In Martin Lieback's book 'A Concise Introduction to Pure Mathematics', he posts an exercise(page 177,Q5):
Prove that exactly half of the $n!$ permutations in $S_n$ are even.
(Hint: Show that if $g$ is an even permutation, then $g(12)$ is odd. Try to use this to define a bijection from the set of odd permutations to the set of even permutations.)
I couldn't understand what is meant by $g(12)$. Does it mean the composite permutation function $g\circ (1 \ \ 2)$ or the function $g\circ (12)$? Also, what would be your thought process for this question?
It is the composition of the permutation $g$ and the transposition $(1\ 2)$. Since $g$ is even and all transpositions are odd, it is elementary that $g (1\ 2) = g \circ (1\ 2)$ is odd, as the product of an even and odd permutation is odd.
(There is a space between $1$ and $2$; this is not the number $12$.)
Edit: To also answer your second question: the hint tells you that to each even permutation $g$ you may associate an odd one $g (1 \; 2)$. Note that this correspondence is injective, because $g_1 (1 \; 2) = g_2 (1 \; 2) \implies g_1 = g_2$ (just multiply by the inverse of $(1 \; 2)$ on the right, which happens to be $(1 \; 2)$ again). Now note that this correspondence is also invertible, simply because $(1 \; 2)$ has an inverse in $S_n$. Therefore, you have a bijection between the set of even permutations and the set of odd permutations, therefore these two sets must have the same number of elements, hence the conclusion.