Another way to prove this is using ordinary character theory and Brauer character theory, though it needs a little more background. I outline a proof, since it illustrates many more general facts in a nice way.
Since $G = G^{\prime},$ every complex irreducible character of $G$ which does not contain $Z(G)$ in its kernel has even degree. If the degrees of the irreducible characters not containing $Z(G)$ in their kernels are $2m_{1},\ldots,2m_{t},$ possibly with repetition, we have $\sum_{i=1}^{t}m_{i}^{2} = 15.$ Hence at least $3$ of the $m_{i}$ are odd, and we may suppose that $m_{1} = 1,$ as claimed.
Now $G$ contains a unique involution, as its two dimensional complex irreducible character is faithful. Hence $G$ has a quaternion Sylow $2$-subgroup of order $8.$ Any element of order $6$ in $G$ is conjugate to its inverse. Hence the value of the irreducible character of degree $2$ on $5$-regular elements is rational.
But an irreducible Brauer character is realizable over the field of its character, and the $2$-dimensional irreducible complex representation of $G$ clearly remains absolutely irreducible on reduction (mod $5$). Hence $G$ is isomorphic to to a subgroup of ${\rm GL}(2,5).$ Since $G = G^{\prime},$ we see that $G$ is isomorphci to a subgroup of ${\rm GL}(2,5)^{\prime} = {\rm SL}(2,5).$ Since $|G| = 120,$ we have $G \cong {\rm SL}(2,5).$

A similar argument using the $4$-dimensional complex representation of a double cover of $A_{7}$ shows that $A_{7}$ is isomorphic to a subgroup of ${\rm GL}(4,2)$, and the index is easily seen to be $8.$ This gives an embedding of ${\rm GL}(4,2)$ into $A_{8},$ which is an isomorphism on consideration of order, thus exhibiting the well-known "exceptional" isomorphism $A_{8} \cong {\rm GL}(4,2).$