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=SymGrpA
pgrpsubgsymgbi.b B=BaseG
pgrpsubgsymg.c F=BaseP
Assertion pgrpsubgsymg AVPGrpFB+P=fF,gFfgFSubGrpG

Proof

Step Hyp Ref Expression
1 pgrpsubgsymgbi.g G=SymGrpA
2 pgrpsubgsymgbi.b B=BaseG
3 pgrpsubgsymg.c F=BaseP
4 1 symggrp AVGGrp
5 simp1 PGrpFB+P=fF,gFfgPGrp
6 4 5 anim12i AVPGrpFB+P=fF,gFfgGGrpPGrp
7 simp2 PGrpFB+P=fF,gFfgFB
8 simp3 PGrpFB+P=fF,gFfg+P=fF,gFfg
9 1 2 symgbasmap fBfAA
10 9 ssriv BAA
11 sstr FBBAAFAA
12 10 11 mpan2 FBFAA
13 resmpo FAAFAAfAA,gAAfgF×F=fF,gFfg
14 13 anidms FAAfAA,gAAfgF×F=fF,gFfg
15 12 14 syl FBfAA,gAAfgF×F=fF,gFfg
16 eqid AA=AA
17 eqid +G=+G
18 1 16 17 symgplusg +G=fAA,gAAfg
19 18 eqcomi fAA,gAAfg=+G
20 19 reseq1i fAA,gAAfgF×F=+GF×F
21 15 20 eqtr3di FBfF,gFfg=+GF×F
22 21 3ad2ant2 PGrpFB+P=fF,gFfgfF,gFfg=+GF×F
23 8 22 eqtrd PGrpFB+P=fF,gFfg+P=+GF×F
24 7 23 jca PGrpFB+P=fF,gFfgFB+P=+GF×F
25 24 adantl AVPGrpFB+P=fF,gFfgFB+P=+GF×F
26 2 3 grpissubg GGrpPGrpFB+P=+GF×FFSubGrpG
27 6 25 26 sylc AVPGrpFB+P=fF,gFfgFSubGrpG
28 27 ex AVPGrpFB+P=fF,gFfgFSubGrpG