Is $G$ isomorphic to $\frac{G}{H}\times H$?

Question

Let $G$ be a group and $H$ a normal subgroup. Is $G$ necessarily isomorphic to $\frac{G}{H}\times H$?

Answer 1

If $G$ is the cyclic group of order $4$, and $H$ is its two-element subgroup, then $H \times \frac{G}{H} \cong H \times H$ and has no element of order $4$, so cannot be isomorphic to $G$.

Answer 2

Not necessarily. $D_8$ has $\langle r \rangle \cong C_4$ as a subgroup where $r$ is an element of order $4$ and $D_8 / \langle r \rangle \cong C_2$, but $\left(D_8 / \langle r \rangle \right) \times \langle r \rangle \cong C_2 \times C_4$ is abelian while $D_8$ is not.

Answer 3

Not necessarily. For instance, $S_3$ has as a normal subgroup $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ with quotient $S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}$. But, the product of these groups is abelian, while $S_3$ is not.

Answer 4

No. The only normal subgroup of $S_3$, the symmetric group on $3$ elements, is $A_3 \cong \mathbb{Z}/3\mathbb{Z}$, the alternating group on three elements. But $$ (S_3/A_3) \times A_3 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z} \not \cong S_3 \text{.} $$

Answer 5

Not only need they not be isomorphic, but $G$ need not even have a subgroup (much less a direct factor) isomorphic to $G/H$. For example, $SL(2,5)$ has a center $Z$ of order 2, and $SL(2,5)/Z$ is isomorphic to $A_5$, but $SL(2,5)$ has no subgroup isomorphic to $A_5$.

Answer 6

This only happens when $G$ is a direct product of two groups to begin with (and even in that situation not for all normal subgroups), i.e. when $G$ is not indecomposable. Some counterexamples:

• The symmetric group $S_n$ for $n\geq 3$
• The quaternion group $Q_8$
• Any semidirect product of indecomposable groups that is not direct
• Any cyclic group $\mathbb{Z}_{p^k}$ for prime $p$ and $k\geq 2$
• The integers $\mathbb{Z}$
• etc.

The situation you describe is actually quite rare. To see that note that for example this doesn't work even for groups that are products, e.g. $G=\mathbb{Z}_4\oplus\mathbb{Z}_2$. Now take the subgroup $H=\mathbb{Z}_2\oplus\mathbb{Z}_2$ and note that

$$H\oplus G/H\simeq \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2\not\simeq G$$