What is wrong with my 'proof' that there is only ever one Borel subgroup of an algebraic group

Question

Let $G$ be an algebraic group. Let $\mathfrak{B}$ be the collection of Borel subgroups of $G$. Let $S$ be a connected, solvable, normal subgroup of $G$. Denote the identity component of the subgroup $H\subset G$ by $H^\circ$.

Which incorrect statement is generating my contradiction?

1. Since the Borel subgroups are maximal with respect to being connected and solvable, $S\subseteq B$ for some $B\in\mathfrak{B}$.
2. The Borel conjugacy theorem tells us $gSg^{-1}\subseteq gBg^{-1}=B'$, for some $B'\in\mathfrak{B}$.
3. Since $S$ is normal, $gSg^{-1}=S\subseteq B'$.
4. We can conclude that $S$ is in every Borel subgroup. $S\subseteq \bigcap_{B\in \mathfrak{B}}B$.
5. The identity component of a group is the unique irreducible component containing the identity.
6. $S$ contains the identity and since it is connected, it is irreducible.
7. From 5. and 6. both $B^\circ$ and $S$ are irreducible and contain the identity. But the uniqueness tells us that $S=B^\circ$. Since $B$ is connected, $B=B^\circ$, and hence $S=B$.
8. This holds for every Borel. So $S$ is equal to every Borel, and hence there is only ever one Borel subgroup. Contradiction.

Of course I don't believe the conclusion, but I was trying to show that $S\subseteq \left(\bigcap_{B\in\mathfrak{B}} B\right)^\circ$, and obviously one of my deductions is false.

Answer 1

The error is in step seven:

7. From 5. and 6. both $B^\circ$ and $S$ are irreducible and contain the identity. But the uniqueness tells us that $S=B^\circ$. Since $B$ is connected, $B=B^\circ$, and hence $S=B$.

As noted in the comments; the components of a subgroup are not necessarily components of the big group. So $S=S^\circ$ is not necessarily a component of $B$, and hence we cannot use the uniqueness of the identity component to deduce that $S=B^\circ$.