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 ) |