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This question already has an answer here:

Lagrange's lets us write the deceptively tidy relation: $$\left|\frac{G}{H}\right|=\frac{|G|}{|H|}$$ and from this we can do neat things like, in the proof of the Orbit-stabiliser theorem, $$\frac{G}{\text{stab}(s)}\cong \text{orb}(s)\implies\left|\frac{G}{\text{stab}(s)}\right|=|\text{orb}(s)|\implies\frac{|G|}{|\text{stab}(s)|}=|\text{orb}(s)|\\\implies|G|=|\text{orb}(s)|\times|\text{stab}(s)|$$ Which leads to my question: is it always/sometimes/ever possible to use naive reasoning and say that $$\frac{G_1}{H}\cong G_2\implies G_1\cong H\times G_2$$

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marked as duplicate by Jack Schmidt, Grigory M, Myself, Davide Giraudo, Potato Jun 16 '13 at 17:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Take the cyclic group $\mathbb{Z}/4\mathbb{Z}$ of order $4$. This has a subgroup of order $2$, let us call it $H$. $H \cong \mathbb{Z}/2\mathbb{Z}$, since this is the only group of order $2$. By order-counting, $G/H \cong \mathbb{Z}/2\mathbb{Z}$, because it has order $2$ as well. But $\mathbb{Z}/4\mathbb{Z}$ is not $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

The idea here is the following: there are many ways to build a new group out of smaller pieces besides for the direct product.

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  • $\begingroup$ what are some of these other ways? my experience in group theory so far is just a first year course! $\endgroup$ – Tim Jun 16 '13 at 17:02
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    $\begingroup$ Try and look up the semi-direct product. $\endgroup$ – Elchanan Solomon Jun 16 '13 at 17:04
  • $\begingroup$ ok thanks! that and the Schur–Zassenhaus theorem seem to come up from a bit of google $\endgroup$ – Tim Jun 16 '13 at 17:07
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No, because $(\mathbb Z / 4\mathbb Z) / (2) \cong \mathbb Z / 2\mathbb Z$, where $(2) \cong \mathbb Z / 2\mathbb Z$, but clearly $\mathbb Z / 4\mathbb Z$ is not isomorphic to $\mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z $.

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  • $\begingroup$ ok, so that's a counterexample, but is this always the case? would there never be any lucky exceptions? $\endgroup$ – Tim Jun 16 '13 at 17:02
  • $\begingroup$ It's not generally true that the existence of a counterexample ensures the proposition fails in all cases. For instance, we have the naive example $e/e \simeq e \simeq e \times e$ where $e$ denotes the identity group. $\endgroup$ – andybenji Jun 16 '13 at 17:24
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Hint:

Try to do this with

$$G=C_4=\text{ the cyclic group of order }\;4\;,\;\;H=C_2\;,\;\;G/H\cong C_2\stackrel?\implies C_4\cong C_2\times C_2$$

Further hint: no, you can't...why?

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  • $\begingroup$ I see how that one doesn't work, but what about an example like $\mathbb{R}/\mathbb{Z}\times\mathbb{Z}$? does that also not work? $\endgroup$ – Tim Jun 16 '13 at 17:06
  • $\begingroup$ I'm not sure I understand: what would you want $\,\Bbb R/\Bbb Z\,$ to be isomorphic with? $\endgroup$ – DonAntonio Jun 16 '13 at 17:29
  • $\begingroup$ sorry, guess it's slightly different, but clearly $\mathbb{R}/\mathbb{Z}$ is isomorphic to itself, so I guess it's a question of if the example I gave is isomorphic to $\mathbb{R}$ $\endgroup$ – Tim Jun 16 '13 at 18:06

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