumption $g^2\in N$, so $(gN)^2=g^2N=N$, which is the identity element of $G/N$, so $gN$ has order 2. Since $gN$ was an arbitrary choice of non-identity element, any non-identity element has order 2.
Let $g\in G, n,n'\in N$. We have: $gngn'=gg(g^{-1}ng)n'$ since $g^{-1}ng\in N$ and $g^2\in N$, we deduce that $gngn'\in N$.