Problem about normal subgroups

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SUMMARY

The discussion centers on proving the equation A ∩ BC = B(A ∩ C) given that A, B, and C are normal subgroups of a group G, with B being a subset of A. The proof begins by establishing that if x is an element of A ∩ BC, then x can be expressed as a product of elements from B and C. The challenge lies in demonstrating that this product can also be represented as a product of an element from B and an element from A ∩ C, thereby confirming the equality. The participants are working through the algebraic manipulation necessary to complete the proof.

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  • Understanding of group theory, specifically normal subgroups.
  • Familiarity with group operations and subgroup properties.
  • Knowledge of algebraic manipulation involving group elements.
  • Experience with proofs in abstract algebra.
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  • Study the properties of normal subgroups in group theory.
  • Learn about the intersection of subgroups and its implications.
  • Explore examples of subgroup proofs in abstract algebra.
  • Review the concept of group homomorphisms and their relevance to subgroup structures.
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Students of abstract algebra, particularly those focusing on group theory, as well as educators and mathematicians seeking to deepen their understanding of subgroup relationships and proof techniques.

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Homework Statement



If A,B and C are normal subgroups of G where B[itex]\subseteq[/itex]A show that
A[itex]\bigcap[/itex]BC=B(A[itex]\bigcap[/itex]C)

Homework Equations



The Attempt at a Solution


Let x[itex]\in[/itex]A[itex]\bigcap[/itex]BC.then x[itex]\in[/itex]A and x[itex]\in[/itex]BC
Now as B[itex]\subseteq[/itex]A thus BA=A.thus left side is BA[itex]\bigcap[/itex]BC

Dont know how to proceed.
 
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So you know that [itex]x\in A[/itex] and [itex]x\in BC[/itex]. The last means that there are [itex]b\in B[/itex] and [itex]c\in C[/itex] such that [itex]x=bc[/itex].

You must prove that [itex]x\in B(A\cap C)[/itex]. So you must write x as a product of something in B and something in [itex]A\cap C[/itex].
 
x=bc [itex]\Rightarrow[/itex] a-1bac[itex]\in[/itex]BC
Also a-1bca[itex]\in[/itex]BC Now a-1b=a1 for some a1[itex]\in[/itex]A[itex]\Rightarrow[/itex]a1ca[itex]\in[/itex]BC
a1(aa-1)ca[itex]\in[/itex]BC
(a1a)c1[itex]\in[/itex]BC for some c1[itex]\in[/itex]C
a2c1[itex]\in[/itex]BC
Cant still figure out
 

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