Galois group of x^10+x^5+1

This was a problem from an old qualifying exam, that I solved today, with a few pointers. First of all, is it reducible? It actually is. Note that x^{15} - 1 = (x^5-1)(x^{10}+x^5+1) = (x^3-1)(x^{12} + x^9 + x^6 + x^3 + 1). 1 + x + x^2, as a prime element of \mathbb{Q}[x] that divides not x^5-1 must divide the polynomial, the Galois group of which we are looking for. The other factor of it corresponds to the multiplicative group of \mathbb{F}_{15}, which has 8 elements. Seeing that it has 3 elements of order 2 and 4 elements of order 4 and is abelian, it must be C_2 \times C_4. Thus, the answer is C_2 \times C_2 \times C_4.

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