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Hi. I'm new to Group Theory and wanted to see if I had the right train of thought for this problem.
Let S be a set with an associative law of composition and with an identity element. Prove that the subset of S consisting of invertible elements is a group.
A group is a set G together with a law of composition that is associative and has an identity element, and such that every element of G has an inverse.
I'll denote my subset of S by G.
I know G has the identity element of S as the identity element is invertible.
The associativity for the law of composition is inherited and the existence of inverses we get for free. All that I'm left with is verifying that G has closure. That is for a,b in G, ab is in G.
It's sufficient to show ab is invertible.
But, [tex](ab)(b^{-1}a^{-1})=e[/tex]
where e is the identity element. So [tex]ab[/tex] is invertible and thus is an element of G and G is a group.
Homework Statement
Let S be a set with an associative law of composition and with an identity element. Prove that the subset of S consisting of invertible elements is a group.
Homework Equations
A group is a set G together with a law of composition that is associative and has an identity element, and such that every element of G has an inverse.
The Attempt at a Solution
I'll denote my subset of S by G.
I know G has the identity element of S as the identity element is invertible.
The associativity for the law of composition is inherited and the existence of inverses we get for free. All that I'm left with is verifying that G has closure. That is for a,b in G, ab is in G.
It's sufficient to show ab is invertible.
But, [tex](ab)(b^{-1}a^{-1})=e[/tex]
where e is the identity element. So [tex]ab[/tex] is invertible and thus is an element of G and G is a group.
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