# What is this (kind of) group product?

In Isaacs' Algebra, question 5.13, he walks the reader through a construction of a non-Abelian group of order $pq$, where $p$ and $q$ are primes such that $q | p -1$.

In particular, we let $P$ be a cyclic group of order $p$, and $Q \subseteq Aut(P) \cong Z/(p-1)Z$ be a subgroup of order $q$. Then the collection of set maps $\phi_{a,\sigma}$, defined by $\phi_{a,\sigma}(x) = \sigma(xa)$, is a subgroup of $Sym(P)$.

This seems like a neat way to make a new group out of an old one, with possibly more applications than just this question. (I think the verification of the group axioms would go through for any group $G$ and subgroup of $Aut(G)$, but it's late and I might have missed something.) Does it have a name?

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Read in some book or google "semidirect product" ... – DonAntonio May 12 '14 at 3:59

It is called as semidirect pruduct.

I am trying to give the intuation behind it.

Let $H,N$ be subgroup of $G$ such that $N$ is normal and $H\cap N =1$. Since $N$ is normal, we can say that $HN$ is a subgroup of $G$.

Notice that $hNh^{-1}=N$ for all $h\in H$ so $H$ act on $N$ by conjugation moreover the map

$\phi_h :n\to hnh^{-1}$ is an isomorphism $\implies$ we have an homomorphism from $H$ to $Aut(N)$.

Now, question is that "Is converse true?". If I have a two group $H$ and $N$ and I have a homomorphism from $H$ to $Aut(N)$, Can I define $HN$ ? like they are subgroup of a group ?

Answer is yes and it is called as semiderict product which is very useful tool.

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You're right, this is a semidirect product when you work out the composition. – AreaMan May 12 '14 at 16:18