This is a question from a practice quiz at my university.
Is the question asking for the cardinality of Σ1 = {a,b} to the power of four?
if that's the case, then the set would still have a cardinality of 2 since elements in a set are unique. It wouldn't be {a,b,a,b,a,b,a,b} right?
This is how you should solve it. Let $\Sigma_1=\Sigma$
We are asked to find: $|\Sigma^ 4|$ Now
$$\Sigma^4 = \Sigma . \Sigma . \Sigma . \Sigma = \{a,b\} . \{a,b\}. \{a,b\} . \{a,b\}$$
$$\text{ Here . means the concatenation operator}$$
So this is equivalent to number of ways of forming strings of length $4$ with symbols from $\Sigma$.
We have $4$ places,
Each of the places have two choices to get filled : either $a$ or $b$. So,
_ _ _ _
2 . 2. 2. 2
So answer is $2^4 =16$
A set "multiplied" with another set is called the cartesian product and is formed of tuples of elements, one from each set you are multiplying.
If you have $S = \{a, b\}$ and $T = \{x, y\}$, then $$S \times T = \{ (a,x), (a,y), (b, x), (b, y)\},$$ and its cardinality is actually $|S\times T| = |S| \cdot |T|$.
