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I know that given a group $G$ and two normal subgroups $H,K \subset G$ then it is not true that:

"if $H \cong K$ then $ \frac{G}{H} \cong \frac{G}{K} $ (the counterexample is quite easy with products of cyclic groups) "

My question is: Is the converse true?

i.e.

Given that $\frac{G}{H} \cong \frac{G}{K}$ then $H \cong K$ ?

I feel that the answer is no, but I can't think of an example.

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    $\begingroup$ This has been answered in this question, within the question itself. For a variation, see here. $\endgroup$ Commented Dec 21, 2015 at 17:42

3 Answers 3

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Let $$G = \mathbb Z/4\mathbb Z\times\mathbb Z/2\mathbb Z$$ and consider the subgroups $$H = \mathbb Z/4\mathbb Z\times \{e\}\\K=\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$$

Then $$G/K\cong G/H\cong\mathbb Z/2\mathbb Z$$ but $H\not\cong K$.

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  • $\begingroup$ How about if $H$ is normal in $G$ and another group $G'$ and $G/H \cong G'/H$. Is it true that $G\cong G'$? $\endgroup$
    – user5826
    Commented Apr 21, 2017 at 12:30
  • $\begingroup$ No. Take $G = \mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$, $G'= \mathbb Z/4\mathbb Z$ and $H = \mathbb Z/2\mathbb Z$. $\endgroup$
    – Mathmo123
    Commented Apr 21, 2017 at 15:03
  • $\begingroup$ But $H$ is not a Cartesian product. How is $H$ a subgroup of $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$? $\endgroup$
    – user5826
    Commented Apr 21, 2017 at 18:45
  • $\begingroup$ The subgroup of elements of the form $ (a,0)$ is isomorphic to $H$. $\endgroup$
    – Mathmo123
    Commented Apr 22, 2017 at 20:21
  • $\begingroup$ So then $H$ is not really a subgroup of both $G$ and $G'$. $\endgroup$
    – user5826
    Commented Apr 23, 2017 at 8:27
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Take, for instance, $G=\mathbb Z/4\mathbb Z \times \mathbb Z/2\mathbb Z$, $H=\mathbb Z/4 \mathbb Z \times \mathbf 0$, and $K=\mathbb Z /2\mathbb Z \times \mathbb Z/2 \mathbb Z$, so that $G/K\cong G/H \cong \mathbb Z/2 \mathbb Z$.

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  • $\begingroup$ The group $K$ should be written as $2\mathbf Z/4\mathbf Z \times \mathbf Z/2\mathbf Z$ to make it naturally be a subgroup of $G$. $\endgroup$
    – KCd
    Commented May 5, 2022 at 22:21
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This is (also) not true. Let $G = \mathbf Z^{(\mathbf N)}$, $H = (0)$, and $K = \mathbf Z\cdot e_1$ (where $e_1 = (1, 0,\ldots)$). Then $G/K \cong G \cong G/(0)$, but $K \ne 0 = H$.

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  • $\begingroup$ Excuse me, but $Z^{(N)}$ stands for? $\endgroup$
    – HaroldF
    Commented Dec 21, 2015 at 15:48
  • $\begingroup$ It is the set of sequences compactly supported (stationary at $0$ sequences) whose coefficients are in $\mathbb{Z}$. $\endgroup$
    – C. Falcon
    Commented Dec 21, 2015 at 18:39

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