Order of a group element ($\mathrm{ord}(g)$)

The order of a group element $g$ is the least positive integer $n$ for which $g^n$ is the identity. If no such integer exists, $g$ has infinite order.

A group element has finite order if some positive integer power equals the identity element. Its order of a group element is the least such positive integer.

For every positive integer $m$, $(g^{-1}{)}^m=(g^m{)}^{-1}$, hence $g^m=e$ if and only if $(g^{-1}{)}^m=e$. The sets of positive powers returning to the identity element are therefore equal. Taking the least positive exponent therefore gives the same finite order of a group element for $g$ and $g^{-1}$; if the sets are empty, both elements have infinite order. Equivalently, $g$ and $g^{-1}$ generate the same cyclic subgroup.

For every positive integer $n$, cancellation gives $(ha{h}^{-1}{)}^n=h{a}^n{h}^{-1}$. Thus the power on one side is the identity element exactly when $a^n=e$, proving preservation of the order of a group element, including infinite order. In particular, $ab$ and $ba$ are conjugate because $a^{-1}(ab)a=ba$, and hence have the same order.

Let commuting elements $a,b$ have finite coprime orders $m,n$. Their cyclic subgroups have trivial intersection by Lagrange's theorem. If $(ab{)}^k=1$, then $a^k=b^{-k}$ belongs to this intersection, so $m\mid k$ and $n\mid k$. Thus $mn\mid k$. Conversely, $(ab{)}^{mn}=1$ because the elements commute. This proves the displayed order formula.

A group element has infinite order when no positive power of it is the identity element.

An involution is a group element $s$ satisfying $s^2=1$. Some conventions require $s\ne 1$.

An involution partitions its set into fixed points and pairs exchanged by the function. If there is exactly one fixed point, every finite model has odd size. Conversely any finite set of odd size admits such a function by pairing all but one element. The first-order sentence $\forall xf(f(x))=x\wedge \exists z(f(z)=z\wedge \forall x(f(x)=x\to x=z))$ consequently has exactly the positive odd finite model sizes. Its arbitrarily large finite models imply an infinite model by the compactness theorem.

Let $F$ be a linear map on a vector space over a field of characteristic different from two, with $F^2=I$. ThenThe projections satisfy $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$, giving existence and uniqueness of the decomposition. In characteristic two, the two signs coincide and the conclusion need not hold.

For complementary vector subspaces $U,W$ of a real vector space, the rule $T(u+w)=u-w$ is a well-defined linear map satisfying $T^2=I$. It fixes $U$ pointwise and reverses $W$. In a basis formed by adjoining bases of $U$ and $W$, its matrix is $\mathrm{diag}(I,-I)$. Conversely every real linear map with $T^2=I$ has this form, with $U=\ker (T-I)$ and $W=\ker (T+I)$, as follows from $v=(v+Tv)/2+(v-Tv)/2$.

In a Euclidean space, such a linear map is an orthogonal transformation if and only if $U$ and $W$ are perpendicular. Necessity follows from $⟨u,w⟩=⟨Tu,Tw⟩=-⟨u,w⟩$; sufficiency follows by expanding the squared Euclidean norm of $u+w$. In three dimensions, a fixed plane with reversed complementary line is a possibly oblique reflection, and a fixed line with reversed complementary plane is a possibly oblique axial symmetry. Their diagonal forms alone do not imply a Euclidean reflection in a hyperplane or a Euclidean half-turn.

For a linear operator with $A^2=I$, every even power is $I$ and every odd power is $A$. Separating the absolutely convergent matrix exponential series into those powers proves the displayed identity without a Hermiticity assumption. If $A$ is also a Hermitian operator, the exponential is a unitary operator.