The first isomorphism theorem

Statement

Let \(G\) and \(H\) be groups and let \(\varphi : G \to H\) be a group homomorphism. Then \(\ker\varphi = \{g \in G : \varphi(g) = e_H\}\) is a normal subgroup of \(G\), \(\operatorname{im}\varphi = \{\varphi(g) : g \in G\}\) is a subgroup of \(H\), and the map \[ \psi : G/\ker\varphi \longrightarrow \operatorname{im}\varphi, \qquad \psi(g\ker\varphi) = \varphi(g) \] is a well-defined group isomorphism. In particular \(G/\ker\varphi \cong \operatorname{im}\varphi\).

Why it matters

This is the single most-used structural result in group theory: it converts every homomorphism into a quotient-and-embed picture. Instead of studying an arbitrary map \(\varphi:G\to H\), we may study the normal subgroup \(\ker\varphi\), the quotient \(G/\ker\varphi\), and the subgroup \(\operatorname{im}\varphi\le H\) — objects built entirely from \(G\) and \(H\) themselves. It is the reason "classify subgroups of \(G\) up to normality" and "classify homomorphic images of \(G\)" are really the same problem.

It is also the template for the analogous theorems in every other algebraic category with a sensible notion of substructure and quotient — rings and ideals, modules, vector spaces, Lie algebras, topological groups — so understanding this proof in full is understanding a pattern used throughout algebra.

Hypotheses

Proof

Result

Corollaries & converses

• Order corollary (finite case). If \(G\) is finite, \(|G| = |\ker\varphi|\cdot|\operatorname{im}\varphi|\), by combining the theorem with Lagrange's theorem applied to \(G/\ker\varphi\).
• Injectivity criterion. \(\varphi\) is injective if and only if \(\ker\varphi=\{e_G\}\) — immediate from Step 7's argument alone, without needing the full isomorphism.
• Every quotient is realised as an image. Conversely, for any normal \(N\trianglelefteq G\), the canonical projection \(\pi:G\to G/N\) has \(\ker\pi=N\) and \(\operatorname{im}\pi=G/N\), so the theorem applied to \(\pi\) recovers \(G/N\cong G/N\) trivially — showing every quotient group arises as the image of some homomorphism from \(G\), i.e. the theorem's construction is exhaustive, not just one direction.
• Converse of the isomorphism statement does not hold as a "recognition" tool. Knowing abstractly that \(G/N \cong K\) for some normal \(N\trianglelefteq G\) and some group \(K\) does not by itself hand you a homomorphism \(G\to K\) with that kernel and image unless one is exhibited; the theorem produces the isomorphism from a given \(\varphi\), it is not an equivalence "\(G/N\cong K\) iff there is a natural \(\varphi\)" — the isomorphism \(\psi\) in the theorem is canonical, but an isolated abstract isomorphism \(G/N\cong K\) need not be.

Fails without

Common errors

• Writing \(G/\ker\varphi \cong H\) instead of \(G/\ker\varphi \cong \operatorname{im}\varphi\) — forgetting that the isomorphism is onto the image, not the whole codomain, unless \(\varphi\) is also surjective.
• Trying to form \(G/K\) for a subgroup \(K\) that has not been shown to be normal, silently assuming coset multiplication is well-defined.
• Proving \(\psi\) is well-defined and injective but forgetting to separately verify \(\psi\) is a homomorphism (Step 6) — well-definedness plus bijectivity of a set map is not the same as being a group isomorphism.
• Confusing "\(\ker\varphi=\{e_G\}\)" (injective) with "\(\ker\varphi=G\)" (the zero/trivial map onto \(\{e_H\}\)) — sign/direction slip when reasoning about extremes.
• Quoting the theorem for rings while using the group-theoretic normal-subgroup argument verbatim, without noting that the ring version needs ideals and a different (though structurally parallel) well-definedness check for the multiplicative part of the ring structure.

Discussion

The first isomorphism theorem is the group-theoretic instance of a completely general categorical phenomenon: given any morphism \(\varphi\) in a category with kernels and cokernels (or, more elementarily, congruences and quotients), \(\varphi\) factors as (surjection onto the coimage) followed by (injection of the image), and in "nice" categories — groups, rings, modules, vector spaces — the coimage and image are canonically isomorphic. This factorization \(G \twoheadrightarrow G/\ker\varphi \xrightarrow{\sim} \operatorname{im}\varphi \hookrightarrow H\) is often drawn as a commutative triangle/square and is the prototype for the "canonical factorization" idea that recurs throughout algebra and later, category theory.

Historically the result crystallized alongside the abstract definition of quotient groups in the late nineteenth and early twentieth century (associated with the structural approach to algebra developed by Dedekind, Hölder, and later formalized in the Noetherian tradition — the theorem is sometimes called "Noether's first isomorphism theorem" after Emmy Noether's systematic treatment of isomorphism theorems for groups, rings, and modules in the 1920s), replacing earlier, more computational treatments of homomorphisms with a structural one centred on kernels and quotients.

The theorem underlies the classification of homomorphic images: to find every group that is a homomorphic image of \(G\), it suffices to enumerate the normal subgroups of \(G\), since each quotient \(G/N\) is (up to isomorphism) exactly the image of some homomorphism out of \(G\), and every homomorphic image arises this way. This is why the lattice of normal subgroups of \(G\) is often called the key to understanding all "views of \(G\) from outside."

A subtlety worth flagging at the more advanced level: the isomorphism \(\psi\) constructed here is canonical (it does not depend on any arbitrary choice — it is literally forced by \(\varphi\)), which is what makes the first isomorphism theorem functorial: it assembles into a natural isomorphism between the functor "coimage" and the functor "image" on the category of groups. This canonicity is exactly what fails in more general categories that lack it (e.g. some categories of topological groups, where a continuous bijective homomorphism need not have a continuous inverse, so the abstract group isomorphism survives but the "correct" categorical isomorphism can fail) — a caution for anyone tempted to port the proof verbatim into a topological or algebraic-geometric setting without re-checking the extra structure.

Common misconceptions. Students often think the theorem produces an isomorphism \(G/\ker\varphi \cong H\); it produces one onto \(\operatorname{im}\varphi \le H\), which equals \(H\) only when \(\varphi\) is surjective. Another misconception is treating the theorem as defining \(G/\ker\varphi\) — normality of \(\ker\varphi\) (Step 2) must be established first, independently, before the quotient group even makes sense as an object to compare against \(\operatorname{im}\varphi\).

Worked examples

Problems

1. Let \(\varphi:(\mathbb{R},+)\to(\mathbb{C}^\times,\times)\) be \(\varphi(t)=e^{2\pi i t}\). Find \(\ker\varphi\) and \(\operatorname{im}\varphi\), and state the resulting isomorphism.
   Solution

   \(\varphi\) is a homomorphism since \(\varphi(s+t)=e^{2\pi i(s+t)}=e^{2\pi is}e^{2\pi it}=\varphi(s)\varphi(t)\). \(\ker\varphi=\{t\in\mathbb{R}: e^{2\pi it}=1\}=\mathbb{Z}\) (exactly the integers, since \(e^{2\pi it}=1\) iff \(t\) is an integer). The image is the unit circle \(\operatorname{im}\varphi = \{z\in\mathbb{C}^\times : |z|=1\} =: \mathbb{T}\), since \(|e^{2\pi it}|=1\) for all real \(t\) and every point on the unit circle is \(e^{2\pi i t}\) for some real \(t\) (surjectivity of the standard parametrisation of the circle). By the first isomorphism theorem, \(\mathbb{R}/\mathbb{Z} \cong \mathbb{T}\).
2. Let \(\varphi: S_n \to \{\pm 1\}\) be the sign homomorphism (\(\{\pm1\}\) under multiplication). Identify \(\ker\varphi\) and use the theorem to compute \(|S_n|/|\ker\varphi|\) for \(n\ge 2\).
   Solution

   \(\ker\varphi\) is precisely the alternating group \(A_n\) (permutations of sign \(+1\)) by definition of the sign map. \(\varphi\) is surjective for \(n\ge2\) since a transposition has sign \(-1\) and the identity has sign \(+1\), so \(\operatorname{im}\varphi=\{\pm1\}\), a group of order 2. By the theorem, \(S_n/A_n \cong \{\pm1\}\), and by the order corollary \(|S_n| = |A_n|\cdot|\{\pm1\}| = 2|A_n|\), so \(|S_n|/|A_n|=2\), i.e. \(A_n\) has index 2 in \(S_n\).
3. Give an explicit example of two groups \(G\), \(H\) and a non-surjective homomorphism \(\varphi:G\to H\) where naively writing "\(G/\ker\varphi \cong H\)" (omitting "image") gives a false statement, and correct it.
   Solution

   Take \(G=(\mathbb{Z},+)\), \(H=(\mathbb{Z},+)\), \(\varphi(k)=2k\). This is an injective (hence \(\ker\varphi=\{0\}\)) but non-surjective homomorphism (image is \(2\mathbb{Z}\), not all of \(\mathbb{Z}\)). Naively, \(G/\ker\varphi = \mathbb{Z}/\{0\}\cong\mathbb{Z}=H\) would seem to hold numerically here by coincidence of both being isomorphic to \(\mathbb{Z}\), but the correct statement is \(G/\ker\varphi \cong \operatorname{im}\varphi = 2\mathbb{Z}\), not \(H\) itself as a matter of the theorem's content — it is only because \(2\mathbb{Z}\cong\mathbb{Z}\cong H\) as abstract groups that the numerically-false-in-general shortcut looks right here; for a case where it visibly fails, use \(\varphi:\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}\), \(\varphi(k)=(k,0)\): \(\ker\varphi=\{0\}\), \(\operatorname{im}\varphi=\mathbb{Z}\times\{0\}\cong\mathbb{Z}\), but \(H=\mathbb{Z}\times\mathbb{Z}\not\cong\mathbb{Z}\), so \(G/\ker\varphi\cong\operatorname{im}\varphi\) is correct while "\(G/\ker\varphi\cong H\)" is false.
4. Let \(N\trianglelefteq G\) be any normal subgroup and let \(\pi:G\to G/N\) be the canonical projection \(\pi(g)=gN\). Prove directly from the definitions (not just by quoting the theorem) that \(\ker\pi = N\), and explain why this shows every normal subgroup is a kernel.
   Solution

   \(\pi\) is a homomorphism since \((gN)(g'N)=(gg')N\) is exactly the group operation on \(G/N\) (Step 3 of the proof), so \(\pi(gg')=\pi(g)\pi(g')\). The identity of \(G/N\) is \(eN=N\). So \(\ker\pi=\{g\in G:\pi(g)=N\}=\{g\in G: gN=N\}\). But \(gN=N\) iff \(g\in N\) (standard coset fact: \(gN=N \iff g=g\cdot e\in N\), since \(N\) is a subgroup containing \(e\), and cosets are either identical or disjoint). Hence \(\ker\pi=N\) exactly. This shows the converse direction of the theorem's machinery: not only does every homomorphism produce a normal subgroup as its kernel, but every normal subgroup \(N\) arises as \(\ker\pi\) for at least one homomorphism (namely its own canonical projection) — so "normal subgroup" and "kernel of some homomorphism" are exactly the same notion for groups.
5. Let \(\varphi: G \to H\) and \(\theta: H \to K\) be group homomorphisms with \(\varphi\) surjective. Show that \(\ker(\theta\circ\varphi) \trianglelefteq G\) contains \(\ker\varphi\), and that \(G/\ker(\theta\circ\varphi) \cong \operatorname{im}(\theta\circ\varphi)\), then express this image in terms of \(\theta\) and \(\operatorname{im}\varphi\).
   Solution

   \(\theta\circ\varphi:G\to K\) is a homomorphism (composite of homomorphisms is a homomorphism: \((\theta\circ\varphi)(ab) = \theta(\varphi(a)\varphi(b)) = \theta(\varphi(a))\theta(\varphi(b))\)), so the first isomorphism theorem applies directly to it, giving \(\ker(\theta\circ\varphi)\trianglelefteq G\) and \(G/\ker(\theta\circ\varphi)\cong\operatorname{im}(\theta\circ\varphi)\) by Steps 1–9 verbatim with \(\varphi\) replaced by \(\theta\circ\varphi\). For containment: if \(g\in\ker\varphi\) then \(\varphi(g)=e_H\), so \((\theta\circ\varphi)(g)=\theta(e_H)=e_K\) (homomorphisms send identity to identity), so \(g\in\ker(\theta\circ\varphi)\); hence \(\ker\varphi\subseteq\ker(\theta\circ\varphi)\). Since \(\varphi\) is surjective, \(\operatorname{im}(\theta\circ\varphi) = \theta(\varphi(G)) = \theta(H) = \operatorname{im}\theta\) (using surjectivity of \(\varphi\) to replace \(\varphi(G)\) with all of \(H\)). So the conclusion refines to \(G/\ker(\theta\circ\varphi)\cong\operatorname{im}\theta\).