I don't know whether or not I am right about $\displaystyle 2^n$ and $\displaystyle n \in G$, where $\displaystyle G = \{-2,-1,0,1,2\}$ being an operation on $\displaystyle (G,*)$. Please tell me.

Suppose that I am correct, then I may say: Since it is a Group, it is a binary operation as well.

Suppose that the group $\displaystyle (G,*)$ is under multiplication.

Then $\displaystyle (n\cdot n)=2^n$ on $\displaystyle (G,\cdot)$. I constructed the group table below. Please verify whether it's correct.

$\displaystyle \begin{array}{c|c|c|c|c|c}\cdot&-2&-1&0&1&2\\

\hline

-2&2^4&2^2&2^0&2^1&2^2\\

\hline

-1&2^2&2^1&2^0&2^1&2^2\\

\hline

0&2^0&2^0&2^0&2^0&2^0\\

\hline

1&2^{-2}&2^{-1}&2^0&2^1&2^2\\

\hline

2&2^{-4}&2^{-2}&2^0&2^1&2^2\end{array}$

Since $\displaystyle (1*n)=2^n$, it follows that $\displaystyle 1$ is the identity of $\displaystyle (G,\cdot)$.

Yes?