First isomorphism theorem: Difference between revisions

Latest revision as of 21:34, 10 April 2026

The first isomorphism theorem is a fundamental result in abstract algebra that describes the relationship between a homomorphism, its kernel, and its image. The theorem appears uniformly across algebraic structures such as groups, rings, and modules, and serves as a prototype for many structural results in algebra. Specifically, given a homeomorphism, the quotient of its domain by its kernel is isomorphic to its image.

Group theory

Statement

Let $G$ and $H$ be groups and $f : g \to H$ a group homomorphism. Then,

The kernel of $f$ , $\ker f ⊴ G$ is a normal subgroup of $G$ .
The image of $f$ , $im f < G$ is a subgroup of $H$ .
$G / \ker f ≅ im f$ .

Proof

Proof of 1

By definition, $\ker f = {g \in G ∣ f (g) = e_{H}}$ where $e_{H}$ is the identity of $H$ . $\ker f$ is a subgroup of $G$ because:

Identity: Since $f$ is a homomorphism, $f (e_{G}) = e_{H}$ . Therefore $e_{G} \in \ker f$ , implying $\ker f$ is non-empty and has an identity.
Closure: Let $a, b \in \ker f$ , then

$f (a b) = f (a) f (b) = e_{H} e_{H} = e_{H} .$ Thus $a b \in \ker f$ .

Inverses: Let $a \in \ker f$ , then $f (a^{- 1}) = f (a)^{- 1} = e_{H}^{- 1} = e_{H} .$

Thus $a^{- 1} \in \ker f$ .

Therefore, $\ker f$ is a subgroup of $G$ .

Let $g \in G$ and $k \in \ker f$ , then $f (g k g^{- 1}) = f (g) f (k) f (g^{- 1}) = f (g) e_{H} f (g^{- 1}) = f (g) f (g^{- 1}) = f (g) f (g)^{- 1} = e_{H},$ thus $g k g^{- 1} \in \ker f$ .

Therefore, $\ker f ⊴ G$ is a normal subgroup.

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Proof of 2

By definition, $im f = {h \in H ∣ \exists g \in G : f (g) = h}$ . $im f$ is a subgroup of $H$ because:

Identity: Since $f$ is a homomorphism, $f (e_{G}) = e_{H}$ . Therefore $e_{H} \in im f$ , implying $im f$ is non-empty and has an identity.
Closure: Let $h_{1}, h_{2} \in im f$ , then by definition, there exists $g_{1}, g_{2} \in G$ such that $f (g_{1}) = h_{1}$ and $f (g_{2}) = h_{2}$ . Thus

$h_{1} h_{2} = f (g_{1}) f (g_{2}) = f (g_{1} g_{2}) .$ Thus $h_{1} h_{2} \in im f$ .

Inverses: Let $h \in im f$ , and $g \in G$ such that $f (g) = h$ . Then $h^{- 1} = {(f (g))}^{- 1} = f (g^{- 1}) .$

Thus $h^{- 1} \in im f$ .

Therefore, $im f < H$ is a subgroup.

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@@ Line 6: / Line 6: @@
 Let <math>G</math> and <math>H</math> be [[Group|groups]] and <math>f\colon g\to H</math> a group homomorphism. Then,
-* The kernel of <math>f</math> is a normal subgroup of <math>G</math>.
+# The kernel of <math>f</math>, <math>\ker f\trianglelefteq G</math> is a normal subgroup of <math>G</math>.
-* The image of <math>f</math> is a subgroup of <math>H</math>.
+# The image of <math>f</math>, <math>\operatorname{im}f<G</math> is a subgroup of <math>H</math>.
+# <math>G/\ker f \cong \operatorname{im} f</math>.
-* <math>G/\ker f \cong \operatorname{im} f</math>.
+=== Proof ===
+{{Proof|title=Proof of 1|proof=
+By definition, <math>\ker f=\{g\in G\mid f(g)=e_H\}</math> where <math>e_H</math> is the identity of <math>H</math>. <math>\ker f</math> is a subgroup of <math>G</math> because:
+* '''Identity''': Since <math>f</math> is a homomorphism, <math>f(e_G)=e_H</math>. Therefore <math>e_G\in \ker f</math>, implying <math>\ker f</math> is non-empty and has an identity.
+* '''Closure''': Let <math>a,b\in \ker f</math>, then
+<math display="block">f(ab)=f(a)f(b)=e_He_H=e_H.</math>
+Thus <math>ab\in \ker f</math>.
+* '''Inverses''': Let <math>a\in\ker f</math>, then <math display="block">f(a^{-1})=f(a)^{-1}=e_H^{-1}=e_H.</math>
+Thus <math>a^{-1}\in \ker f</math>.
+Therefore, <math>\ker f</math> is a subgroup of <math>G</math>.
+Let <math>g\in G</math> and <math>k\in \ker f</math>, then
+<math display="block">f\left(gkg^{-1}\right)=f(g)f(k)f\left(g^{-1}\right)=f(g)e_Hf\left(g^{-1}\right)=f(g)f\left(g^{-1}\right)=f(g)f(g)^{-1}=e_H,</math>
+thus <math>gkg^{-1}\in \ker f</math>.
+Therefore, <math>\ker f\trianglelefteq G</math> is a normal subgroup.
+}}
+{{Proof|title=Proof of 2|proof=
+By definition, <math>\operatorname{im} f=\{h\in H\mid \exists g\in G: f(g)=h\}</math>. <math>\operatorname{im} f</math> is a subgroup of <math>H</math> because:
+* '''Identity''': Since <math>f</math> is a homomorphism, <math>f(e_G)=e_H</math>. Therefore <math>e_H\in \operatorname{im} f</math>, implying <math>\operatorname{im} f</math> is non-empty and has an identity.
+* '''Closure''': Let <math>h_1,h_2\in \operatorname{im} f</math>, then by definition, there exists <math>g_1,g_2\in G</math> such that <math>f(g_1)=h_1</math> and <math>f(g_2)=h_2</math>. Thus
+<math display="block">h_1h_2=f(g_1)f(g_2)=f(g_1g_2).</math>
+Thus <math>h_1h_2\in \operatorname{im} f</math>.
+* '''Inverses''': Let <math>h\in\operatorname{im}f</math>, and <math>g\in G</math> such that <math>f(g)=h</math>. Then <math display="block">h^{-1}=\left(f(g)\right)^{-1}=f\left(g^{-1}\right).</math>
+Thus <math>h^{-1}\in \operatorname{im} f</math>.
+Therefore, <math>\operatorname{im} f< H</math> is a subgroup.
+}}