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Description: The symmetric group on a singleton has cardinality 1 . (Contributed by AV, 9-Dec-2018)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | symg1bas.1 | ⊢ 𝐺 = ( SymGrp ‘ 𝐴 ) | |
| symg1bas.2 | ⊢ 𝐵 = ( Base ‘ 𝐺 ) | ||
| symg1bas.0 | ⊢ 𝐴 = { 𝐼 } | ||
| Assertion | symg1hash | ⊢ ( 𝐼 ∈ 𝑉 → ( ♯ ‘ 𝐵 ) = 1 ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | symg1bas.1 | ⊢ 𝐺 = ( SymGrp ‘ 𝐴 ) | |
| 2 | symg1bas.2 | ⊢ 𝐵 = ( Base ‘ 𝐺 ) | |
| 3 | symg1bas.0 | ⊢ 𝐴 = { 𝐼 } | |
| 4 | snfi | ⊢ { 𝐼 } ∈ Fin | |
| 5 | 3 4 | eqeltri | ⊢ 𝐴 ∈ Fin |
| 6 | 1 2 | symghash | ⊢ ( 𝐴 ∈ Fin → ( ♯ ‘ 𝐵 ) = ( ! ‘ ( ♯ ‘ 𝐴 ) ) ) |
| 7 | 5 6 | ax-mp | ⊢ ( ♯ ‘ 𝐵 ) = ( ! ‘ ( ♯ ‘ 𝐴 ) ) |
| 8 | 3 | fveq2i | ⊢ ( ♯ ‘ 𝐴 ) = ( ♯ ‘ { 𝐼 } ) |
| 9 | hashsng | ⊢ ( 𝐼 ∈ 𝑉 → ( ♯ ‘ { 𝐼 } ) = 1 ) | |
| 10 | 8 9 | eqtrid | ⊢ ( 𝐼 ∈ 𝑉 → ( ♯ ‘ 𝐴 ) = 1 ) |
| 11 | 10 | fveq2d | ⊢ ( 𝐼 ∈ 𝑉 → ( ! ‘ ( ♯ ‘ 𝐴 ) ) = ( ! ‘ 1 ) ) |
| 12 | fac1 | ⊢ ( ! ‘ 1 ) = 1 | |
| 13 | 11 12 | eqtrdi | ⊢ ( 𝐼 ∈ 𝑉 → ( ! ‘ ( ♯ ‘ 𝐴 ) ) = 1 ) |
| 14 | 7 13 | eqtrid | ⊢ ( 𝐼 ∈ 𝑉 → ( ♯ ‘ 𝐵 ) = 1 ) |