Metamath Proof Explorer


Theorem pgrpsubgsymgbi

Description: Every permutation group is a subgroup of the corresponding symmetric group. (Contributed by AV, 14-Mar-2019)

Ref Expression
Hypotheses pgrpsubgsymgbi.g 𝐺 = ( SymGrp ‘ 𝐴 )
pgrpsubgsymgbi.b 𝐵 = ( Base ‘ 𝐺 )
Assertion pgrpsubgsymgbi ( 𝐴𝑉 → ( 𝑃 ∈ ( SubGrp ‘ 𝐺 ) ↔ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) )

Proof

Step Hyp Ref Expression
1 pgrpsubgsymgbi.g 𝐺 = ( SymGrp ‘ 𝐴 )
2 pgrpsubgsymgbi.b 𝐵 = ( Base ‘ 𝐺 )
3 2 issubg ( 𝑃 ∈ ( SubGrp ‘ 𝐺 ) ↔ ( 𝐺 ∈ Grp ∧ 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) )
4 3anass ( ( 𝐺 ∈ Grp ∧ 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ↔ ( 𝐺 ∈ Grp ∧ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) )
5 3 4 bitri ( 𝑃 ∈ ( SubGrp ‘ 𝐺 ) ↔ ( 𝐺 ∈ Grp ∧ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) )
6 1 symggrp ( 𝐴𝑉𝐺 ∈ Grp )
7 ibar ( 𝐺 ∈ Grp → ( ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ↔ ( 𝐺 ∈ Grp ∧ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) ) )
8 7 bicomd ( 𝐺 ∈ Grp → ( ( 𝐺 ∈ Grp ∧ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) ↔ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) )
9 6 8 syl ( 𝐴𝑉 → ( ( 𝐺 ∈ Grp ∧ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) ↔ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) )
10 5 9 syl5bb ( 𝐴𝑉 → ( 𝑃 ∈ ( SubGrp ‘ 𝐺 ) ↔ ( 𝑃𝐵 ∧ ( 𝐺s 𝑃 ) ∈ Grp ) ) )