Editing Inverse Galois problem (section)

===Worked example: the cyclic group of order three===
For {{math|1=''n'' = 3}}, we may take {{math|1=''p'' = 7}}. Then {{math|Gal('''Q'''(''μ'')/'''Q''')}} is cyclic of order six. Let us take the generator {{mvar|η}} of this group which sends {{mvar|μ}} to {{math|''μ''<sup>3</sup>}}. We are interested in the subgroup {{math|1=''H'' = {1, ''η''<sup>3</sup>}}} of order two. Consider the element {{math|1=''α'' = ''μ'' + ''η''<sup>3</sup>(''μ'')}}. By construction, {{mvar|α}} is fixed by {{mvar|H}}, and only has three conjugates over <math>\mathbb{Q}</math>:

: {{math|1=''α'' = ''η''<sup>0</sup>(''α'') = ''μ'' + ''μ''<sup>6</sup>}}, 
: {{math|1=''β'' = ''η''<sup>1</sup>(''α'') = ''μ''<sup>3</sup> + ''μ''<sup>4</sup>}}, 
: {{math|1=''γ'' = ''η''<sup>2</sup>(''α'') = ''μ''<sup>2</sup> + ''μ''<sup>5</sup>}}.

Using the identity:

:{{math|1=1 + ''μ'' + ''μ''<sup>2</sup> + ⋯ + ''μ''<sup>6</sup> = 0}},

one finds that

: {{math|1=''α'' + ''β'' + ''γ'' = −1}},
: {{math|1=''αβ'' + ''βγ'' + ''γα'' = −2}},
: {{math|1=''αβγ'' = 1}}.

Therefore {{mvar|α}} is a [[root of a polynomial|root]] of the polynomial

:{{math|1=(''x'' − ''α'')(''x'' − ''β'')(''x'' − ''γ'') = ''x''<sup>3</sup> + ''x''<sup>2</sup> − 2''x'' − 1}},

which consequently has Galois group {{math|'''Z'''/3'''Z'''}} over <math>\mathbb{Q}</math>.