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The identity and three half-turns form a normal Klein four-group $V_4$ inside the proper rotational tetrahedral symmetry group $A_4$. Conjugation preserves double-transposition cycle type, establishing normality. The quotient group has order three and is therefore cyclic. If $r$ is any 3-cycle, every element is uniquely $r^{j}k$ with $j\in \{0,1,2\}$ and $k\in V_4$. The formula $r^{j}k\mapsto e^{2\pi ij/3}$ defines a surjective group homomorphism with kernel $V_4$. Normality allows the factors from $V_4$ to be moved past powers of $r$, making exponents add modulo three and proving the homomorphism property directly.

For a subgroup $H$ of finite index, choose representatives $t$ of right cosets $Ht$. Write $tg=h(t,g)t_g$ with $h(t,g)\in H$. The transfer is $V_{G,H}(g)={\prod }_{t}h(t,g)(\mathrm{mod}{H}^{\prime })$. Replacing $t$ by $a_{t}t$ multiplies each factor by $a_t$ on the left and $a_{t_g}^{-1}$ on the right; these cancel in the abelianization because $t\mapsto t_g$ is a permutation. The identity $h(t,g_1{g}_2)=h(t,g_1)h(t_{g_1},g_2)$ proves the group homomorphism property after multiplying all factors.

The trivial group has exactly one element, its identity. Every group homomorphism to it is trivial, and a group is trivial precisely when all elements in a generating set equal the identity.

Let $p$ be prime, $a\in {\mathbb{F}}_p^{\times }$, $x,y\in \mathbb{Z}/(p-1)\mathbb{Z}$ and $u,v\in {\mathbb{F}}_p$. Fermat's little theorem makes the displayed multiplication well defined. Its identity is $(0,0)$ and inverse is $(-x,-a^{-x}u)$; associativity follows by expanding both bracketings. Projection onto the first coordinate is a group homomorphism with normal cyclic kernel of order $p$. Under the change $U=a^{-x}u$, multiplication becomes $(x,U)(y,V)=(x+y,U+a^{-x}V)$, the semidirect product of the additive prime field by the cyclic group of order $p-1$. The group is abelian precisely when $a=1$; $a$ need not be a generator of the multiplicative field group.

Let $E:y^2=f(x)$ with $f$ monic and separable, and let $T=(\theta ,0)$. Over $K$ containing $\theta$, the map ${\alpha }_T(P)=[x(P)-\theta ]$ for $P\ne O,T$, completed by ${\alpha }_T(O)=1$ and ${\alpha }_T(T)=[f^{\prime }(\theta )]$, is a group homomorphism to the square-class group of a field. A nonvertical chord has the product of its three $x-\theta$ values a square; a chord through $T$ instead has the product of the other two values equal to $f^{\prime }(\theta )$. These identities include tangencies.

The Lie algebra of a uniform pro-p group construction and BCH convergence on a powerful Lie lattice are mutually inverse on objects and morphisms. On the group side morphisms are continuous group homomorphisms; on the lattice side they are linear Lie algebra homomorphisms over the p-adic integers. A group homomorphism respects powers, roots and limits, and a Lie homomorphism respects every convergent Baker--Campbell--Hausdorff formula term.

The Lie algebra of a uniform pro-p group construction and BCH convergence on a powerful Lie lattice are mutually inverse on objects and morphisms. On the group side morphisms are continuous group homomorphisms; on the lattice side they are linear Lie algebra homomorphisms over the p-adic integers. A group homomorphism respects powers, roots and limits, and a Lie homomorphism respects every convergent Baker--Campbell--Hausdorff formula term.

For every abelian group $A$ and group homomorphism $f:G\to A$, there is a unique group homomorphism $\bar{f}:G^{\mathrm{ab}}\to A$ such thatThis is a universal property. To see why it holds, compute $f([x,y])=[f(x),f(y)]=e_A$, using commutativity in $A$. Hence $G^{\prime }\subseteq \ker f$, and $f$ is constant on each coset of $G^{\prime }$. The displayed formula is therefore well-defined and preserves multiplication. It is unique because $q$ is surjective. This is an instance of factoring a group homomorphism through a quotient group; see Conrad, Universal mapping properties, theorem 3.4.

In particular, every abelian quotient $G/N$ comes from a surjective group homomorphism $G^{\mathrm{ab}}\to G/N$, $g{G}^{\prime }\mapsto gN$. For $S_3$, any group homomorphism to an abelian group is determined by the common image $a$ of its three transpositions, with $a^2=e_A$. Every even permutation must map to $e_A$. Thus sign of a permutation captures all possible information of this kind, including maps with infinite abelian groups as their targets.

For every function $f:S\to G$ from a set into a group, there is a unique group homomorphism $\bar{f}:F(S)\to G$ whose restriction to $S$ is $f$.

Given group homomorphisms $f_A:A\to H$ and $f_B:B\to H$, there is exactly one group homomorphism $A\ast B\to H$ extending both. Its value on an alternating normal form is the product of the factor-map values. Thus the free product is the coproduct in the category of groups.

For group homomorphisms ${\varphi }_j:G_0\to G_j$, a group homomorphism from their pushout of groups to $H$ is uniquely a pair $u_j:G_j\to H$ with $u_1{\varphi }_1=u_2{\varphi }_2$. Indeed extend the pair to the free product and factor through its quotient group: the agreement kills the normal closure of all ${\varphi }_1(g){\varphi }_2(g{)}^{-1}$. Uniqueness follows because the images of the two factors generate the pushout of groups.

Let $N$ be a normal subgroup of $G$, and let $\pi :G\to G/N$ be the quotient map of groups. If a group homomorphism $\phi :G\to H$ has group kernel containing $N$, there is exactly one group homomorphism $\bar{\phi }:G/N\to H$ with $\phi =\bar{\phi }\circ \pi$. Its value is forced: $\bar{\phi }(gN)=\phi (g)$. Every measurement by a group homomorphism that ignores $N$ can thus be performed on the quotient group directly. This is the universal property of the construction.