# Why don't semi-direct products determine a group uniquely?

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 $f$?

• 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 – Derek Holt Aug 30 '14 at 16:20
• 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
• @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

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