Product of Representations of Lorentz Group

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Main Question or Discussion Point

How to prove that direct product of two rep of Lorentz group ##(m,n)⊗(a,b)=(m⊗a,n⊗b)## ?

Let ##J\in {{J_1,J_2,J_3}}##
Then we have :
##[(m,n)⊗(a,b)](J)=(m,n)(J)I_{(a,b)}+I_{(m,n)}⊗(a,b)(J)=##
##=I_m⊗J_n⊗I_a⊗I_b+J_m⊗I_n⊗I_a⊗I_b+I_m⊗I_n⊗J_a⊗I_b+I_m⊗I_n⊗I_a⊗J_b##
and
##(m⊗a,n⊗b)(J)=I_{(m⊗a)}⊗J_{(n⊗b)}+J_{(m⊗a)}⊗I_{(n⊗b)}=##
##=I_m⊗I_a⊗(I_n⊗J_b+J_n⊗I_b)+(I_m⊗J_a+J_m⊗I_a)I_n⊗I_b=##
##=I_m⊗I_a⊗I_n⊗J_b+I_m⊗I_a⊗J_n⊗I_b+J_m⊗I_a⊗I_n⊗I_b+I_m⊗J_a⊗I_n⊗I_b##, and we have that

##[(m,n)⊗(a,b)](J)\neq [(m,n)⊗(a,b)](J)##

Where ##I_a## is unit matrix ##(2a+1) (2a+1)##, matrix, identical is for ##J_b##

How this is work ?
 

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