By "Euclidean transformation", do you mean a map , such that , where is an orthogonal matrix, and is some vector?

If so, to show that is a subgroup of the group of Euclidean transformations, you need to show that is closed under composition of maps, under inverses and that it contains the neutral element.

Closed under composition:

Take , and say and , where and are orthogonal matrices of determinant 1. Then:

and provided that is an orthogonal matrix of determinant 1, this composition is indeed in . But this is true, as you may try to show.

Closed under inverses.

Let . We need to prove that the inverse is also an element of Suppose is of the form , where is orthogonal of determinant 1. Then the inverse of is . Since is again an orthogonal matrix of determinant 1 (which you may check), the inverse is an element of as needed.

The neutral element is there.

The neutral element is the map given by , where is the identity matrix. This is certainly an element of .

As for the question "Determine which of the isometries are in ", it depends on your definition of isometry. Are you isometries linear or are they just supposed to preserve the distance in ?

All of the Euclidean transformations preserve the distance in , and so does in particular. If you are thinking in geometric terms, then consists of the maps, which are rotations followed by a translation.