Subgroup Isomorphic but Quotient Group Not Isomorphic

The following is a slightly shocking counterexample for beginning students of Group Theory: If G is a group, and H\cong K are normal subgroups of G, it may be possible that G/H\not\cong G/K!

Counter-example: Take G=\mathbb{Z}/4\times\mathbb{Z}/2, H=\langle (0,1)\rangle, K=\langle (2,0)\rangle.

Note that H\cong K\cong Z/2, but G/H=\{(0,0), (1,0), (2,0), (3,0)\}\cong Z/4 while G/K=\{(0,0), (0,1), (1,0), (1,1)\}\cong Z/2\times Z/2!

(By \mathbb{Z}/n we mean \mathbb{Z}/n\mathbb{Z}.)

Author: mathtuition88

4 thoughts on “Subgroup Isomorphic but Quotient Group Not Isomorphic”

      1. the following is the literature (printed books) that I have from Amazon India on counter examples in analysis and topology:

        1) Counterexamples in Topology by Steen and Seebach (Dover)
        2) Experiments in Topology by Barr (Dover)
        3) Counterexamples in Analysis by Bernard R. Gelbaum and John M. H. Olmsred (Dover)
        4) Theorems and Counterexamples in Mathematics by Bernard r. Gelbaum and John M. H. Olmsted (Springer International Edition) — this book does have sections on Algebra (also), Analysis, Geometry/Topology, Complex Variable theory, probability theory and logic/set theory

        Liked by 1 person

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