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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 $f$?

John
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  • 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 http://math.stackexchange.com/questions/527800 – Derek Holt Aug 30 '14 at 16:20
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    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 $K$. – Jessica B Aug 30 '14 at 16:26
  • @JessicaB What do you mean by "you don't know that $h^{-1}kh$ is"? – John Aug 30 '14 at 16:53
  • @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. – Jessica B Aug 30 '14 at 16:56
  • @JessicaB Are you saying that $H\rtimes_fK$ need not be isomorphic to $H'\rtimes_fK'$? How is that a problem? – John Aug 30 '14 at 17:05
  • @John What is the $f$ you use in these two cases? You cannot define the same homomorphism on different groups (although you could define one, plus isomorphisms and use composition, but isomorphisms are not unique). – Jessica B Aug 30 '14 at 17:07
  • @JessicaB So $H\rtimes_f K\cong H'\rtimes_gK'$ but $g$ can be multiple different maps? – John Aug 30 '14 at 17:25
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    @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 $g$. – Jessica B Aug 30 '14 at 17:29

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$\newcommand{\Aut}[0]{\mathrm{Aut}}$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.

Andreas Caranti
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