New poster, apologies if the formatting/conventions are incorrect.
I am reading Serre's book Galois Cohomology. My question is about exactly how Serre's inductive principle (chapter 1 prop. 39 cor. 3) is used in chapter 3, Section 4.3 Theorem 4 which says for fields $k$ with property $(F)$ and linear algebraic groups $B$ defined over $k$, $H^1(k,B)$ is finite.
Fixing notation and rewriting chapter 1 prop. 39 cor. 3:
Let $k$ be a field and $\bar{k}$ be it's separable closure. By $H^1(k,G)$ I mean $H^1(\text{Gal}(\bar{k}/k),G(\bar{k}))$ where $G(\bar{k})$ are the $\bar{k}$ points of an algebraic group $G$ with an action of $\text{Gal}(\bar{k}/k)$ (this is usually the usual Galois action). Let $$1 \to A \to B \to C \to 1$$ be an exact sequence of algebraic groups defined over $k$ each with an action of $\text{Gal}(\bar{k}/k)$. This induces a long exact sequence in cohomology:
$$1 \to A^{\text{Gal}(\bar{k}/k)} \to B^{\text{Gal}(\bar{k}/k)} \to C^{\text{Gal}(\bar{k}/k)} \to H^1(k,A) \to H^1(k,B) \to H^1(k,C)$$
Here is Corollary 3 to Proposition 39 in Chapter 1:
In order that $H^1(k,B)$ be countable (resp. finite, resp. reduced to a single element), it is necessary and sufficient that the same be true for its image in $H^1(k,C)$, and for all the quotients $H^1(k,bA)/(bC)^{\text{Gal}(\bar{k}/k)}$,for $b \in Z^1(k,B)$.
The proof of the finiteness of $H^1(k,B)$ for $k$ with property $(F)$ goes roughly as follows: The above corollary is used to reduce to the case where $B$ is connected. Then the corollary is again applied to a composition series for $B$ dealing with various terms in cases. I do not see how the reduction to the connected case takes place.
Let $B_0$ be the connected component of $B$. Then $B/B_0$ is finite.
To reduce to the connected case, Serre applies the corollary to the short exact sequence $$1 \to B_0 \to B \to B/B_0 \to 1.$$ For $H^1(k,B)$ to be finite, it is sufficient that $H^1(k,B/B_0)$ be finite (This is true since $B/B_0$ is finite and $k$ has condition (F), see prop 8. ch 4.) and $H^1(k,bB_0)$ be finite for all $b \in Z^1(k,B)$.
Clearly, twisting by the trivial cocycle $b = e$ reduces to the connected case i.e. showing that $H^1(k, B_0)$ is finite. For nontrivial cocycles $b$ I see that $bB_0$ is still the same group as $B_0$ (and hence connected) but it is equipped with a different action of $\text{Gal}(\bar{k}/k)$.
I am confused because I think the text is implying that we can reduce to the case of showing that $H^1(k,B)$ is finite where $B$ is connected and $B$ has the usual (i.e. not twisted) Galois action of $\text{Gal}(\bar{k}/k)$.
I think that one way to possibly answer this is that $H^1(k, bB_0)$ is isomorphic as a pointed set to $H^1(k, bE)$ where $bE$ is a $k$-form of $B_0$ associated to $b$ and where $H^1(k, bE)$ makes use of the usual Galois action of $\text{Gal}(\bar{k}/k)$ on $bE(\bar{k})$. Should it be clear that $k$-forms of a connected group $B_0$ need to be connected?
Are the twists $bB_0$ of $B_0$ by $b \in Z^1(k,B)$ being considered as the connected group $B_0$ with a twisted action of $\text{Gal}(\bar{k}/k)$ or is there some way to see that $H^1(k, bB_0)$ is isomorphic as a pointed set to $H^1(k, bE)$ where $bE$ is some connected linear algebraic group over $k$ with the usual Galois action of $\text{Gal}(\bar{k}/k)$?
In Serre's Corollary 3 to Theorem 3 of section 2.4 of chapter 3, he does this reduction explicitly. Under the assumption that the field has cohomological dimension $\leq 1$ and is perfect, Theorem 1' of the previous section shows that connected linear algebraic groups $B_0$ have trivial $H^1(k,B_0)$. Let $B_0$ be the connected component of $B$. He concludes that the map $H^1(k,B) \to H^1(k,B/B_0)$ is injective by using the above inductive principle and says explicitly that since all twists $bB_0$ are connected and linear, $H^1(k, bB_0)=0$.