While reading some group theory notes I came up to this fact:

Proposition: If $G$ is the inner semi direct product of $H,K$ ($G=HK$, $H\cap K=\left\{1\right\}$ and $H\unrhd G$) then $G\cong H\rtimes_fK$ where $f$ is expicitely given by $k\mapsto f_k(h)=h^{-1}kh$.

Now I had no problem with this until I read the Classification of finite groups with order $pq$ for prime $p,q$ where $p\equiv 1\mod q$.

After application of the Sylow Theorems one obtains such $H,K$. But then, the writer turned to the problem of counting the homomorphisms in $f:K\to Aut(H)$. This seems the way all other group theory books deal with this but it is never explained why this is needed.

Unless I am confused, the Proposition implies that $G$, up to isomorphism is determined by $H,K,f$ and $f$ is explicitely given. If $H\rtimes_gK\not\cong H\rtimes_fK$ then it can't be that $G\cong H\rtimes_gK$ as that would contradict the transitivity of $\cong$. If we only had the existence of $f$ in Proposition, then I agree we would need to determine all such $f$, as $G$ would be the product of $H,K$ and one of these $f$. So why do we need to do this even when we have a specific $f$?

  • $\begingroup$ I cannot understand what you are asking. When analysing groups of order $pq$ you are not given a specific $f$, so you have to consider all possible $f$. But it is possible for distinct $f$ to define isomorphic semidirect products. The question of when two of these are isomorphic is discussed in math.stackexchange.com/questions/527800 $\endgroup$ – Derek Holt Aug 30 '14 at 16:20
  • 2
    $\begingroup$ I'm not certain, but I think the point is that you don't know what $h^{-1}kh$ is until you know the map $f$ or the group $G$. Then the Proposition does not say that $G$ is completely determined by the abstract groups $H$ and $K$. $\endgroup$ – Jessica B Aug 30 '14 at 16:26
  • $\begingroup$ @JessicaB What do you mean by "you don't know that $h^{-1}kh$ is"? $\endgroup$ – John Aug 30 '14 at 16:53
  • $\begingroup$ @John I mean that given two groups $H$ and $K$, there is no a priori definition of multiplying an element of one by an element of the other. If you already know that they are specific subgroups of a larger group $G$ then you have a definition there. But then that doesn't apply to new groups $H'$ and $K'$ that are isomorphic to $H$ and $K$ respectively. $\endgroup$ – Jessica B Aug 30 '14 at 16:56
  • 1
    $\begingroup$ @John I'm not certain, but I think $H\rtimes_f K$ and $H\rtimes_g K$ could be isomorphic or not depending on the choices of $f$ and $g$. $\endgroup$ – Jessica B Aug 30 '14 at 17:29

$\newcommand{\Aut}[0]{\mathrm{Aut}}$In case this is what you are interested in, given $H$ and $K$, two different $f$ may well lead to isomorphic groups $G$.

For instance, in the case you mentioned, consider the primes $p = 7$ and $q = 3$. Let $K = \langle b \rangle$ and $H = \langle a \rangle$. Consider the homomorphisms $f, g : K \to \Aut(H)$ determined by $$ f : b \mapsto (a \mapsto a^{2}) $$ and $$ f : b \mapsto (a \mapsto a^{4}). $$ Then $$ H \rtimes_{f} K \cong H \rtimes_{g} K. $$

Actually, when $p \equiv 1 \pmod{q}$, you should have seen that there are two isomorphism classes of groups of order $pq$. One, the cyclic group, corresponds to the trivial homomorphism $K \to \Aut(H)$, the other to all other homomorphisms.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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