# Proof that a semidirect product with trivial action is the direct product

I'm learning abstract algebra from Cameron's book on Introduction to Algebra, and this is one of the assignments from there. However, I can't seem to wrap my head around the idea of how semidirect and direct products relate to action homomorphism from B to Aut(A). Hence, I don't know where to start the proof.

Any help (and explanations of this fairly basic concept) greatly appreciated. I'm in a tutorial system, so there's no lectures/class notes I can look for help in. Google has also been unhelpful in this area...

Thanks again!

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As for accusation against Google not telling you about this, if you asked him properly, he would have told you. He tells me a few links that tell me this, but, anyway, I am writing up an answer here. :) – user21436 Feb 8 '12 at 16:29

I am not sure how you have defined semi-direct product, but any way, I'll write a complete answer, recalling the definitions, in the process, we'll also establish notation.

I learnt this from Rotman, so, there'll certainly be a flavour of his notation here.

Definition 1

A group $G$ is a semi-direct product of $K$ by $Q$, if the following is true:

1. $K \lhd G$ (Atleast one of them is good.)
2. $KQ=G$ (The subgroups are fat enough!)
3. $K \cap Q=\{e_G\}$ (They are fat in a nice way.)

This is the important abstract thing going on here:

Theorem 2

Given a group $K$, $Q$ and a homomorphism $\theta:K \to\operatorname{Aut} H$, there is a way to build a new group, which is a semi-direct product of $K$ and $Q$ in the following way:

The set $G=\{(h,k)|h \in H; k \in K\}$ and the operation in the group is twisted by the element that $\theta$ points to.

$$(h,k) \ast (h',k')=(h,[\theta(k)](h),kk')$$ $$(h,k)^{-1}=([\theta(k^{-1})](h),k^{-1})$$

Prove that the above structure is a group. (Or so you'd have done this in your class.)

The difficulty will then lie in understanding the following:

What is a trivial action? All elements of $K$ are so good (bad?) that they do nothing to any element of $H$. Then, all elements in $K$ fix all elements of $H$. The, $K \to \operatorname{Aut} H$ is the trivial map.

So, no new twist, then what happens?

This means $\theta(k)$ is the identity permutation for all $k \in K$. Then, $[\theta(k)](h)=h \forall h \in H$. Then, the operations just become that of the usual direct product.

I know, I have not used mathematical language, but this is how I keep this in my head! And, it is not difficult to put this into Mathematical formalism.

P.S.:

Think of group actions as following:

Suppose $G$ is a group, and $S$ is a set. Let us say $G$ acts on $S$. Then, this is what that means: Every element in $G$ takes an element of $S$ to some other element of $S$. And, identity, "is identity" on the elements of $S$, I mean it does not take any element in $S$ to a new element. And, if $g$ takes $s$ to $s_1$ and $h$ takes $s_1$ to $s_2$, then $gh$ takes $s$ to $s_2$.

This means that every element in $G$, in short, permutes $S$.

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The first thing I would try and completely solidify would be the idea of a group action, *, of $B$ on $A$ and how it corresponds to a homomorphism $\phi$ from $B$ to $Aut(A)$. With that in hand, consider the definition of a semidirect product $A \rtimes_{\phi} B$. Similarly to a direct product, it is a set of ordered pairs $(a,b), a \in A, b\in B$, but now our composition law depends on our action homomorphism: $(a_1,b_1)$ multiplied with $(a_2,b_2)$ is defined as $(a_1 [\phi(b_1)](a_2), b_1 b_2)$, or directly in terms of the action *, $(a_1 b_1*a_2, b_1 b_2)$. As usual, concatenation denotes composition in our original group.

Now, if our action is trivial, $b*a = [\phi(b)]a = a, \forall a \in A, \forall b \in b.$ What would our composition law for $A \rtimes_{\phi} B$ look like in this case?

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