# $G_1/H\cong G_2\implies G_1\cong H\times G_2$? [duplicate]

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 $$\frac{G_1}{H}\cong G_2\implies G_1\cong H\times G_2$$

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

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.

• what are some of these other ways? my experience in group theory so far is just a first year course! – Tim Jun 16 '13 at 17:02
• Try and look up the semi-direct product. – Elchanan Solomon Jun 16 '13 at 17:04
• ok thanks! that and the Schur–Zassenhaus theorem seem to come up from a bit of google – Tim Jun 16 '13 at 17:07

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 $\mathbb Z / 2\mathbb Z \times \mathbb Z / 2\mathbb Z$.

• ok, so that's a counterexample, but is this always the case? would there never be any lucky exceptions? – Tim Jun 16 '13 at 17:02
• 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. – andybenji Jun 16 '13 at 17:24

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?

• 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? – Tim Jun 16 '13 at 17:06
• I'm not sure I understand: what would you want $\,\Bbb R/\Bbb Z\,$ to be isomorphic with? – DonAntonio Jun 16 '13 at 17:29
• 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 $\mathbb{R}$ – Tim Jun 16 '13 at 18:06