Metamath Proof Explorer


Theorem subgnm

Description: The norm in a subgroup. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Hypotheses subgngp.h 𝐻 = ( 𝐺s 𝐴 )
subgnm.n 𝑁 = ( norm ‘ 𝐺 )
subgnm.m 𝑀 = ( norm ‘ 𝐻 )
Assertion subgnm ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 𝑀 = ( 𝑁𝐴 ) )

Proof

Step Hyp Ref Expression
1 subgngp.h 𝐻 = ( 𝐺s 𝐴 )
2 subgnm.n 𝑁 = ( norm ‘ 𝐺 )
3 subgnm.m 𝑀 = ( norm ‘ 𝐻 )
4 eqid ( Base ‘ 𝐺 ) = ( Base ‘ 𝐺 )
5 4 subgss ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 𝐴 ⊆ ( Base ‘ 𝐺 ) )
6 5 resmptd ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → ( ( 𝑥 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) ) ↾ 𝐴 ) = ( 𝑥𝐴 ↦ ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) ) )
7 1 subgbas ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 𝐴 = ( Base ‘ 𝐻 ) )
8 eqid ( dist ‘ 𝐺 ) = ( dist ‘ 𝐺 )
9 1 8 ressds ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → ( dist ‘ 𝐺 ) = ( dist ‘ 𝐻 ) )
10 eqidd ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 𝑥 = 𝑥 )
11 eqid ( 0g𝐺 ) = ( 0g𝐺 )
12 1 11 subg0 ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → ( 0g𝐺 ) = ( 0g𝐻 ) )
13 9 10 12 oveq123d ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) = ( 𝑥 ( dist ‘ 𝐻 ) ( 0g𝐻 ) ) )
14 7 13 mpteq12dv ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑥𝐴 ↦ ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) ) = ( 𝑥 ∈ ( Base ‘ 𝐻 ) ↦ ( 𝑥 ( dist ‘ 𝐻 ) ( 0g𝐻 ) ) ) )
15 6 14 eqtr2d ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑥 ∈ ( Base ‘ 𝐻 ) ↦ ( 𝑥 ( dist ‘ 𝐻 ) ( 0g𝐻 ) ) ) = ( ( 𝑥 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) ) ↾ 𝐴 ) )
16 eqid ( Base ‘ 𝐻 ) = ( Base ‘ 𝐻 )
17 eqid ( 0g𝐻 ) = ( 0g𝐻 )
18 eqid ( dist ‘ 𝐻 ) = ( dist ‘ 𝐻 )
19 3 16 17 18 nmfval 𝑀 = ( 𝑥 ∈ ( Base ‘ 𝐻 ) ↦ ( 𝑥 ( dist ‘ 𝐻 ) ( 0g𝐻 ) ) )
20 2 4 11 8 nmfval 𝑁 = ( 𝑥 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) )
21 20 reseq1i ( 𝑁𝐴 ) = ( ( 𝑥 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑥 ( dist ‘ 𝐺 ) ( 0g𝐺 ) ) ) ↾ 𝐴 )
22 15 19 21 3eqtr4g ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 𝑀 = ( 𝑁𝐴 ) )