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If G is a finite group and N is a normal subgroup in G , then can we say G $\cong$ G/N $\times$ N always? Is it true for like normal nilpotent or normal solvable or any such special classes. I couldn't help but ask for it. What if G is infinite? Please tell me if you can, the concept I am skipping here. If you can avoid using exact sequences, It will be much clearer to me. I have seen an answer which still is not so clear to me. Thanks!

Bhaskar Vashishth
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    $\mathbb{Z}$ has no non-identity elements of finite order. But what about $n\mathbb{Z} \times (\mathbb{Z}/n\mathbb{Z})$? As a finite example, the cyclic group $C_{4}$ has a normal subgroup of order $2$, with quotient of order $2$, but $C_{2} \times C_{2}$ has no elements of order 4. – Alastair Litterick Nov 13 '14 at 22:15
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    If $N$ is a normal subgroup of a finite group $G$ and the order $|N|$ and index $[G:N] = |G|/|N|$ are relatively prime then $G$ is isomorphic not necessarily to the direct product of $N$ and $G/N$, but to some semidirect product of $N$ and $G/N$. This is called the Schur--Zassenhaus theorem. – KCd Nov 13 '14 at 22:29
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    No. See http://en.wikipedia.org/wiki/Group_extension. – Qiaochu Yuan Nov 13 '14 at 23:51
  • @AlastairLitterick Ok that is fine that it is not for direct product but will there must be extension always , right that is why we only need all finite simple groups to know all the finite grouos (which is done CFSG) if we can study all the extension (which is not done yet). Right? – Bhaskar Vashishth Nov 14 '14 at 04:58
  • Such questions are asked here only because the OPs have considered no single example before asking. – Martin Brandenburg Nov 14 '14 at 09:19
  • Group theory is interesting because the property is false. – Martín-Blas Pérez Pinilla Nov 14 '14 at 10:49
  • @Martín-BlasPérezPinilla: Hence, linear algebra is not interesting? ;) In any case, even if there is some isomorphism $G \cong N \times G/N$, there will be no canonical one, i.e. compatible with homomorphisms. This makes linear algebra still interesting because this is not just about classification of vector spaces, but rather about linear maps between them. Although vecor spaces split up, linear maps don't split up. – Martin Brandenburg Nov 14 '14 at 10:55
  • @MartinBrandenburg, you are right, the election of word can be unfortunate. In any case group theory is much harder than linear algebra, and my "interesting" meant to say "difficult". – Martín-Blas Pérez Pinilla Nov 14 '14 at 11:07

3 Answers3

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Hints: $S_3$ has order $6$. It has a subgroup $N$ of index 2, so $N$ is normal. Then prove that $S_3\not\cong \mathbb{Z}_2\times\mathbb{Z}_3$, which is contrary to your claim.

user1729
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The claim fails even more profoundly. While a finite abelian group always has a subgroup isomorphic to its any quotient, this is not true in general. (As said before, the original claim is still false for abelian groups.)

The smallest counterexample is the quaternion group $Q_8$. Its center is $Z(Q) \cong \mathbb{Z}_2$ and the quotient $Q/Z(Q)$ is isomorphic to the Klein four-group, but the subgroups of order $4$ of $Q_8$ are cyclic.

Leppala
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I've intended to ask a similar question. Fortunately, the community has avoided a duplicate. Moreover, I've had more information to better my proposition, which is written below.

The proposition: The necessary and sufficient conditions of group $G$ and its normal subgroup $N$ (regardless of commutativity and finiteness) to satisfy $G \sim N \times G / N$ is that:

$i) \; \; N \le Z(G)$ (center of group $G$), and

$ii)$ There is a way to choose representative elements for classes of $G/N$ so that those elements would make up a subgroup of $G$.

I think I proved this, but I'm not so sure.

Vincent J. Ruan
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