Metamath Proof Explorer


Theorem cayleylem1

Description: Lemma for cayley . (Contributed by Paul Chapman, 3-Mar-2008) (Revised by Mario Carneiro, 13-Jan-2015)

Ref Expression
Hypotheses cayleylem1.x 𝑋 = ( Base ‘ 𝐺 )
cayleylem1.p + = ( +g𝐺 )
cayleylem1.u 0 = ( 0g𝐺 )
cayleylem1.h 𝐻 = ( SymGrp ‘ 𝑋 )
cayleylem1.s 𝑆 = ( Base ‘ 𝐻 )
cayleylem1.f 𝐹 = ( 𝑔𝑋 ↦ ( 𝑎𝑋 ↦ ( 𝑔 + 𝑎 ) ) )
Assertion cayleylem1 ( 𝐺 ∈ Grp → 𝐹 ∈ ( 𝐺 GrpHom 𝐻 ) )

Proof

Step Hyp Ref Expression
1 cayleylem1.x 𝑋 = ( Base ‘ 𝐺 )
2 cayleylem1.p + = ( +g𝐺 )
3 cayleylem1.u 0 = ( 0g𝐺 )
4 cayleylem1.h 𝐻 = ( SymGrp ‘ 𝑋 )
5 cayleylem1.s 𝑆 = ( Base ‘ 𝐻 )
6 cayleylem1.f 𝐹 = ( 𝑔𝑋 ↦ ( 𝑎𝑋 ↦ ( 𝑔 + 𝑎 ) ) )
7 eqid ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) = ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) )
8 1 2 7 gaid2 ( 𝐺 ∈ Grp → ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) ∈ ( 𝐺 GrpAct 𝑋 ) )
9 oveq12 ( ( 𝑥 = 𝑔𝑦 = 𝑎 ) → ( 𝑥 + 𝑦 ) = ( 𝑔 + 𝑎 ) )
10 ovex ( 𝑔 + 𝑎 ) ∈ V
11 9 7 10 ovmpoa ( ( 𝑔𝑋𝑎𝑋 ) → ( 𝑔 ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) 𝑎 ) = ( 𝑔 + 𝑎 ) )
12 11 mpteq2dva ( 𝑔𝑋 → ( 𝑎𝑋 ↦ ( 𝑔 ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) 𝑎 ) ) = ( 𝑎𝑋 ↦ ( 𝑔 + 𝑎 ) ) )
13 12 mpteq2ia ( 𝑔𝑋 ↦ ( 𝑎𝑋 ↦ ( 𝑔 ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) 𝑎 ) ) ) = ( 𝑔𝑋 ↦ ( 𝑎𝑋 ↦ ( 𝑔 + 𝑎 ) ) )
14 6 13 eqtr4i 𝐹 = ( 𝑔𝑋 ↦ ( 𝑎𝑋 ↦ ( 𝑔 ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) 𝑎 ) ) )
15 1 4 14 galactghm ( ( 𝑥𝑋 , 𝑦𝑋 ↦ ( 𝑥 + 𝑦 ) ) ∈ ( 𝐺 GrpAct 𝑋 ) → 𝐹 ∈ ( 𝐺 GrpHom 𝐻 ) )
16 8 15 syl ( 𝐺 ∈ Grp → 𝐹 ∈ ( 𝐺 GrpHom 𝐻 ) )