## 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 $\mathbb{Z}/n\mathbb{Z}$.) http://mathtuition88.com
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### 4 Responses to Subgroup Isomorphic but Quotient Group Not Isomorphic

1. nkpithwa says:

Hey, this is certainly very interesting !! BTW, I have some literature on counter examples in Analysis and Topology, but not in Algebra. Thanks, for a nice example 🙂 🙂 🙂

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• mathtuition88 says:

Thanks! Can you mention what literature you are referring to? I am interested to read those books.

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• nkpithwa says:

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

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• mathtuition88 says:

Thanks a lot!

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