# Prove that if $[G:H] = 2$ then $G/H \cong \{1,-1\}$

I know that if $H < G$ is an index two subgroup, then it must also be normal in $G$.

How do I show that $G/H \cong \{1,-1\}$

My thought is to define the function $\phi: G/H \to \{1,-1\}$ where $\phi(H) = 1$ and $\phi(gH) = -1$ where $g$ is such that $gH \not = H$.

This map is clearly surjective and injective.

My trouble is in verifying that $\phi$ is a homomorphism. Specifically

$\checkmark \phi(H \cdot H) = \phi(H) = 1 = \phi(H) \cdot \phi(H)$

$\checkmark \phi(gH \cdot H) = \phi(gH) = -1 = \phi(gH) \cdot \phi(H)$

Similarly, one can compute $\phi(H \cdot gH)$.

How do I show that

$\phi(gH \cdot gH) = \phi(g^2H) = 1 = \phi(gH) \cdot \phi(H)$

Essentially, I cannot show that if $g$ is such that $gH \not = H$ then $g^2H = H$.

As $H$ is normal, $Hg=gH$ so $gHgH=g^2HH=g^2H$. All you have to do is show that $g^2\in H$. If $g^2\notin H$, then $g^2\in gH$ which implies $g\in H$....
If $[G:H]=2$, then $|G/H| = 2$. Now by Lagrange's theorem, the order of any $kH \in G/H$ must divide $2$. Hence $k^2H = (kH)^2 = H$.
$G/H$ is a group of order $2$. Can you show that any group with two elements is isomorphic to $\{\pm 1\}$?