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The Isomorphism Theorems

Although it is not evident at first, factor groups correspond exactly to homomorphic images, and we can use factor groups to study homomorphisms.

The Isomorphism Theorems

Although it is not evident at first, factor groups correspond exactly to homomorphic images, and we can use factor groups to study homomorphisms. We already know that with every group homomorphism \(\phi: G \rightarrow H\) we can associate a normal subgroup of \(G\), \(\ker \phi\). The converse is also true; that is, every normal subgroup of a group \(G\) gives rise to homomorphism of groups.

Let \(H\) be a normal subgroup of \(G\). Define the natural or canonical homomorphism \[\begin{aligned}\end{aligned}\] by \[\begin{aligned}\end{aligned}\]. This is indeed a homomorphism, since \[\begin{aligned}\end{aligned}\]. The kernel of this homomorphism is \(H\). The following theorems describe the relationships between group homomorphisms, normal subgroups, and factor groups.

Theorem: First Isomorphism Theorem

If \(\psi : G \rightarrow H\) is a group homomorphism with \(K =\ker \psi\), then \(K\) is normal in \(G\). Let \(\phi: G \rightarrow G/K\) be the canonical homomorphism. Then there exists a unique isomorphism \(\eta: G/K \rightarrow \psi(G)\) such that \(\psi = \eta \phi\).

Proof

We already know that \(K\) is normal in \(G\). Define \(\eta: G/K \rightarrow \psi(G)\) by \(\eta(gK) = \psi(g)\). We first show that \(\eta\) is a well-defined map. If \(g_1 K =g_2 K\), then for some \(k \in K\), \(g_1 k=g_2\); consequently, \[\begin{aligned}\end{aligned}\]. Thus, \(\eta\) does not depend on the choice of coset representatives and the map \(\eta: G/K \rightarrow \psi(G)\) is uniquely defined since \(\psi = \eta \phi\). We must also show that \(\eta\) is a homomorphism. Indeed, \[\begin{aligned}\eta( g_1K g_2K ) & = \eta(g_1 g_2K) \\ & = \psi(g_1 g_2) \\ & = \psi(g_1) \psi(g_2) \\ & = \eta( g_1K) \eta( g_2K )\end{aligned}\]. Clearly, \(\eta\) is onto \(\psi( G)\). To show that \(\eta\) is one-to-one, suppose that \(\eta(g_1 K) = \eta(g_2 K)\). Then \(\psi(g_1) = \psi(g_2)\). This implies that \(\psi( g_1^{-1} g_2 ) = e\), or \(g_1^{-1} g_2\) is in the kernel of \(\psi\); hence, \(g_1^{-1} g_2K = K\); that is, \(g_1K =g_2K\).

Mathematicians often use diagrams called commutative diagrams to describe such theorems. The following diagram commutes since \(\psi = \eta \phi\).

Notice that in the course of the proof of , we have also proved the following theorem.

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

Энди сизлар Буни ҳисоблаш учун ҳисоблагич керак эмас, лекин унинг қисмлари ҳисоблаш мумкин. Қуйида бирор нарсани синаб кўринг ёки ўзингиз ёзинг.

Ўзингиз ишлашда давом этинг

Бепул ҳисоб ҳар бир дарсга ёзув қўшиб беради, сиз тугатган нарсалар ва ҳал қилинган масалалар бир жойда қайд этилади ва сиз бу саҳифа ҳақида устоздан сўрашингиз мумкин. Математика ўзида очиқ бўлиб, унга кириш ёки киришни истамаган ҳар кимга очиқ.

Ройхатдан ўтиш Кириш

Бу ерда қўлланилган белгилар

Барча белгилар учун тўлиқ таъриф, сурат ва ҳар бир ҳарфнинг маъносини кўриш учун белгини босинг.

Одамлар сўрайдиган саволлар

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Бу саҳифанинг қисмлари Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). (Ушбу оятда Аллоҳ таоло Пайғамбаримиз Муҳаммадга (с. а. в.) Қуръони Каримни туширишни амр қилмоқда.)

Кўпроқ Abstract Algebra

Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula