Group

Criterion for 𝑎𝑖 =𝑎𝑗 in a group

For a group element 𝑎 𝐺, 𝑎𝑖 =𝑎𝑗 iff either: #m/thm/group

Proof

If 𝑎 has infinite order there exists no nonzero 𝑛 such that 𝑎𝑛 =𝑒, and since 𝑎𝑖 =𝑎𝑗 implies 𝑎𝑖𝑗 =𝑒, it follows 𝑖 =𝑗.

If |𝑎| =𝑛 then we again have the implication 𝑎𝑖𝑗 =𝑒. By the division algorithm 𝑖 𝑗 =𝑝𝑞 +𝑟, with 0 𝑟 <𝑛. Then 𝑒 =𝑎𝑖𝑗 =𝑎𝑝𝑛+𝑟 =𝑎𝑟 =𝑒, and since 𝑛 is the lowest positive integer such that 𝑎𝑛 =𝑒, it follows that 𝑟 =0. Hence 𝑛 (𝑖 𝑗).

Corollary

It immediately follows that 𝑎𝑘 =𝑎 implies |𝑎| divides 𝑘.


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