Q_{2,3} is a subset of Q, by the definition of Q_{2,3}, and since 2^{n} x 3^{m} does not equal 0 for any n and m, Q_{2,3} is a subset of Q/{0}. So to show that Q_{2,3} is a group, I will show that Q_{2,3} is a subgroup of Q/{0} using multiplication as the operation.

Identity:

2^{0} x 3^{0} = 1 x 1 = 1

Therefore 1 is an element of Q_{2,3}.

Inverses:

Let 2^{n} x 3^{m} be an element of Q_{2,3}.

(2^{n} x 3^{m}) x (2^{-n} x 3^{-m}) = 2^{n} x 2^{-n} x 3^{m} x 3^{-m}

because Q/{0} is abelian under multiplication.

2^{n} x 2^{-n} x 3^{m} x 3^{-m} = 2^{n-n} x 3^{m-m} = 2^{0} x 3^{0} = 1

The same thing can be shown for 2^{-n} x 3^{-m} x 2^{n} x 3^{m} = 1

Therefore (2^{n} x 3^{m})^{-1} = 2^{-n} x 3^{-m}

Closure:

Let x,y be elements of Q_{2,3}. I will show that xy is also and element of Q_{2,3}

Since x and y are elements of Q_{2,3}, we can say that x = 2^{n} x 3^{m} and y = 2^{p} x 3^{q}, with n,m,p,q all being elements of Z.

Therefore, by substitution we get

xy = (2^{n} x 3^{m}) x (2^{p} x 3^{q})

and because Q/{0} is abelian

2^{n} x 2^{p} x 3^{m} x 3^{q}

which can be shown to be equal to

2^{n+p} x 3^{m+q}.

Since n,p,m,q are elements of Z, and Z is closed under addition, n+p and m+q are elements of Z as well. Therefore, by definition of Q_{2,3}, xy is an element of Q_{2,3}.

Therefore, because Q_{2,3} is a subset of Q/{0}, and Q_{2,3} contains the identity, inverses, and is closed, Q_{2,3} is a subgroup of Q/{0} and by definition of subgroup is a group.