(Part of Math and Wellness Month.)

Groups! A group is a set with an associative binary operation such that there exists an identity element and inverse elements! And my *favorite* thing about groups is that all the time that you spend thinking about groups, is time that you're *not* thinking about pain, betrayal, politics, or moral uncertainty!

Groups have subgroups, which you can totally guess just from the name are subsets of the group that themselves satisfy the group axioms!

The *order* of a finite group is its number of elements, but this is not to be confused with the order of an *element* of a group, which is the smallest integer such that the element raised to that power equals the identity! Both senses of "order" are indicated with vertical bars like an absolute value (|*G*|, |*a*|).

Lagrange proved that the order of a subgroup divides the order of the group of which it is a subgroup! History remains ignorant of how often Lagrange cried.

To show that a nonempty subset *H* of a group is in fact a subgroup, it suffices to show that if *x*, *y* ∈ *H*, then *xy*⁻¹ ∈ *H*.

Exercise #6 in §2.1 of Dummit and Foote *Abstract Algebra* (3rd ed'n) asks us to prove that if *G* is a commutative ("abelian") group, then the *torsion subgroup* {*g* ∈ *G* | |g| < ∞} is in fact a subgroup. I argue as follows: we need to show that if *x* and *y* have finite order, then so does *xy*⁻¹, that is, that (*xy*⁻¹)^*n* equals the identity. But (*xy*⁻¹)^*n* equals (*xy*⁻¹)(*xy*⁻¹)...(*xy*⁻¹), "*n* times"—that is, pretend *n* ≥ 3, and pretend that instead of "..." I wrote zero or more extra copies of "(*xy*⁻¹)" so that the expression has *n* factors. (I usually dislike it when authors use ellipsis notation, which feels so icky and informal compared to a nice Π or Σ, but let me have this one.) Because group operations are associative, we can drop the parens to get *xy*⁻¹ *xy*⁻¹ ... *xy*⁻¹. And because we said the group was commutative, we can reörder the factors to get *xxx*...*y⁻¹y⁻¹y*⁻¹, and *then* we can consolidate into powers to get *x*^*n* y^(−*n*)—but that's the identity if *n* is the least common multiple of |*x*| and |*y*|, which means that *xy*⁻¹ has finite order, which is what I've been trying to tell you this entire time.