Recipe
Compute the order of an element
Use when: you have one specific element a of a group and need the smallest n with a^n = e
Goal: the order of a, a single positive integer, or the conclusion that a has infinite order
Taught in Order of an element
Quick recipe
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1
Compute a, then a \ast a, then a \ast a \ast a, and so on, applying the group's own operation and simplifying at each step.
Example
In (\mathbb{Z}_6, +), start from 2: the running result is 2, then 2+2=4.
Why: the order is the SMALLEST exponent that works, so the search has to start from the smallest candidates and move up.
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2
Track how many combinations you've performed; the moment the running result equals e, stop.
Example
Continuing 2 in \mathbb{Z}_6: 4+2=6\equiv0, reached on the third combination, not sooner.
- the group is finite — the chain is guaranteed to reach the identity eventually, so the search always terminates.
- the group is infinite and no combination ever produces the identity — a has infinite order, and the direct search never terminates on its own.
Why: the order is specifically the FIRST return to the identity, so continuing past it or stopping too early both give the wrong count.
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3
The number of combinations it took to first reach the identity is |a|, the order of a.
Example
0 was reached on the third combination above, so |2|=3 in \mathbb{Z}_6.
Why: this count is exactly what the definition names, the smallest positive n with a^n=e.
Common mistakes
- Continuing to combine past the first return to the identity and reporting a multiple of the true order.
- Confusing an element's own order with the whole group's order.
- Stopping the search too early, before actually reaching the identity, and reporting a step count that doesn't satisfy a^n = e.
- Assuming every element in an infinite group automatically has infinite order, without checking whether its own chain reaches e.