Metamath Proof Explorer


Theorem ablfaclem1

Description: Lemma for ablfac . (Contributed by Mario Carneiro, 27-Apr-2016) (Revised by Mario Carneiro, 3-May-2016)

Ref Expression
Hypotheses ablfac.b 𝐵 = ( Base ‘ 𝐺 )
ablfac.c 𝐶 = { 𝑟 ∈ ( SubGrp ‘ 𝐺 ) ∣ ( 𝐺s 𝑟 ) ∈ ( CycGrp ∩ ran pGrp ) }
ablfac.1 ( 𝜑𝐺 ∈ Abel )
ablfac.2 ( 𝜑𝐵 ∈ Fin )
ablfac.o 𝑂 = ( od ‘ 𝐺 )
ablfac.a 𝐴 = { 𝑤 ∈ ℙ ∣ 𝑤 ∥ ( ♯ ‘ 𝐵 ) }
ablfac.s 𝑆 = ( 𝑝𝐴 ↦ { 𝑥𝐵 ∣ ( 𝑂𝑥 ) ∥ ( 𝑝 ↑ ( 𝑝 pCnt ( ♯ ‘ 𝐵 ) ) ) } )
ablfac.w 𝑊 = ( 𝑔 ∈ ( SubGrp ‘ 𝐺 ) ↦ { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑔 ) } )
Assertion ablfaclem1 ( 𝑈 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑊𝑈 ) = { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑈 ) } )

Proof

Step Hyp Ref Expression
1 ablfac.b 𝐵 = ( Base ‘ 𝐺 )
2 ablfac.c 𝐶 = { 𝑟 ∈ ( SubGrp ‘ 𝐺 ) ∣ ( 𝐺s 𝑟 ) ∈ ( CycGrp ∩ ran pGrp ) }
3 ablfac.1 ( 𝜑𝐺 ∈ Abel )
4 ablfac.2 ( 𝜑𝐵 ∈ Fin )
5 ablfac.o 𝑂 = ( od ‘ 𝐺 )
6 ablfac.a 𝐴 = { 𝑤 ∈ ℙ ∣ 𝑤 ∥ ( ♯ ‘ 𝐵 ) }
7 ablfac.s 𝑆 = ( 𝑝𝐴 ↦ { 𝑥𝐵 ∣ ( 𝑂𝑥 ) ∥ ( 𝑝 ↑ ( 𝑝 pCnt ( ♯ ‘ 𝐵 ) ) ) } )
8 ablfac.w 𝑊 = ( 𝑔 ∈ ( SubGrp ‘ 𝐺 ) ↦ { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑔 ) } )
9 eqeq2 ( 𝑔 = 𝑈 → ( ( 𝐺 DProd 𝑠 ) = 𝑔 ↔ ( 𝐺 DProd 𝑠 ) = 𝑈 ) )
10 9 anbi2d ( 𝑔 = 𝑈 → ( ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑔 ) ↔ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑈 ) ) )
11 10 rabbidv ( 𝑔 = 𝑈 → { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑔 ) } = { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑈 ) } )
12 fvex ( SubGrp ‘ 𝐺 ) ∈ V
13 2 12 rabex2 𝐶 ∈ V
14 13 wrdexi Word 𝐶 ∈ V
15 14 rabex { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑈 ) } ∈ V
16 11 8 15 fvmpt ( 𝑈 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑊𝑈 ) = { 𝑠 ∈ Word 𝐶 ∣ ( 𝐺 dom DProd 𝑠 ∧ ( 𝐺 DProd 𝑠 ) = 𝑈 ) } )