# Group homomorphisms and semidirect products

Suppose $\phi: G \rightarrow \overline{G}$ is a homomorphism, onto, with kernel $N$. Then, $G/N \cong \overline{G}$. Can we also conclude that $G$ is isomorphic to the semidirect product $N \rtimes \overline{G}$? It is true that the orders agree ($|G| = |N| |\overline{G}|$) and that $N$ is normal. Does $G$ have a subgroup that is isomorphic to $\overline{G}$ and that has trivial intersection with $N$?

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As many examples are showing, if $G$ is a finite $p$-group of order greater than $p$ with a unique subgroup $N$ of order $p$ then you can't write $G$ as a semidirect product using $N$ because the complementary factor's subgroup of order $p$ would have to be $N$. For instance, this is the case if $G$ is a cyclic $p$-group of order greater than $p$. See math.stackexchange.com/questions/81607/… – KCd Mar 10 '13 at 17:46
Moreover, if $p$ is odd then any finite $p$-group with a unique subgroup of order $p$ must be cyclic. So if you want nonabelian examples then $p = 2$, and in that case $G$ must be a generalized quaternion group. See Theorem 4.7 of math.uconn.edu/~kconrad/blurbs/grouptheory/genquat.pdf. – KCd Mar 10 '13 at 17:50

If $G = C_4$ is a cyclic group of order $4$, then there is an epimorphism $G \to H$, where $H = C_2$, with kernel $N$ which is the unique subgroup of $G$ of order $2$. So $G$ cannot be possibly a semidirect product.

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+ Thanks for the help – Babak S. Mar 10 '13 at 17:53

No. Let $G=\mathbb Z/4\mathbb Z$ and $N=2\mathbb Z/4\mathbb Z\cong \mathbb Z/2\mathbb Z$. Then $G/N\cong \mathbb Z/2\mathbb Z$ and the only semidirect product of $\mathbb Z/2\mathbb Z$ and $\mathbb Z/2\mathbb Z$ is the direct product, as $\mathbb Z/2\mathbb Z$ has trivial automorphism group, but $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\not\cong G$.

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It is not always true that $G$ has a subgroup isomorphic to $G/N$. The quaternion group $Q_8$ is the smallest example of such a group, but I cannot remember exactly which subgroup $N$ you take. It only has $8$ elements though so if you're willing to look it shouldn't be too hard to find.

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Take $N = Z(Q_8) = \{\pm 1\}$. Then $\overline{G}$ is abelian (in fact a product of two groups of order 2) and it's impossible to write $G = HN$ where $H$ and $N$ have trivial intersection, because $H$ must have a subgroup of order 2 and $N$ is the only subgroup of $G$ with order 2. – KCd Mar 10 '13 at 17:44