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For fixed $h\in G$, the conjugation map is the compositefollowed by multiplication. Inversion, constant maps, diagonal maps, and multiplication are continuous in a topological group, so $c_h(g)=gh{g}^{-1}$ is continuous.

Write an element of the Integer Heisenberg group as $(x,y,z)$. Matrix multiplication givesThe subgroupis normal and isomorphic to ${\mathbb{Z}}^2$. If $t=(1,0,0)$, thenThus, on the coordinate column $(y,z{)}^T$, conjugation by $t$ is the linear map with matrixEvery element has a unique expression $(0,y,z)t^x$, so

The commutators $[t,(0,y,z)]$ fill the central subgroup of matrices $(0,0,z)$, while the quotient by this subgroup is generated freely and abelianly by the images of $(1,0,0)$ and $(0,1,0)$. Equivalently, $(A-I){\mathbb{Z}}^2$ is the second coordinate axis. Therefore the abelianization is

Write $\bar{a}=f(a)$, $\bar{t}=f(t)$, and $\bar{b}={\bar{a}}^{\bar{t}}={\bar{t}}^{-1}\bar{a}\bar{t}$. The elements $\bar{a}$ and $\bar{b}$ are conjugate group elements, so they have the same order of a group element, say $n$. The defining relation becomesIterating conjugation gives ${\bar{b}}^j\bar{a}{\bar{b}}^{-j}={\bar{a}}^{2^j}$. Since ${\bar{b}}^n=1$, taking $j=n$ yields $\bar{a}={\bar{a}}^{2^n}$, and henceThe stated consequence of Fermat's little theorem implies $n=1$. Thus $\bar{a}=1$, and the surjective group homomorphism $f$ shows that $Q$ is generated by $\bar{t}$. Therefore $Q$ is a cyclic group.

Conjugation preserves order of a group element, because $(gx{g}^{-1}{)}^k=g{x}^k{g}^{-1}$, and preserves the sign of a permutation, because sign is a group homomorphism into $\{1,-1\}$. These invariants do not determine cycle type. In $S_6$, $(123)$ and $(123)(456)$ both have order three and positive sign, but have three and zero fixed points respectively. Conjugation relabels cycles, so these permutations are not conjugate.

The nonnegative cone determines the partial order by $a\le b⟺a^{-1}b\in G_{\ge 1}$. It contains the identity element, is closed under products and conjugation, and intersects its inverse only in the identity. Conversely these conditions define a two-sided invariant partial order. Its strictly positive part omits the identity; conventions for $G^{+}$ differ, so specify which is intended.

Over ${\mathbb{F}}_5$, putThese determinant-one matrices satisfy $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. Conjugation by $r$ cycles $i,ij,j$. Thus $⟨i,j⟩$ is a quaternion group of order eight, normalized by an element of order three. Their generated subgroup has order $24$ in $S{L}_2(5)$, hence order $12$ after quotienting by $\{\pm I\}$. Its index-five coset action, together with simplicity and perfectness, identifies $PS{L}_2(5)$ with the alternating group $A_5$.

Let $G$ be a connected compact Lie group, $T$ a maximal torus, and $W$ its Weyl group. With normalized Haar measures and $D(t)={\prod }_{\alpha >0}(1-e^{-\alpha }(t))$, every continuous class function satisfies ${\int }_{G}f(g)dg=|W{|}^{-1}{\int }_{T}f(t)|D(t){|}^{2}dt$. For a general continuous function, replace $f(t)$ by its average ${\int }_{G/T}f(gt{g}^{-1})d(gT)$. Conjugation from $G/T\times T$ covers the regular elements $|W|$ times, and its real Jacobian determinant is $|D(t){|}^2$.

The word metric measures how many allowed moves are needed to change one group element into another. For a finite symmetric generating set of a group $S$ with $1\notin S$, it is the graph distance between vertices of the Cayley graph, equivalentlyHere word length counts letters in a shortest representative of $g^{-1}h$. The path metric on the entire Cayley graph also permits points inside edges; the word metric is its restriction to the group itself.^[ref]

The word metric is left invariant:This corresponds to the label-preserving group action on the Cayley graph. Symmetry follows by reversing a shortest graph path; concatenating routes gives the triangle inequality.^[ref]

Right invariance can fail in a non-abelian group, because right multiplication replaces $g^{-1}h$ by $a^{-1}{g}^{-1}ha$ through conjugation. In the free group $F(x,y)$ with $S=\{x,x^{-1},y,y^{-1}\}$, we have $d_S(1,y)=1$, whereas $d_S(x,yx)=|x^{-1}yx{|}_S=3$: the latter freely reduced word has three letters. In an abelian group, conjugation fixes each element, so right invariance does hold. Cayley graphs of the integers and word metric balls in the integer lattice give examples where the word metric can be calculated explicitly.