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 $2$.

Question

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.

Answer 1

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 $gngn'\in N$.

Answer 2

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.

Answer 3

Here's an answer along the lines that you tried. As $N$ is a subgroup, $NN = N$. As $N$ is normal, $gNg^{-1} = N$ for every $g \in G$. As $g^2 \in N$ for every $g \in G$, $g^2N = N$ for every $g \in G$. So, for every $g \in G$:

$$ (gN)^2 = (gN)(gN) = (gNg^{-1})(g^2N) = NN = N $$