While I was doing an exercise about the classification of groups of order 56, I had some problems concerning the semidirect product.

Let $G$ a group of order 56 and let us suppose that the 7-Sylow is normal (let's call it $H$). Then we want to construct the non abelian group whose 2-Sylow $S$ is $S \cong \mathbb Z_8$.

First of all, we have to determine the homomorphism $\phi \colon \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)$. It is known that $\text{Aut}(\mathbb Z_7) \cong \mathbb Z_6$ and the isomorphism is given by $$ \begin{split} & \mathbb Z_6 \to \text{Aut}(\mathbb Z_7) \\ & a \mapsto \psi_a \colon \mathbb Z_7 \ni n \mapsto an \in \mathbb Z_7 \end{split} $$

So we can start by finding the homomorphism $\mathbb Z_8 \to \mathbb Z_6$. There are exacly $(6,8)=2$ such homomorphism. Who are they? Simply the one who sends everything to $0$ and the multiplication by $3$ (which is the only element in $\mathbb Z_6$ whose order - 2 - divides 8). In multiplicative terms, they are the homomorphism which sends everything to $1$ and the homomorphism which sends $n \mapsto 6^n=(-1)^n$.

So, by composition, we have the two homomorphism $$ \begin{split} \phi_1 \colon & \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)\\ & n \mapsto \text{id} \end{split} $$ and $$ \begin{split} \phi_2 \colon & \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)\\ & n \mapsto \psi_n \colon \mathbb Z_7 \ni x \mapsto 6^nx = (-1)^nx \in \mathbb Z_7 \end{split} $$

Am I right?

Now, if we take $\phi_1$ we simply get the direct product. What if we take $\phi_2$?

For sake of simplicity, let's assume additive notation (this is stupid, I know but it has helped me somehow to understand). If I'm not wrong, we obtain that $H \rtimes_{\phi_2} \mathbb Z_8$ is the set $H \times \mathbb Z_8$ with the operation given by $$ (a,b) + (c,d) = (a+(-1)^bc,b+d) $$

Now if I do $(0,k)+(h,0)-(0,k) = ((-1)^k h, 0) = \phi_k(h)$ which is exactly what I want.

Now I must pass to the much more confortable multiplicative notation: so let's $C_7=\langle s \rangle$ and $C_8=\langle r \rangle$ be the cyclic groups of order 7 and 8. Then we define the automorphisms $$ \begin{split} \phi_n \colon & C_7 \to C_7 \\ & x \mapsto x^{(-1)^n} \end{split} $$ and the homomorphism $$ \begin{split} \psi \colon & C_8 \to \text{Aut}(C_7) \\ & n \mapsto \phi_n \end{split} $$

In other words, we can simply say that $\psi$ is the homomorphism which sends the generator $r$ to the inversion $x^{-1}$. Am I right so far?

Well, now $C_7 \rtimes_{\psi} C_8$ is the set $C_7 \times C_8$ with the operation given by $$ (a,b)(c,d) = (ac^{(-1)^b},bd) $$

I do again the calculation $(1,k)(h,1)(1,k)^{-1}=(h^{(-1)^k},1) = \phi_k(h)$ which is exactly what I want (also according to ineff's answer).

Are there any mistakes? May I ask one more question? Who is this mysterious group I've built up? Is it isomorphic to some other (simpler) group? How can I do to write down a presentation?

I thank you in advance for your kind help.

  • $\begingroup$ Perhaps it's only me, but the above question is toooooo long and with so many sub-questions in between that it is pretty hard to follow. I'd advice you to edit this question and focus in one single thing, say semidirect product with Sylow sbgp $\,\cong\Bbb Z_8\,$ , and then, in case of need, write new questions for other such related questions. $\endgroup$
    – DonAntonio
    Aug 5, 2012 at 3:53
  • $\begingroup$ @DonAntonio I follow your advice and edited the OP. Thanks a lot for your comment. $\endgroup$
    – Romeo
    Aug 5, 2012 at 10:00

1 Answer 1


I suppose your confusion is due to the double representation of the semidirect product: internal vs external.

The semidirect product's theorem states that if you have a group $G$ having two subgroups $H,K < G$ such that $H$ is normal in $G$, $H \cap K = \{1_G\}$ and $G=HK$ then there's an isomorphism $G \cong H \rtimes_\psi K$, for a certain $\psi \colon K \to \text{Aut}(H)$.

By the theorem we can represent every element of $G$ as a pair $(h,k) \in H \times K$ (which is the support of the group $H \rtimes K$).

Consider the two subgroups $\bar H = \{(h,1_K) | h \in H\} \leq H \rtimes K$ and $\bar K = \{(1_H,k)|k \in K\} \leq H \rtimes K$, these subgroups correspond, via the isomorphism, to the subgroups $H$ and $K$ of $G$.

In $H \rtimes K$ we have that for every $h \in H$ and $k \in K$ $$(1_H,k) * (h,1_K) *(1_H,k)^{-1} = (\psi_k(h),1_K)$$ if we identify every $h \in H$ with its corresponding element $(h,1_K)$ and every $k \in K$ with $(1_H,k)$ then this equality become (internally in $G$) $$k*h*k^{-1}=\psi_k(h)$$

The $\psi$ which determine the operation in the semidirect product is exactly the homomorphism sending every $k \in K$ in the (restriction to $H$ of the) automorphism $\psi_k \colon H \to H$ which send $h \in H$ in $khk^{-1}$ (this is clearly well defined because $H$ is normal in $G$.

Hope this help.

  • $\begingroup$ Thank you for your kind explanation, you've been very clear. Anyway I've edited the OP, I hope I've understood something. Would you mind giving a look at it? Thanks. $\endgroup$
    – Romeo
    Aug 4, 2012 at 20:23
  • $\begingroup$ @Romeo I see you have found your group multiplication by yourself and I'm not aware of any known group isomorphic to $C_7 \rtimes C_8$, so I don't know what else I could do, anyway if you need feel free to ask :). $\endgroup$ Aug 5, 2012 at 20:04

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