Show that the set: {I, -I, X, -X, Y, -Y, Z, -Z} form a group under matrix multiplication.?

To show closure, you have to do a bunch of grunt work. Multiply all possible pairs of matrices (although you can cut down the work by factoring out negative signs)

so for example:

XY= - Z

X(-Y)= Z

XX=I

YY=I

ZZ=I

X(-X)= - I

etc. I won't do all of them.

As for associativity, you can appeal to the fact that multiplication is associative for ALL 3x3 matrices, so in particular it is associative for THESE 3x3 matrices.

the set contains an identity element, names the identity matrix I.

And while you check closure, you end up seeing that every element is its own inverse. that is, every element multiplied by itself gives I.

so you have a group.

an easier way to show closure, without doing a lot of "grunt work"

notice all matrices are diagonal. the product of two diagonal matrices is again diagonal, the entries on the diagonal of the product being the products of the diagonal entries.

so every product is of the form diag(a,b,c), where a,b, and c = ±1.

but these 8 matrices form every possible matrix of that form:

I = diag(1,1,1)

X = diag(-1,1,1)

Y = diag(1,-1,1)

Z = diag(1,1,-1)

-X = diag(1,-1,-1)

-Y = diag(-1,1,-1)

-Z = diag(-1,-1,1)

-I = diag(-1,-1,-1)

so the set is closed under multiplication.

but this is a subset of the group GL(3,F) (in any field for which -1 ≠ 1), and since it is finite, is a subgroup of that group (the general linear group of degree 3 over F).