Metamath Proof Explorer


Theorem conjsubgen

Description: A conjugated subgroup is equinumerous to the original subgroup. (Contributed by Mario Carneiro, 18-Jan-2015)

Ref Expression
Hypotheses conjghm.x X = Base G
conjghm.p + ˙ = + G
conjghm.m - ˙ = - G
conjsubg.f F = x S A + ˙ x - ˙ A
Assertion conjsubgen S SubGrp G A X S ran F

Proof

Step Hyp Ref Expression
1 conjghm.x X = Base G
2 conjghm.p + ˙ = + G
3 conjghm.m - ˙ = - G
4 conjsubg.f F = x S A + ˙ x - ˙ A
5 subgrcl S SubGrp G G Grp
6 eqid x X A + ˙ x - ˙ A = x X A + ˙ x - ˙ A
7 1 2 3 6 conjghm G Grp A X x X A + ˙ x - ˙ A G GrpHom G x X A + ˙ x - ˙ A : X 1-1 onto X
8 5 7 sylan S SubGrp G A X x X A + ˙ x - ˙ A G GrpHom G x X A + ˙ x - ˙ A : X 1-1 onto X
9 f1of1 x X A + ˙ x - ˙ A : X 1-1 onto X x X A + ˙ x - ˙ A : X 1-1 X
10 8 9 simpl2im S SubGrp G A X x X A + ˙ x - ˙ A : X 1-1 X
11 1 subgss S SubGrp G S X
12 11 adantr S SubGrp G A X S X
13 f1ssres x X A + ˙ x - ˙ A : X 1-1 X S X x X A + ˙ x - ˙ A S : S 1-1 X
14 10 12 13 syl2anc S SubGrp G A X x X A + ˙ x - ˙ A S : S 1-1 X
15 12 resmptd S SubGrp G A X x X A + ˙ x - ˙ A S = x S A + ˙ x - ˙ A
16 15 4 eqtr4di S SubGrp G A X x X A + ˙ x - ˙ A S = F
17 f1eq1 x X A + ˙ x - ˙ A S = F x X A + ˙ x - ˙ A S : S 1-1 X F : S 1-1 X
18 16 17 syl S SubGrp G A X x X A + ˙ x - ˙ A S : S 1-1 X F : S 1-1 X
19 14 18 mpbid S SubGrp G A X F : S 1-1 X
20 f1f1orn F : S 1-1 X F : S 1-1 onto ran F
21 19 20 syl S SubGrp G A X F : S 1-1 onto ran F
22 f1oeng S SubGrp G F : S 1-1 onto ran F S ran F
23 21 22 syldan S SubGrp G A X S ran F