well forming a bijection is trivial, since both groups as sets clearly have 4 elements.

one could check all 16 possible products, and conclude that f(ab) = f(a)f(b) for every possible product. it's pretty clear that since both groups are abelian, we really only have to check 10 products, and of those, the ones involving the identity won't be very exciting (since f(e) = 1). so we can "get away" with only checking the following products:

o*o

o*t

o*ot

t*t

t*ot

ot*ot

you may, if you wish, check these 6 products yourself. it's good exercise. but there is another way:

the relations a^2 = b^2 = e, ab = ba DEFINE the Viergruppe, in the sense that any group with 4 elements that satisfies them is isomorphic to it.

(what is happening here is we are defining V as a quotient group of the free group on 2 generators, via the normal subgroup R generated by the relations).

so all we really NEED to do is verify two things:

t*o = ot, and (1 3)(2 4)(1 2)(3 4) = (1 2)(3 4)(1 3)(2 4) = (1 4)(2 3), since it's clear that t,o (1 3)(2 4), and (1 2)(3 4) are all of order 2.

since D4 is dihedral, we have that t*o = o^-1*t. but o^-1 = o, since a rotation of 180 degrees (clockwise OR counter-clockwise) is of order 2.

so t*o = o*t = ot. that's the easy part (you can also express this yourself more concretely if you wish, by writing t and o as 2x2 real matrices).

now (1 3)(2 4)(1 2)(3 4) is the composition:

1-->2-->4

2-->1-->3

3-->4-->2

4-->3-->1, which is the same as the double transposition (1 4)(2 3). similarly, (1 2)(3 4)(1 3)(2 4) is the composition:

1-->3-->4

2-->4-->3

3-->1-->2

4-->2-->1, which we see is also equal to (1 4)(2 3).