If $G$ is a finite group, it is not true in general that $G$ is the semidirect product of a normal subgroup $N$ and the quotient group $G/N$. It is also not true in general that there is a subgroup of $G$ isomorphic to $G/N$.

But if $N=[G,G]$, the commutator subgroup of $G$, then I believe both these statements should be true. Are they?

My intuition is that, since the elements of $N$ are products of non-trivial commutators of $G$, whereas $G/N$ is abelian, then if $G/N$ were isomorphic to a subgroup of $G$, we would clearly have $N \bigcap G/N=\langle e \rangle$. Further, since $G/N$ is the "albelianization" of $G$, it seems intuitive that $G$ should contain a subgroup $H$ isomorphic to $G/N$. That would naturally lead to all elements of $G$ as products of elements of $N$ and $H$, which gives us $G$ as a semidirect product of $N$ and $H$. But I do not know how to prove that such an $H$ exists, or if it may not.

I ask this question because, if true, it would lead to a nice intuitive characterization of nonabelian solvable groups as nonabelian only with respect to the semidirect product of otherwise abelian factors. Thanks in advance!

EDIT: As I note in my comment, the counterexample $Q_8$ may not be valid, as $Q_8/\mathbb Z_2 \cong \mathbb Z_2 \times \mathbb Z_2$, not $\mathbb Z_4$. But please let me know if this is irrelevant.

EDIT 2: I have asked a related question about this as a group extension here.


2 Answers 2


It's false. Take the quaterions $Q_8$. Here $[G,G]=Z(G)=\langle -1\rangle \cong \mathbb Z_2$, but $Q_8$ is not the semidirect product of $\mathbb Z_2$ by $\mathbb Z_4$. To see this, simply note that every element of $Q_8$ of order $4$ squares to $-1$.

EDITED: I should have said $Q_8$ is not the semidirect product of $\mathbb Z_2$ by a group of order $4$.

  • $\begingroup$ But isn't $Q_8 / {\mathbb Z_2} \cong {\mathbb Z_2} \times {\mathbb Z_2}$, not $\mathbb Z_4$? Or does this not matter? $\endgroup$
    – user452
    Dec 3, 2013 at 1:49
  • $\begingroup$ Note that $G$ is the semi direct product of $H$ and $K$ if $G=HK$, $H$ is normal in $G$, and $H\cap K=1$. If you propose to have $H=\langle -1\rangle$, then $K$, whatever it may be, wouldn't be a subgroup or it wouldn't satisfy $H\cap K=1$. $\endgroup$
    – JMag
    Dec 3, 2013 at 2:26
  • $\begingroup$ Just affirming that J.Maglione is 100% correct. $\endgroup$
    – Doc
    Dec 3, 2013 at 4:05
  • $\begingroup$ @trb456: You are right, $Q_8 / Z(Q_8) \cong {\mathbb Z_2} \times {\mathbb Z_2}$. You could argue like J. Maglione does, or note that $Q_8$ does not contain a subgroup isomorphic to ${\mathbb Z_2} \times {\mathbb Z_2}$. $\endgroup$ Dec 3, 2013 at 7:15
  • 6
    $\begingroup$ The easy way to see that $Q_8$ cannot be a (nontrivial) semidirect product is to note that $Q_8$ has a unique element of order $2$. If we had $Q_8\cong K\rtimes H$, then $K$ and $H$ would both have to be $2$-groups, so each would have an element of order $2$ by Cauchy. But we know $Q_8$ only has one element of order $2$, so it can't work. $\endgroup$
    – Alexander Gruber
    Dec 3, 2013 at 7:40

For another counterexample, consider the free group on two generators, $G = \langle x,y\rangle$. The abelianization of $G$ (i.e., $G/[G:G]$) is $\mathbb Z \times \mathbb Z$, and in order to express $G$ as a semidirect product, you need to find a complementary subgroup to $[G,G]$ which is isomorphic to $\mathbb Z \times \mathbb Z$. However, the centralizer of every element of $G$ is cyclic (see if you can figure out why!), so this is not possible.


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