Semidirect products via $\rho\colon B\to \mathrm{Aut}(A)$ Show that for every homomorphism $ρ$ , ($a_1,b_1)\cdot (a_2, b_2)= (a_1ρ(b_1)(a_2),b_1b_2)$ defines a group structure on product $A\times B$ and $B$ normalizes $A$ in this product. I don't really get what the question needs from me. Neither do i know if my attempt to prove this is right.
My attempt: Let $G: A\times B\to C$ where $(a,b)\to ab$ and let $A$, $B$ be subgroups of $C$. Then $AB= \{ab :a \in A, b\in B\}$ is a subgroup of $C$. Since $ρ$ is a group homomorphism we define $ρ(b)(a) = bab^{-1}$ for every $a\in A$ and $b\in B.$
Then $(a_1,b_1)=(a_1\rho(a_1)(a_1),(b_1b_2)$ and since the product $(a_1,b_1)\cdot(a_2,b_2)=(a_1\cdot a_2,b_1\cdot b_2)$
then $a_1b_1a_2b_2=a_1b_1a_2b^{-1}b_1b_2$ and therefore this product belongs to $A\times B$ and so closure is proved. There are other axioms of groups that i will show once i am sure that's the correct prove.
For the second part I tried to prove that $B$ normalizes $A$ using normalization definition but I got stuck because the product is defined a bit complexly. Any corrections, methods for proving or explanations are appreciated.

We have a binary operation $$(a_1,b_1)\cdot (a_2, b_2)= (a_1ρ(b_1)(a_2),b_1b_2)$$ to complete a group we need

*

*identity

*associativity of the operation

*inverses

The identity is easy, just take $(1,1)$ and show that it is the identity for the binary operation here.
Associativity is also quite easy but may involve a lot of writing, just expand out the definition of the binary operation for both:

*

*$$((a_1,b_1)\cdot (a_2, b_2))\cdot (a_3, b_3)$$

*$$(a_1,b_1)\cdot ((a_2, b_2)\cdot (a_3, b_3))$$
and then use associativity of your underlying groups to prove these two terms equal.
For the inverse, given $(a_1,b_1)$ I will let you come up with some $x$ such that $$(a_1,b_1)\cdot x = 1.$$
