Metamath Proof Explorer


Theorem grplmulf1o

Description: Left multiplication by a group element is a bijection on any group. (Contributed by Mario Carneiro, 17-Jan-2015)

Ref Expression
Hypotheses grplmulf1o.b B=BaseG
grplmulf1o.p +˙=+G
grplmulf1o.n F=xBX+˙x
Assertion grplmulf1o GGrpXBF:B1-1 ontoB

Proof

Step Hyp Ref Expression
1 grplmulf1o.b B=BaseG
2 grplmulf1o.p +˙=+G
3 grplmulf1o.n F=xBX+˙x
4 1 2 grpcl GGrpXBxBX+˙xB
5 4 3expa GGrpXBxBX+˙xB
6 eqid invgG=invgG
7 1 6 grpinvcl GGrpXBinvgGXB
8 1 2 grpcl GGrpinvgGXByBinvgGX+˙yB
9 8 3expa GGrpinvgGXByBinvgGX+˙yB
10 7 9 syldanl GGrpXByBinvgGX+˙yB
11 eqcom x=invgGX+˙yinvgGX+˙y=x
12 simpll GGrpXBxByBGGrp
13 10 adantrl GGrpXBxByBinvgGX+˙yB
14 simprl GGrpXBxByBxB
15 simplr GGrpXBxByBXB
16 1 2 grplcan GGrpinvgGX+˙yBxBXBX+˙invgGX+˙y=X+˙xinvgGX+˙y=x
17 12 13 14 15 16 syl13anc GGrpXBxByBX+˙invgGX+˙y=X+˙xinvgGX+˙y=x
18 eqid 0G=0G
19 1 2 18 6 grprinv GGrpXBX+˙invgGX=0G
20 19 adantr GGrpXBxByBX+˙invgGX=0G
21 20 oveq1d GGrpXBxByBX+˙invgGX+˙y=0G+˙y
22 7 adantr GGrpXBxByBinvgGXB
23 simprr GGrpXBxByByB
24 1 2 grpass GGrpXBinvgGXByBX+˙invgGX+˙y=X+˙invgGX+˙y
25 12 15 22 23 24 syl13anc GGrpXBxByBX+˙invgGX+˙y=X+˙invgGX+˙y
26 1 2 18 grplid GGrpyB0G+˙y=y
27 26 ad2ant2rl GGrpXBxByB0G+˙y=y
28 21 25 27 3eqtr3d GGrpXBxByBX+˙invgGX+˙y=y
29 28 eqeq1d GGrpXBxByBX+˙invgGX+˙y=X+˙xy=X+˙x
30 17 29 bitr3d GGrpXBxByBinvgGX+˙y=xy=X+˙x
31 11 30 syl5bb GGrpXBxByBx=invgGX+˙yy=X+˙x
32 3 5 10 31 f1o2d GGrpXBF:B1-1 ontoB