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Prove that ordda | ordma, when d|m.
Some conditions are 1 ≤ d, 1 ≤ m, and gcd(a,d)=1.
What I have so far:
let x=ordma, which gives us ax[itex]\equiv[/itex] 1 (mod m) [itex]\Rightarrow[/itex] ax=mk+1 for some k[itex]\in[/itex]Z
Let m=m'd. Then ax=mk+1=d(m'k)+1
Some conditions are 1 ≤ d, 1 ≤ m, and gcd(a,d)=1.
What I have so far:
let x=ordma, which gives us ax[itex]\equiv[/itex] 1 (mod m) [itex]\Rightarrow[/itex] ax=mk+1 for some k[itex]\in[/itex]Z
Let m=m'd. Then ax=mk+1=d(m'k)+1