# Thread: semidirect product (sorry the text looks messy)

1. ## semidirect product (sorry the text looks messy)

I am working on the semidirect product in group theory. And I got stuck on a question which is not difficult. Here is the question:

If Q, K are groups and h: Q -->Aut(K) is a homomorphism
Prove that the semidirect product is a group ( sorry, I can't type the notation of semidirect product). So, I have to let G = (K x Q, *, (1,1))

The group operation * is
(k_1,q_1)*(k_2,q_2) =(k_1h(q_1)(k_2), q_1q_2)

This is my attempt.

First show that the group is associative:
[(k_1,q_1)*(k_2,q_2)]*(k_3,q_3) = (k_1h(q_1)(k_2) , q_1q_2)*(k_3,q_3)
= (k_1h(q_1)(k_2)h(q_1q_2)(k_3),q_1q_2q_3)
= ...???
then I dont know how to simplify this. Can we use the fact that h is homomorphism.
And:
(k_1,q_1)*[(k_2,q_2)*(k_3,q_3)] = (k_1,q_1)*(k_2h(q_2)(k_3),q_2q_3)
=(k_1h(q_1)(k_2)h(q_2)(k_3),q_1q_2q_3)
= ......???

Second, show that the group G has an identity,

If: (1,1)*(k,q) = (1.h(1)k,1.q) = (k,q)
and (k,q)*(1,1) = (k.h(q)(1),q.1) = (k.h(q),q) = (???,q)

Finally, G has an inverse if (k,q)*(k^-1,q^-1) = (1,1)
Thus, (k,q)*(k^-1,q^-1) = (k.h(q)(k^-1),q.q^-1) =(h(q), 1)

Then G is a group

I dont know what h(q) is equal to, so that I can simplify.

If you find any mistake please correct it.

Thank you so much

P.S: Sorry about the text, it's a bit hard to read if you write down on paper it will be easier.

2. Originally Posted by knguyen2005
first show that the group is associative:
\begin{aligned}{}[(k_1,q_1)*(k_2,q_2)]*(k_3,q_3) &= (k_1h(q_1)(k_2) , q_1q_2)*(k_3,q_3) \\ &= (k_1h(q_1)(k_2)h(q_1q_2)(k_3),q_1q_2q_3)\end{align ed}
...
then i don't know how to simplify this.
Correct. No simplification needed.

Originally Posted by knguyen2005
and:
\begin{aligned}(k_1,q_1)*[(k_2,q_2)*(k_3,q_3)] &= (k_1,q_1)*(k_2h(q_2)(k_3),q_2q_3) \\ &=(k_1h(q_1){\color{red}\bigl(}(k_2)h(q_2)(k_3) {\color{red}\bigr)},q_1q_2q_3) \end{aligned}

Remember that h takes an element of Q to an automorphism of K. In this case, $h(q_1)$ acts on the element
$k_2h(q_2)(k_3)$. And because it is an automorphism, it is multiplicative. Therefore $h(q_1)\bigl(k_2.h(q_2)k_3\bigr) = h(q_1)k_2.h(q_1)\bigl(h(q_2)k_3\bigr)$. Also, $h(q_1)h(q_2) = h(q_1q_2)$. Thus \begin{aligned}(k_1,q_1)*[(k_2,q_2)*(k_3,q_3)] &= (k_1h(q_1)(k_2)h(q_1q_2)(k_3),q_1q_2q_3) \\ &= [(k_1,q_1)*(k_2,q_2)]*(k_3,q_3)\end{aligned}

Originally Posted by knguyen2005
second, show that the group g has an identity,

if: (1,1)*(k,q) = (1.h(1)k,1.q) = (k,q)
and (k,q)*(1,1) = (k.h(q)(1),q.1) = (k.h(q),q) = (???,q)

You can't write h(q) on its own here; h(q) is an automorphism, and it has to act on an element of K. In this case, it acts on the identity element of K, and an automorphism always takes the identity to the identity. Therefore h(q)(1) = 1, and
(k.h(q)(1),q.1) = (k.1,q.1) = (k,q), as required.

Originally Posted by knguyen2005
finally, g has an inverse if (k,q)*(k^-1,q^-1) = (1,1)
thus, (k,q)*(k^-1,q^-1) = (k.h(q)(k^-1),q.q^-1) =(h(q), 1)
Careful here! The inverse of (k,q) has to be an element (j,p) such that (j,p)*(k,q) = (1,1). But (j,p)*(k,q) = (j.h(p)k, pq). If this is equal to (1,1) then pq = 1 (in Q), and so $p=q^{-1}$. But j.h(p)k = 1 (in K), and so $j = \bigl(h(p)k\bigr)^{-1} = \bigl(h(q^{-1})k\bigr)^{-1} = h(q^{-1})k^{-1}$ (because an automorphism takes inverses to inverses). Therefore $(k,q)^{-1} = (h(q^{-1})k^{-1}, q^{-1})$. (In other words, the inverse of (k,q) is not $(k^{-1},q^{-1})$.)

Afterthought: I think you would find this whole calculation easier to follow if you change the notation for an automorphism. Instead of writing $h(q)(k)$, write it as $h_q(k)$. That will remind you that $h_q$ is an operator, that has to act on an element of K. It satisfies the conditions $h_q(k_1k_2) = h_q(k_1)h_q(k_2)$, $\bigl(h_q(k)\bigr)^{-1} = h_q(k^{-1})$, and $h_{q_1q_2}(k) = h_{q_1}(k)h_{q_2}(k)$.