# If the quotient group $G/N$ be isomorphic to $G_2$, is $G$ isomorphic to $G_1\times G_2$ where $N$ is isomorphic to $G_1$?

Respected all

Suppose that $G$ is given group and $N$ be a normal subgroup of $G$ and $N$ is isomorphic to $G_1$ as groups. If $G/N$ is isomorphic to the group $G_2$, can we establish that $G$ is isomorphic to the group $G_1\times G_2$ ?

• Have you looked at a few examples? – Tobias Kildetoft Sep 11 '15 at 11:01

Hint: Try $G=S_3$ and $N=A_3$.

Solution:

$A_3 \cong C_3$, $S_3/A_3 \cong C_2$ but $S_3 \not\cong C_3 \times C_2 \cong C_6$, because $S_3$ is not abelian.

No. Counterexample: Let $G = \Bbb Z$ and $N = n\Bbb Z$ for some $n > 1$.

Any three groups in this relationship ($G$ a big group, $N$ a normal subgroup and $G/N$ their quotient) can be put into a sequence called short exact, looking like this: $$0\to N \to G \to G/N\to 0$$ where, if $G \cong N \times G/N$, it is called a split exact sequence, and is immediately a lot nicer (if the groups aren't assumed to be abelian, then splitness is a bit wider, as pointed out in a comment below).

• a short exact sequence in the category of groups is usually said to split as long as the last map admits a section, in which case the group need not be a direct product (just a semidirect one). – Tobias Kildetoft Sep 11 '15 at 11:09
• @TobiasKildetoft True, but my answer has the added bonus of being more or less abelian, in which case it doesn't matter. – Arthur Sep 11 '15 at 11:13

Not necessarily. If $G$ is the dyhedral group $D_3$ and $N$ is the normal subgroup of index $2$ which is generated by the rotation $\beta$, then $D_3/N$ is isomorphic to $\Bbb Z_2$. However, $D_3$ is not isomorphic to a direct product $N\times\Bbb Z_2$, as it is not Abelian.

Note that we have short exact sequence $$0 \to \langle\beta\rangle \to D_3 \to \langle\alpha\rangle \to 0$$ where $\alpha$ is the reflection. There is also an inclusion $s:\langle\alpha\rangle \to D_3$. The reason for the map $r:D_3\to\langle\alpha\rangle$ to exist is that $D_3$ is a semidirect product $\langle\beta\rangle \ltimes \langle\alpha\rangle$, that means $\langle\beta\rangle$ is a normal subgroup and $\langle\alpha\rangle$ is a (mere) subgroup such that each $D_3=\langle\beta\rangle\langle\alpha\rangle$ and $\langle\beta\rangle\cap\langle\alpha\rangle = \{e\}$. Then each element in $D_3$ can be written uniquely as a product $ba$ where $b\in\langle\beta\rangle$ and $a\in\langle\alpha\rangle$. Since $\langle\beta\rangle$ is of index $2$ and thus normal, we have $$b_1a_1b_2a_2 = b_1\underbrace{a_1b_2a_1^{-1}}_{\in\langle\beta\rangle}a_1a_2$$ so the map $ba\mapsto a$ is a homomorphism $D_3\to\langle\alpha\rangle$ which is the identity on $\langle\alpha\rangle$ and whose kernel is $\langle\beta\rangle$.
On the other hand, given a subgroup $H$ of a group $G$ and a retraction $r:G\to H$, it can be shown that $G = \text{ker}(r)\ltimes H$.

What prevents $D_3$ from being an (inner) direct product is that $\langle\alpha\rangle$ is not normal in $D_3$, as $\beta\alpha\beta^{-1} = \beta^2\alpha \notin \langle\alpha\rangle$.