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#REDIRECT [[Subgroup#Subgroup tests]]
In [[Abstract Algebra]], the one-step subgroup test is a theorem that states that for any group, a [[subset]] of that [[group]] is itself a group if the inverse of any element in the subset multiplied with any other element in the subset is also in the subset.
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Or more formally let <math>G\,</math> be a group and let <math>H\,</math> be a nonempty a subset of <math>G\,</math>. If <math>\forall{ a, b \in H}, ab^{-1} \in H</math> then <math>H\,</math> is a subgroup of <math>G\,</math>.
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A corollary of this theorem is the two-step subgroup test which state that a nonempty subset of a group is itself a group if the subset is [[Closure (mathematics)|closed]] under the operation as well as under the taking of inverses.
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=Proof=
To prove that <math>H\,</math> is a subgroup of <math>G\,</math> we must show that <math>H\,</math> is nonempty, associative, has an identity, has an inverse for every element, and is closed under the operation.
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Let <math>G\,</math> be a group, let <math>H\,</math> be a nonempty subset of <math>G\,</math> and assume that <math>\forall{ a, b \in H}, ab^{-1} \in H</math>.
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Since the operation of <math>H\,</math> is the same as the operation of <math>G\,</math>, the operation is associative since <math>G\,</math> is a group.
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Next we show that the identity, <math>e\,</math>, is in <math>H\,</math>. Since <math>H\,</math> is not empty there exists an <math>x \in H</math>. Letting <math>a = x\,</math> and <math>b = x\,</math>, we have that <math>e = xx^{-1} = ab^{-1} \in H</math>, so <math>e \in H</math>.
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We now show that every element in <math>H\,</math> has an inverse in <math>H\,</math>. Let <math>x \in H</math>. Since <math>e \in H</math> it follows that <math>ex^{-1} = x^{-1} \in H</math>, so <math>x^{-1} \in H</math>
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Finally we show that <math>H\,</math> is closed under the operation. Let <math>x, y \in H</math>, then since <math> y \in H</math> it follows that <math>y^{-1} \in H</math>. Hence <math>x{(y^{-1})}^{-1} = xy \in H</math> and so <math>H\,</math> is closed under the operation.
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Thus <math>H\,</math> is a subgroup of <math>G\,</math>.
[[Category:Mathematical theorems]]
[[Category:Proofs]]
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