Metamath Proof Explorer
Description: The only subgroup of a trivial group is itself. (Contributed by Rohan
Ridenour, 3-Aug-2023)
|
|
Ref |
Expression |
|
Hypotheses |
trivsubgd.1 |
⊢ 𝐵 = ( Base ‘ 𝐺 ) |
|
|
trivsubgd.2 |
⊢ 0 = ( 0g ‘ 𝐺 ) |
|
|
trivsubgd.3 |
⊢ ( 𝜑 → 𝐺 ∈ Grp ) |
|
|
trivsubgd.4 |
⊢ ( 𝜑 → 𝐵 = { 0 } ) |
|
|
trivsubgd.5 |
⊢ ( 𝜑 → 𝐴 ∈ ( SubGrp ‘ 𝐺 ) ) |
|
Assertion |
trivsubgd |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
trivsubgd.1 |
⊢ 𝐵 = ( Base ‘ 𝐺 ) |
2 |
|
trivsubgd.2 |
⊢ 0 = ( 0g ‘ 𝐺 ) |
3 |
|
trivsubgd.3 |
⊢ ( 𝜑 → 𝐺 ∈ Grp ) |
4 |
|
trivsubgd.4 |
⊢ ( 𝜑 → 𝐵 = { 0 } ) |
5 |
|
trivsubgd.5 |
⊢ ( 𝜑 → 𝐴 ∈ ( SubGrp ‘ 𝐺 ) ) |
6 |
1
|
subgss |
⊢ ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 𝐴 ⊆ 𝐵 ) |
7 |
5 6
|
syl |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
8 |
7 4
|
sseqtrd |
⊢ ( 𝜑 → 𝐴 ⊆ { 0 } ) |
9 |
2
|
subg0cl |
⊢ ( 𝐴 ∈ ( SubGrp ‘ 𝐺 ) → 0 ∈ 𝐴 ) |
10 |
5 9
|
syl |
⊢ ( 𝜑 → 0 ∈ 𝐴 ) |
11 |
10
|
snssd |
⊢ ( 𝜑 → { 0 } ⊆ 𝐴 ) |
12 |
8 11
|
eqssd |
⊢ ( 𝜑 → 𝐴 = { 0 } ) |
13 |
12 4
|
eqtr4d |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |