Metamath Proof Explorer


Theorem conjsubg

Description: A conjugated subgroup is also a subgroup. (Contributed by Mario Carneiro, 13-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 conjsubg S SubGrp G A X ran F SubGrp G

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 1 subgss S SubGrp G S X
6 5 adantr S SubGrp G A X S X
7 df-ima x X A + ˙ x - ˙ A S = ran x X A + ˙ x - ˙ A S
8 resmpt S X x X A + ˙ x - ˙ A S = x S A + ˙ x - ˙ A
9 8 4 syl6eqr S X x X A + ˙ x - ˙ A S = F
10 9 rneqd S X ran x X A + ˙ x - ˙ A S = ran F
11 7 10 syl5eq S X x X A + ˙ x - ˙ A S = ran F
12 6 11 syl S SubGrp G A X x X A + ˙ x - ˙ A S = ran F
13 subgrcl S SubGrp G G Grp
14 eqid x X A + ˙ x - ˙ A = x X A + ˙ x - ˙ A
15 1 2 3 14 conjghm G Grp A X x X A + ˙ x - ˙ A G GrpHom G x X A + ˙ x - ˙ A : X 1-1 onto X
16 13 15 sylan S SubGrp G A X x X A + ˙ x - ˙ A G GrpHom G x X A + ˙ x - ˙ A : X 1-1 onto X
17 16 simpld S SubGrp G A X x X A + ˙ x - ˙ A G GrpHom G
18 simpl S SubGrp G A X S SubGrp G
19 ghmima x X A + ˙ x - ˙ A G GrpHom G S SubGrp G x X A + ˙ x - ˙ A S SubGrp G
20 17 18 19 syl2anc S SubGrp G A X x X A + ˙ x - ˙ A S SubGrp G
21 12 20 eqeltrrd S SubGrp G A X ran F SubGrp G