The powers of $2$ modulo 11;
\begin{array}{c|rrrrrrrrrrrrrrrr}
x& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \hline
2 ^x \bmod 11& 2& 4& 8& 5& 10& 9& 7& 3& 6& 1
\end{array}
We don't need the above 10 due to the Little Fermat Theorem; for a prime $p$ and $p\not| a$ then
$$a^{p-1} \equiv 1 \pmod p$$ We can use this theorem as
$$a^x \equiv a^{x \bmod{p-1}} \pmod p$$
Therefore your friend's power $12$ is equal to $2$; $12 \equiv 2 \pmod{10}$. In power;
$$ 2^{12} = 2^{12 \bmod{10}} = 2^2 = 4 \bmod 11.$$
A little theory;
The non-zero elements of modulus $11$ form a cyclic multiplicative group. The order of the group is given by $\varphi(p) = p-1$ where $\varphi$ is the Euler's totient function, so it has order $10$.
By the Lagrange theorem, the order of the subgroups must divide the order of the group. The converse, in general, is not true, that is for a divisor of the order of the group, there may not be a subgroup.
$2$ is a multiplicative generator, $\langle 2 \rangle = \mathbb{Z}_{11}^*$ and we can see this in the above table. Besides, $3$ is not a generator, since $\langle 3 \rangle = \{1,3,4,5,9\}$, with $1=3^5,3=3^1,4 = 3^4,5 = 3^3,9 =3^3$. As we can see, the order of $3$ is $ord(3)=5 | 10$.
What about order 2? It is generated by $\langle 10 \rangle = \{1,10\}$. The reason is $10 = -1 \bmod 11$
Note that for a prime $p$ forms a field $\mathbb{F}_{p}$ (also written as $GF(P)$) with integer addition and multiplication.
