Let $Q,K$ be groups and $\theta: Q \to Aut(K)$ be a homomorphism. Let us define on the direct product set $K \times Q$ the following multiplication:
$(k_1,q_1)(k_2,q_2):=(k_1k_2^{\theta(q_1)}, q_1q_2)$.
Prove that this is associative!
Determine the unit element and determine the inverse of each element.
Prove that the group constructed this way is the split extension of $(K,1) \simeq K$ by $(1,Q) \simeq Q$.
Calculate the product: $(1,q)(k,1)(1,q)^{-1}$.
My attempt:
For associativity, we have the following:
$$[(a_1,b_1)(a_2,b_2)](a_3,b_3)=(a_1 a_2^{\theta(b_1)},b_1b_2)(a_3,b_3)=(a_1 a_2^{\theta(b_1)}a_3^{\theta(b_2)},b_1b_2b_3).$$
$$(a_1,b_1)[(a_2,b_2)(a_3,b_3)]=(a_1,b_1)(a_2a_3^{\theta (b_2)},b_2b_3)=(a_1a_2a_3^{\theta(b_2) \theta(b_1)},b_1b_2b_3).$$
I am not entirely sure how to prove that they are the same, or if I have calculated this right.
The unit element should be $(1_K, 1_Q)$, where $1_K \in K$ is the unit element from $K$, and $1_Q$ is the unit element from $Q$, since, for given $(k_1,q_1)$:
$$(k_1,q_1)(1_K, 1_Q)=(k_1 1_K^{\theta(1_K)},q_1 1_Q)=(k_1,q_1)$$
$$(1_K, 1_Q)(k_1,q_1)=(1_K k_1^{\theta(k_1)}, 1_Q q_1)=(k_1^{\theta(k_1)},q_1)$$
Once again, I have no idea what to do with this exponent.
The inverse of $(k_1,q_1)$ should be $(k_1^{-1}, q_1^{-1})$ for similar reasons.
The product
$$(1,q)(k,1)(1,q)^{-1}=(k^{\theta(k)},q)=(k,q)(1,q)^{-1}=(k,q)(1,q^{-1})=(k,1).$$
If you can help me with any of the $4$ problems, it would be very appreciated.
For the associativity, you did not calculate it correctly. $$ [(a_1,b_1)(a_2,b_2)](a_3,b_3)=(a_1 a_2^{\theta(b_1)},b_1b_2)(a_3,b_3)=(a_1 a_2^{\theta(b_1)}a_3^{\theta(b_1 b_2)},b_1b_2b_3). $$ $$ (a_1,b_1)[(a_2,b_2)(a_3,b_3)]=(a_1,b_1)(a_2a_3^{\theta (b_2)},b_2b_3)=(a_1(a_2a_3^{\theta(b_2)})^ { \theta(b_1)},b_1b_2b_3) =(a_1a_2^{\theta(b_1)}(a_3^{\theta(b_2)})^ { \theta(b_1)},b_1b_2b_3). $$ Since $(a_3^{\theta(b_2)})^ { \theta(b_1)}=a_{3}^{\theta(b_{1})\circ \theta(b_2)}=a_{3}^{\theta(b_{1}b_{2})}$, the associativity holds.
For the identity, $$ (1_K, 1_Q)(k_1,q_1)=(1_K k_1^{\theta(1_Q)}, 1_Q q_1)=(k_1^{1},q_1)=(k_1,q_1). $$
For inverses, you can set $(k,q)(x,y)=(1,1)$ and solve for $x,y$.
I'm not sure about the third question.
For the last one, the first step is not correct.