Prove that the subgroups of G have order 2 
Let $G = \{1_G, g, h, k\}$ be a group with $4$ elements and suppose $G$ is not cyclic.  Using Lagrange’s Theorem show that $g$, $h$ and $k$ all have order $2$ and write down a table for the group operation.

I am not really sure how I am supposed to prove it. I was thinking that since $|G| =  4$, using Lagrange's theorem we could say that the order of any subgroup of $G$ has to divide the order of $G$ and therefore the subgroups $\{1_G, g\}$, $\{1_G, h\}$, and $\{1_G, k\}$ have order $2$.
Secondly, what does it mean to write down a table for the group operation?
Any ideas?

Since $G$ is not cyclic, we have $\langle g \rangle \neq G$. This means that $\#\langle g \rangle \neq 4$. Since $\#\langle g \rangle \mid 4$ by Lagrange, we must have $\# \langle g \rangle = 2$, so $\langle g \rangle = \{e, g \}$ and $g^2=e$. Similarly, $h$ and $k$ have order $2$.
We now find
$$
\begin{array}{c|cccc}
\times & e & g & h & k \\
\hline
e & e & g & h & k \\
g & g & e & {\color{red}?} & ? \\
h & h & ? & e & ? \\
k & k & ? & ? & e
\end{array}
$$
for the multiplication table. Note that for the red "?" we must have $gh=k$, since $gh=g$ (which implies $h=e$), $gh=h$ (which implies $g=e$) and $gh=e$ (which implies $g=h$) are impossible. Similarly, we find the other elements of the multiplication table:
$$
\begin{array}{c|cccc}
\times & e & g & h & k \\
\hline
e & e & g & h & k \\
g & g & e & k & h \\
h & h & k & e & g \\
k & k & h & g & e
\end{array}
$$
Remark: this group is known as the Klein group.
