Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $N$?

share|cite|improve this question
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… – 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 – KCd Mar 10 '13 at 17:50
up vote 3 down vote accepted

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.

share|cite|improve this answer
+ 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 $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z\not\cong G$.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.