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Given a group $G$ and a normal subgroup $H \leq G$ , is it natural / accurate to say that $$G \cong H \oplus (G / H)$$

This is what I feel is true intuitively, but I'm sure a group theorist could argue otherwise.

Aloizio Macedo
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gdepaul
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  • This was asked just recently but I can't find it. – rschwieb Apr 17 '19 at 18:44
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    Please ask follow up questions as separate posts, especially after accepting an answer to your original question. – Shaun Apr 17 '19 at 18:45
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    No, in general there is no structure you can name with $H$ and $G/H$ that will make the statement true. If $N\triangleleft G$, then $G$ is an extension of $N$ by $G/N$ (caveat: it is also called an extension of $G/N$ by $N$). There is a construction that will always contain $G$ as a proper subgroup, which is the regular wreath product, $N\wr G/N$. – Arturo Magidin Apr 17 '19 at 18:51
  • For your follow up question, try this: $$|G|=|H|\times |G/H|,$$ where $|K|$ is the order of the group $K$. – Shaun Apr 17 '19 at 18:51
  • It's better use $\times$ than $\oplus$. It's a direct product. By the way, isn't this a multi-duplicate? – YCor Apr 17 '19 at 19:09

3 Answers3

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Let $G=\Bbb Z$ and $H=2\Bbb Z$. Then $H\lhd G$ and $$\begin{align}H\oplus(G/H)&\cong 2\Bbb Z\oplus (\Bbb Z/2\Bbb Z)\\ &\cong \Bbb Z\oplus\Bbb Z_2\\ &\not\cong\Bbb Z. \end{align}$$

Shaun
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Take $G=\mathbb{Z_4}$. It has a normal subgroup isomorphic to $\mathbb{Z_2}$, and the quotient group is $\mathbb{Z_2}$ as well. But $\mathbb{Z_4}$ is not isomorphic to $\mathbb{Z_2}\times\mathbb{Z_2}$.

Mark
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Let’s take a look at the sequence

$$1\to H \xrightarrow{i} G \xrightarrow{\pi} G/H \to 1$$

where $i$ is inclusion and $\pi$ is natural projection.

What you want to be true, that $G\cong H\oplus G/H$, is what it means to say that the sequence above splits. It is equivalent to the existence of a homomorphism $\phi: G\to H$ such that $\phi\circ i$ is the identity map on $H$.

Let’s consider the sequence that Mark gives in his answer:

$$1\to \Bbb{Z}/2\Bbb{Z} \xrightarrow{i} \Bbb{Z}/4\Bbb{Z} \xrightarrow{\pi} \Bbb{Z}/2\Bbb{Z}\to 1.$$

There are two possible choices for $\phi$: either $\phi(1) = 1$ or $\phi(1)=0$. In the first case, we have

$$(\phi\circ i)(1) = \phi(2) = \phi(1) + \phi(1) = 0$$

and in the second case we have

$$(\phi\circ i)(1) = \phi(2) = 0.$$

Thus no such $\phi$ exists, so the sequence does not split. Clearly, in Mark’s example it is much easier to see that the sequence doesn’t split, but this sort of argument is common when talking about more complicated examples.

Follow up:

Note that in the above case, you are assuming that everything in $H$ and what you identify as the quotient $G/H$ within $G$ commute. This might not always be the case.

If you can find a map $\phi: G/H \to G$ so that $(\pi\circ \phi)$ is the identity on $G/H$, then it’s true that

$$G\cong H\rtimes G/H,$$

where the symbol denotes the semidirect product. For example, it’s true that $D_{2n}\cong \Bbb{Z}/n\Bbb{Z}\rtimes \Bbb{Z}/2\Bbb{Z}$.

Santana Afton
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    There’s a lot to be said on this topic, but note that your intuition is exactly correct for linear transformations $\Bbb{R}^n\to\Bbb{R}^n$ — your claim is precisely the rank-nullity theorem. – Santana Afton Apr 17 '19 at 18:55
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    Okay, one last comment. The big question here is this: if $A$ and $B$ are groups, what sorts of groups $G$ are there so that they fit into the exact sequence $$1\to A\to G\to B\to 1.$$ The direct product and the semidirect product are examples of such a $G$, but in general you could have a wealth of possible $G$ that fit here. This is what Arturo Magidin was referencing when they mentioned that $G$ is an extension of $B$ by $A$. – Santana Afton Apr 17 '19 at 19:06