Cosets and quotients · Definition

First isomorphism theorem

modern algebragroups

Before you start you need to know

It is the bridge between homomorphisms and quotients: whatever a homomorphism «sees» of a group is a quotient of that group.

Definition

Every group homomorphism \(\varphi:G\to H\) induces an isomorphism between the quotient of \(G\) by the kernel of \(\varphi\) and the image of \(\varphi\).

\[G/\ker\varphi\;\cong\;\varphi(G),\qquad a\ker\varphi\mapsto\varphi(a)\]

The idea is that a homomorphism \(\varphi:G\to H\) identifies precisely those elements that differ by an element of the kernel: \(\varphi(a)=\varphi(b)\) is equivalent to \(a^{-1}b\in\ker\varphi\), that is, to \(a\) and \(b\) lying in the same coset. Hence the assignment \(a\ker\varphi\mapsto\varphi(a)\) is well defined and injective; it is a homomorphism because \(\varphi\) is, and it is onto the image by construction.

In practice the theorem is used in two ways. To identify a quotient: if you find a surjective homomorphism \(G\to H\) with kernel \(N\), then \(G/N\cong H\). Thus \(\det:GL_2(\mathbb{R})\to\mathbb{R}^*\) gives \(GL_2(\mathbb{R})/SL_2(\mathbb{R})\cong\mathbb{R}^*\), and reduction \(\mathbb{Z}\to\mathbb{Z}_n\) gives \(\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}_n\). Or to count: \(|\operatorname{im}\varphi|=|G|/|\ker\varphi|\), so the image of \(x\mapsto 3x\) in \(\mathbb{Z}_{12}\) has \(12/3=4\) elements.

A limiting case helps fix ideas. If \(\varphi\) is injective, the kernel is trivial and \(G\cong\operatorname{im}\varphi\): \(G\) lives as a subgroup inside \(H\). If \(\varphi\) is trivial, the kernel is all of \(G\) and the quotient \(G/G\) is trivial, like the image. In general the image is a subgroup of \(H\), not necessarily \(H\): an isomorphism with \(H\) requires \(\varphi\) to be surjective.

The typical mistake is forgetting that the theorem relates the quotient to the image, not to the codomain. Another is believing \(G\) and \(H\) must have the same size; only \(|G|/|\ker\varphi|\) and \(|\operatorname{im}\varphi|\) agree. This result has versions for rings and modules, and it is the first of a family of isomorphism theorems that organize the structure of quotients.

Think about it

If \(\varphi:\mathbb{Z}_{18}\to H\) is surjective with \(|H|=6\), how many elements does its kernel have?

By the theorem, \(H\cong\mathbb{Z}_{18}/\ker\varphi\), so \(|\ker\varphi|=18/6=3\).

Why does \(a\ker\varphi\mapsto\varphi(a)\) not depend on the representative \(a\)?

If \(a\ker\varphi=b\ker\varphi\), then \(b=ak\) with \(k\in\ker\varphi\), and \(\varphi(b)=\varphi(a)\varphi(k)=\varphi(a)\).

Exercises 10

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §3.3
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 10
  • I. N. Herstein, Abstract Algebra, 3rd edition, Wiley, 1996. §2.7
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §14