Metamath Proof Explorer


Theorem pgrpsubgsymg

Description: Every permutation group is a subgroup of the corresponding symmetric group. (Contributed by AV, 14-Mar-2019) (Revised by AV, 30-Mar-2024)

Ref Expression
Hypotheses pgrpsubgsymgbi.g G = SymGrp A
pgrpsubgsymgbi.b B = Base G
pgrpsubgsymg.c F = Base P
Assertion pgrpsubgsymg A V P Grp F B + P = f F , g F f g F SubGrp G

Proof

Step Hyp Ref Expression
1 pgrpsubgsymgbi.g G = SymGrp A
2 pgrpsubgsymgbi.b B = Base G
3 pgrpsubgsymg.c F = Base P
4 1 symggrp A V G Grp
5 simp1 P Grp F B + P = f F , g F f g P Grp
6 4 5 anim12i A V P Grp F B + P = f F , g F f g G Grp P Grp
7 simp2 P Grp F B + P = f F , g F f g F B
8 simp3 P Grp F B + P = f F , g F f g + P = f F , g F f g
9 1 2 symgbasmap f B f A A
10 9 ssriv B A A
11 sstr F B B A A F A A
12 10 11 mpan2 F B F A A
13 resmpo F A A F A A f A A , g A A f g F × F = f F , g F f g
14 13 anidms F A A f A A , g A A f g F × F = f F , g F f g
15 12 14 syl F B f A A , g A A f g F × F = f F , g F f g
16 eqid A A = A A
17 eqid + G = + G
18 1 16 17 symgplusg + G = f A A , g A A f g
19 18 eqcomi f A A , g A A f g = + G
20 19 reseq1i f A A , g A A f g F × F = + G F × F
21 15 20 eqtr3di F B f F , g F f g = + G F × F
22 21 3ad2ant2 P Grp F B + P = f F , g F f g f F , g F f g = + G F × F
23 8 22 eqtrd P Grp F B + P = f F , g F f g + P = + G F × F
24 7 23 jca P Grp F B + P = f F , g F f g F B + P = + G F × F
25 24 adantl A V P Grp F B + P = f F , g F f g F B + P = + G F × F
26 2 3 grpissubg G Grp P Grp F B + P = + G F × F F SubGrp G
27 6 25 26 sylc A V P Grp F B + P = f F , g F f g F SubGrp G
28 27 ex A V P Grp F B + P = f F , g F f g F SubGrp G