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Suppose the cubing map $T:G\to G$ is a group homomorphism. Expanding $(ab{)}^3=a^3{b}^3$ and cancelling gives $(ba{)}^2=a^2{b}^2$. Alsoso $g^2$ commutes with every cube. For $x,y\in T(G)$, therefore, $x^{2}y=y{x}^2$ and $yxyx=x^2{y}^2$. Multiplying the latter identity on the left by $y^{-1}$ yields $xyx=x^{2}y$; cancelling $x$ gives $yx=xy$. Thus the image of a group homomorphism $T(G)$ is an abelian group.

An endomorphism is a morphism $f:X\to X$: its source and target are the same object, with the same specified structure. For a group it is an operation-preserving group homomorphism back into that group; for a vector space it is a linear endomorphism; for a ring it is a ring endomorphism. A self-map must satisfy the chosen compatibility conditions to qualify. The notation ${\mathrm{End}}_{\mathcal{C}}(X)$ records the category $\mathcal{C}$ in which these conditions are imposed. See Riehl, Category Theory in Context, Definition 1.1.10 and the following paragraph.

An endomorphism can lose information or miss part of its target. On the additive integers, $d(k)=2k$ preserves addition and is injective, but is not surjective: no integer maps to $1$. On the vector space ${\mathbb{R}}^2$ over the real numbers, the linear endomorphism $P(x,y)=(x,0)$ loses the second coordinate, so it is neither injective nor surjective. Neither is an automorphism. The same doubling rule becomes a group isomorphism when its target is the additive even integers, as explained under isomorphism.

The finite cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ offers a small example of information loss that can be drawn completely. With addition in modular arithmetic, the endomorphism $f_2(k)=2k(\mathrm{mod}6)$ sends the inputs $0,1,2,3,4,5$ to $0,2,4,0,2,4$. Its image of a group homomorphism is $\{0,2,4\}$ and its kernel of a group homomorphism is $\{0,3\}$. Each reached output has two inputs; the other three outputs are missed.

For an endomorphism of a group or ring with a finite underlying set, injectivity is equivalent to surjectivity by counting; either condition makes it an automorphism. The additive integers example shows why the finiteness assumption matters. A linear endomorphism of a finite-dimensional vector space also has equivalent injectivity and surjectivity, this time by the rank-nullity theorem even when its underlying set is infinite.

Self-maps can be iterated without changing the object: $f^2=f\circ f$, $f^3=f\circ f\circ f$, and so on. Composition in a category and the identity morphism make the endomorphisms of a fixed object in a locally small category into a monoid. This packages repeated compatible operations, including irreversible ones, into a single structure.

The general language has a documented historical milestone in the 1945 paper of Samuel Eilenberg and Saunders Mac Lane. They defined categories of objects and compatible maps, and distinguished maps admitting inverses within that category. Their examples included groups, vector spaces, and topological spaces. This gave a common framework for compatible self-maps and their invertible cases. See Eilenberg and Mac Lane, General Theory of Natural Equivalences, pp. 234 and 237–239.

Every endomorphism of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$, for a positive integer $n$, has the form $f_a(k)=ak(\mathrm{mod}n)$, determined by $a=f_a(1)$. This is the self-map case of homomorphism from a finite cyclic group: repeated addition forces the value at every element. Conversely, each choice of $a$ respects addition and gives a well-defined map on congruence classes.

Let $d=\gcd (n,a)$, the greatest common divisor. The image of a group homomorphism is the cyclic subgroup generated by $a$, with size $n/d$ by order of an element of a finite cyclic group. The kernel of a group homomorphism isIndeed, $ak=0(\mathrm{mod}n)$ is equivalent to $n/d$ dividing $k$, after cancelling $d$ and using coprimality. Consequently $f_a$ is a group automorphism exactly when $a$ and $n$ are coprime. In that case a modular inverse $b$ of $a$ gives $f_a^{-1}=f_b$. This connects reversible self-maps directly to the generators of a finite cyclic group.^[ref]

A group homomorphism sends identity to identity. For $\theta (g),\theta (h)$ in its image,which is still in the image; hence the image of a group homomorphism is a subgroup of $H$. Let $N=\ker \theta$. It is a subgroup by the same identity, and for $k\in N$ and $g\in G$,Thus $gN{g}^{-1}\subseteq N$, and replacing $g$ by $g^{-1}$ gives equality, proving that $N$ is a normal subgroup.

Define a map from the quotient group byIf $gN=hN$, then $h^{-1}g\in N$ and $\theta (g)=\theta (h)$, so the map is well defined. Conversely equality of the images implies $h^{-1}g\in N$, giving injectivity. It is onto by the definition of image, and $\Phi ((gN)(hN))=\theta (gh)=\theta (g)\theta (h)$. This proves the first isomorphism theorem for groups in this case:An orthogonal matrix $Q$ has $Q^{\mathsf{T}}Q=I$, so $(\det Q{)}^2=1$. The determinant is therefore a group homomorphism $O(3)\to \{1,-1\}$, with image both signs because $\mathrm{diag}(-1,1,1)$ has determinant $-1$. Its kernel is the special orthogonal group $SO(3)$. Hence $SO(3)$ is normal in $O(3)$, andFor the last request, the determinant-twist projection in odd dimension gives an explicit example:Indeed $\det \pi (Q)=(\det Q{)}^4=1$, and scalar factors commute, so $\pi (QR)=\pi (Q)\pi (R)$. If $\pi (Q)=I$, then $Q=(\det Q)I$, making $Q=I$ or $Q=-I$. Both lie in the kernel, which is therefore exactly $\{I,-I\}$, of order two. The map also fixes every member of $SO(3)$, so it is onto.

A subgroup $K$ is a normal subgroup of $G$ when $gK{g}^{-1}=K$ for every $g\in G$.

For a group homomorphism $\varphi :G\to H$, its kernel contains the identity and, for $x,y$ in the kernel, $\varphi (x{y}^{-1})=\varphi (x)\varphi (y{)}^{-1}=e$. Hence the kernel of a group homomorphism is a subgroup. Also $\varphi (gk{g}^{-1})=\varphi (g)e\varphi (g{)}^{-1}=e$ for $k$ in the kernel; applying this to $g$ and $g^{-1}$ gives equality under conjugation. Thus the kernel is always normal.

The image contains the identity and is closed under products and inverses, so the image of a group homomorphism is always a subgroup of the target. It need not be normal: include $C_2=\{e,(12)\}$ into $S_3$. Conjugation by $(123)$ takes $(12)$ to $(23)$, outside that image. Thus the image is not always normal.

For matrices over either $\mathbb{Z}$ or ${\mathbb{F}}_2={\mathbb{Z}}_2$, multiplication stays in the same coefficient ring, and determinants multiply, so the determinant-one matrices are closed. Associativity is inherited from matrix multiplication, the identity has determinant one, and the inverse isIts entries are still in the same ring and its determinant is one. This proves that both sets are special linear groups under multiplication. Over ${\mathbb{F}}_2$, minus and plus agree, and the same formula applies.

Reduction of integers modulo two preserves addition and multiplication. Therefore reducing entries preserves the determinant condition and satisfies $\varphi (AB)=\varphi (A)\varphi (B)$, proving the stated map is a group homomorphism.

Its target acts on the three nonzero vectors of ${\mathbb{F}}_2^2$. The action is faithful: a matrix fixing all three vectors fixes the two standard basis vectors and so is the identity. Thus its image, and in particular the image of $\varphi$, is isomorphic to a permutation subgroup of $S_3$. In fact SL2 over F2 as a permutation group identifies the entire image. There are $3$ choices for the first nonzero column and $2$ for a second independent column, hence six invertible matrices. Every invertible matrix has determinant one over ${\mathbb{F}}_2$, so the faithful action is onto $S_3$.

To show the integer reduction map reaches all of them, takeOn $\{e_1,e_2,e_1+e_2\}$, the reduction of $S$ exchanges $e_1,e_2$ and fixes their sum, while the reduction of $T$ fixes $e_1$ and exchanges the other two. These two transpositions generate $S_3$. Hence

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

For $h\in H$, define $\pi (h)(n)=hn{h}^{-1}$. Since $N$ is a normal subgroup, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an group automorphism of $N$. MoreoverThus $\pi :H\to \mathrm{Aut}(N)$ is a group homomorphism, induced by the group action of conjugation.

If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Henceis a group homomorphism, since $(n_1{h}_1)(n_2{h}_2)=n_1{n}_2{h}_1{h}_2$. Its image of a group homomorphism is a subgroup containing both $N$ and $H$, and every product $nh$ lies in $⟨N,H⟩$, so its image is exactly $⟨N,H⟩$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives

Choose a normal subgroup $N$ of prime order $p\in \{5,7\}$ as in ii, and choose a subgroup $H$ of the other prime order $q$. Their intersection is trivial by Lagrange's theorem. Both are cyclic groups, and ii gives $|\mathrm{Aut}(N)|=p-1$.

The image of a group homomorphism $\pi :H\to \mathrm{Aut}(N)$ from iii has order dividing $q$, because its kernel of a group homomorphism partitions $H$ into equally sized cosets. Its order also divides $p-1$ by Lagrange's theorem. Butso $\pi$ is trivial in either case. By iii, $⟨N,H⟩\cong C_p\times C_q$ has order $pq=35$ and therefore equals $G$.

If $n$ and $h$ generate $N$ and $H$, the element $(n,h)$ has order $\mathrm{lcm}(p,q)=pq=35$, since its $k$th power is the identity exactly when both $p$ and $q$ divide $k$. Thus this direct product of groups is cyclic, and

Assume $T$ is a group homomorphism. For any $a,b\in G$, expanding $(ab{)}^3=a^3{b}^3$ and cancelling the initial $a$ and final $b$ gives the useful identityWe also need a consequence of compatibility with conjugation. Calculate the cube of $gz{g}^{-1}$ in two ways:Cancelling shows $g^2{z}^3=z^3{g}^2$. Thus the square of every element of $G$ commutes with every element of the image of a group homomorphism $J=T(G)$.

Now take any $x,y\in J$. In particular $x^{2}y=y{x}^2$. Applying (1) to $x,y$ and then using this commutation givesThis proves the abelian image of a cubing homomorphism property. Since the image of a group homomorphism is a subgroup, $\boxed{T(G)\text{ is an abelian group}}$.

The image of a group homomorphism contains $e_{G_2}=\varphi (e_{G_1})$. If $u=\varphi (g)$ and $v=\varphi (h)$ lie in the image, then $u{v}^{-1}=\varphi (g{h}^{-1})$ also lies there. The subgroup test proves $\mathrm{im}\varphi \le G_2$.

The kernel of a group homomorphism is $K=\{g:\varphi (g)=e_{G_2}\}$. It is nonempty and closed under $k{h}^{-1}$ for $k,h\in K$, so it is a subgroup. For any $g\in G_1$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Replacing $g$ by $g^{-1}$ gives the reverse inclusion, provingFinally, if $K=\{e\}$ and $\varphi (g)=\varphi (h)$, then $\varphi (h^{-1}g)=e$. Thus $h^{-1}g\in K$ forces $g=h$, establishing injectivity. Conversely an injective homomorphism has trivial kernel, soNone of these arguments requires either group to be finite.