Metamath Proof Explorer


Theorem gapm

Description: The action of a particular group element is a permutation of the base set. (Contributed by Jeff Hankins, 11-Aug-2009) (Proof shortened by Mario Carneiro, 13-Jan-2015)

Ref Expression
Hypotheses gapm.1 X=BaseG
gapm.2 F=xYA˙x
Assertion gapm ˙GGrpActYAXF:Y1-1 ontoY

Proof

Step Hyp Ref Expression
1 gapm.1 X=BaseG
2 gapm.2 F=xYA˙x
3 1 gaf ˙GGrpActY˙:X×YY
4 3 ad2antrr ˙GGrpActYAXxY˙:X×YY
5 simplr ˙GGrpActYAXxYAX
6 simpr ˙GGrpActYAXxYxY
7 4 5 6 fovcdmd ˙GGrpActYAXxYA˙xY
8 3 ad2antrr ˙GGrpActYAXyY˙:X×YY
9 gagrp ˙GGrpActYGGrp
10 9 ad2antrr ˙GGrpActYAXyYGGrp
11 simplr ˙GGrpActYAXyYAX
12 eqid invgG=invgG
13 1 12 grpinvcl GGrpAXinvgGAX
14 10 11 13 syl2anc ˙GGrpActYAXyYinvgGAX
15 simpr ˙GGrpActYAXyYyY
16 8 14 15 fovcdmd ˙GGrpActYAXyYinvgGA˙yY
17 simpll ˙GGrpActYAXxYyY˙GGrpActY
18 simplr ˙GGrpActYAXxYyYAX
19 simprl ˙GGrpActYAXxYyYxY
20 simprr ˙GGrpActYAXxYyYyY
21 1 12 gacan ˙GGrpActYAXxYyYA˙x=yinvgGA˙y=x
22 17 18 19 20 21 syl13anc ˙GGrpActYAXxYyYA˙x=yinvgGA˙y=x
23 22 bicomd ˙GGrpActYAXxYyYinvgGA˙y=xA˙x=y
24 eqcom x=invgGA˙yinvgGA˙y=x
25 eqcom y=A˙xA˙x=y
26 23 24 25 3bitr4g ˙GGrpActYAXxYyYx=invgGA˙yy=A˙x
27 2 7 16 26 f1o2d ˙GGrpActYAXF:Y1-1 ontoY