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The key theorem involved is:

Suppose $G$ is a group with subgroups $H$ and $k$ such that $H$ is normal in $G$ and $H \cap K=1$. Let $\phi: K \to Aut(H)$ be the homomorphism defined by mapping $k \in K$, TO THE automorphism by conjugation by $k$. Then if $G = HK$, we have $G$ is the semidirect product of $H$ and $K$ with respect to $\phi$.

I am just confused about how to use this to classify groups of small orders. Say for instance I have a noncommutative group with order $pq$, $p$ prime, $q$ prime, $p < q$. Then Let $H$ be the $q-$sylow subgroup of $G$, $K$ be a $p-$sylow subgroup of $G$. Then $H \cap K = 1$, and $H$ is normal in $G$. It is also not hard to show that $G = HK$. Then by the theorem shouldn't we naturally have that $G$ is the semidirect product of $H$ and $K$ with respect to the homomorphism induced by conjugation?

However, all illustrations on this example start by describing all the possible homomorphisms from $Z_p \to Z_{q-1}$. I am totally lost here, why should we care about homomorphisms other than the one induced by the conjugation?

Alex
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2 Answers2

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The semidirect product can be defined with respect to any homomorphism $\phi$ from $K$ to the group of automorphisms of $H$. The trivial homomorphism (everything goes to the identity) gives the direct product, which always exists. The automorphisms of $H$ in the image of $K$ need not all be inner automorphisms, and even if the image of the homomorphism consists of inner automorphisms, $\phi$ need not be the homomorphism defined by conjugation.

It therefore has to be proved that, where non-abelian groups of order $pq$ exist, they are all isomorphic. In this simple case this is not hard to do.

Mark Bennet
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  • So the only place where this technique is useful is when all of them are isomorphic? – Alex Nov 11 '18 at 16:42
  • @Vanstar I think it is more useful than that. You are seeing this from a particular perspective - but a bigger question is how you can extend one group by another, and semidirect products are important in that context. When groups get larger, the possibilities increase. – Mark Bennet Nov 11 '18 at 16:52
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We're trying to describe all groups of order $pq$. We don't know anything about the group $G$ other than it has order $pq$, so the homomorphism induced by conjugation could be any homomorphism. So we have to consider them all to make sure we exhaust all possibilities for groups of order $pq$.

Ted
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  • Okay, in this case it happens to be that all homomorphisms other than the trivial one give us the same group. But what if different homorphisms produce different products? How will you know which is the one given by the conjugation? – Alex Nov 10 '18 at 23:18
  • @Vanstar I don't understand your question. By the definition of a semidirect product, any homomorphism can define a semidirect product, and that homomorphism becomes conjugation within the semidirect product. The real problem is to determine which homomorphisms give isomorphic groups. – Ted Nov 12 '18 at 05:56