# Is the Standard Model an invariant subgroup of $SU(5)$?

It is well known that the Standard Model (SM) gauge group is a subgroup of $SU(5)$: $$SU(3) \times SU(2)\times U(1) ~\subset~SU(5)$$ This can be easily checked using the method of Dynkin diagrams. Is this subgroup an invariant subgroup such that, $$g _{SU(5)} g _{SM} g _{SU(5)} = g _{SM} ' \,,$$ where $g_{SU(5)}$ ($g_{SM}$) is an element of $SU(5)$ ($SM$)?

Background: The reason I'm interested in this is because then its necessarily true that the non-SM gauge group generators of $SU(5)$ can be written as solely off-diagonal matrices and the SM as solely diagonal (this is easy to see by writing the matrices in block diagonal form), which simplifies calculations.

Actually, $$G~:=~SU(3) \times SU(2)\times U(1)$$ is not a subgroup of $SU(5)$, but $G/\mathbb{Z}_6$ is a subgroup of $SU(5)$, cf. e.g. this Phys.SE post and Ref. 1.

We interprete OP's question (v3) as essentially asking

Is $G/\mathbb{Z}_6$ a normal subgroup of $SU(5)$?

Or in terms of the corresponding Lie algebras,

Is $su(3) \oplus su(2)\oplus u(1)$ an ideal of $su(5)$?

Here we identify $su(5)$ with anti-Hermitian traceless $5\times 5$ matrices; $su(3)$ with anti-Hermitian traceless $3\times 3$ block matrices in rows/columns 1,2,3; and $su(2)$ with the anti-Hermitian traceless $2\times 2$ block matrices in rows/columns 4,5; while $u(1)$ is generated by the diagonal traceless matrix ${\rm diag}(-2,-2,-2,3,3)$ times an imaginary number.

It is not hard to see that if we consider the Lie bracket (or equivalently commutator) $C=[A,B]$ between e.g. an $su(3)$ matrix $A$ with non-zero entries 12 and 21, and an $su(5)$ matrix $B$ with non-zero entries 25 and 52, then $C$ would have non-zero elements 15 and 51, and hence cannot belong to $su(3) \oplus su(2)\oplus u(1)$.

We conclude that $su(3) \oplus su(2)\oplus u(1)$ is not an ideal of $su(5)$.

References:

1. J.C. Baez, Calabi-Yau Manifolds and the Standard Model, arXiv:hep-th/0511086.