Group order

Order of powers of a group element

Given a group element 𝑎 ∈𝐺 with |𝑎| =𝑛, then ⟨𝑎𝑘⟩ =⟨𝑎gcd(𝑛,𝑘)⟩ and ∣𝑎𝑘∣ =𝑛𝑔𝑐𝑑(𝑛,𝑘). group

Proof

Let 𝑑 =gcd(𝑛,𝑘) and 𝑘 =𝑑𝑟. Since 𝑎𝑘 =(𝑎𝑑)𝑟 ∈⟨𝑎𝑑⟩ by closure ⟨𝑎𝑘⟩ ⊆⟨𝑎𝑑⟩. By Bézout’s lemma, there exist 𝑠,𝑡 ∈ℤ such that 𝑑 =𝑛𝑠 +𝑘𝑡, so that 𝑎𝑑 =𝑎𝑛𝑠𝑎𝑘𝑡 =(𝑎𝑛)𝑠(𝑎𝑘)𝑡 =(𝑎𝑘)𝑡 ∈⟨𝑎𝑘⟩. Hence by closure ⟨𝑎𝑑⟩ ⊆⟨𝑎𝑘⟩ and therefore ⟨𝑎𝑘⟩ =⟨𝑎𝑑⟩.

Keeping in mind 𝑑 is a divisor of 𝑛, it is clear that (𝑎𝑑)𝑛/𝑑 =𝑎𝑛 =𝑒, implying ∣𝑎𝑑∣ ≤𝑛𝑑. But if 𝑖 <𝑛𝑑 then 𝑖𝑑 <𝑛 and therefore (𝑎𝑑)𝑖 =𝑎𝑖𝑑 ≠𝑒 by the definition of order. Hence ∣𝑎𝑘∣ =𝑛𝑑.

Using this technique, computing the cyclic group generated by some power of a basic element becomes simple.¹

Corollaries

Order of elements in finite cyclic groups

It immediately follows that the order of an element in a finite cyclic group divides the order of the group. group

Criterion for ‹𝑎ⁱ› = ‹𝑎ʲ› and |𝑎ⁱ| = |𝑎ʲ| in a group

Given a group element 𝑎 ∈𝐺 with |𝑎| =𝑛, then ⟨𝑎𝑖⟩ =⟨𝑎𝑗⟩ iff gcd(𝑛,𝑖) =gcd(𝑛,𝑗). Likewise ∣𝑎𝑖∣ =∣𝑎𝑗∣ iff gcd(𝑛,𝑖) =gcd(𝑛,𝑗). group

Proof

From the above theorem, ⟨𝑎𝑖⟩ =⟨𝑎𝑗⟩ iff ⟨𝑎gcd(𝑛,𝑖)⟩ =⟨𝑎gcd(𝑛,𝑗)⟩. Clearly gcd(𝑛,𝑖) =gcd(𝑛,𝑗) implies ⟨𝑎gcd(𝑛,𝑖)⟩ =⟨𝑎gcd(𝑛,𝑗)⟩. On the other hand, ∣𝑎gcd(𝑛,𝑖)∣ =∣𝑎gcd(𝑛,𝑗)∣ implies 𝑛/gcd(𝑛,𝑖) =𝑛/gcd(𝑛,𝑗) and thence gcd(𝑛,𝑖) =gcd(𝑛,𝑗).

It follows immediately that gcd(𝑛,𝑖) =gcd(𝑛,𝑗) implies ∣𝑎𝑖∣ =∣𝑎𝑗∣. From the above theorem, ∣𝑎𝑖∣ =∣𝑎𝑗∣ =𝑛/gcd(𝑛,𝑖) =𝑛/gcd(𝑛,𝑗) ⟺ gcd(𝑛,𝑖) =gcd(𝑛,𝑗).

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Footnotes

1. 2017, Contemporary Abstract Algebra, p. 79 ↩