*4. prove the following: if (g, *, e) is a finite group with even order, then there exists g fe in g such that g*g=e.

As (G, *, e) is a group with 2n elements, we have 2n-1 elements different from e. Let us assume that there aren't elements of (G, *, e) such that g*g=e. Now, take pairs of elements of (G, *, e) as , i.e., let us associate each element (different from e) with its multiplicative inverse. Let us remark that we are making no distinction between and , because they essentially the same pair.

Now, how many different pairs can we form with 2n-1 elements? Only n-1, which give us 2n-2 elements. Then, there is a ‘‘missing’’ element of that list. That element must fulfill that g*g=e, because it cannot be paired with any other element.

shift 4 units up, and 1 unit left.

it will translate f(x) = 3^x to g(x) = 3^(x+1) + 4

c) yes, because of asa or aas