Subgroup test: Difference between revisions

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In [[Abstract Algebra]], the one-step subgroup test is a theorem that states that for any group, a [[subset]] of that [[Group_%28mathematics%29|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. The two-step subgroup test is a similar theorem which requires the subset to be closed under the operation and taking of inverses.

=One Step Subgroup Test=

Or more formally let <math>G\,</math> be a group and let <math>H\,</math> be a nonempty 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>.

OrLet more formally let <math>G\,</math> be a group and let <math>H\,</math> be a nonempty subset of <math>G\,</math>. If <math>\forall{for all a, and b \in H}, ab^{^-1} is \in H</math>, then <math>H\,</math> is a subgroup of <math>G\,</math>.

===Proof===

ToLet proveG be a group, let H be a nonempty subset of G and assume that <math>for all a and b in H\, ab<sup>-1</mathsup> is in H. To prove that H 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. So,

* 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.

* Since H is not empty there exists an element x in H. Letting a = x and b = x, we have that the identity e = xx^-1 = ab^-1 which is in H, so e is in H.

* Let x be an element of H. Since the identity e is in H it follows that ex^-1 = x^-1 in H, so the inverse of an element in H is in H.

* Finally we show that <math>H\,</math> islet closedx underand they operation.be Letelements <math>x, y \in H</math>, then since <math> y \is in H</math> it follows that y<mathsup>y^{-1}</sup> is \in H</math>. Hence <math>x{(y^{^-1})}^{^-1} = xy \is in H</math> and so <math>H\,</math> is closed under the operation.

Thus <math>H\,</math> is a subgroup of <math>G\,</math>.

=Two Step Subgroup Test=

A corollary of this theorem is the two-step subgroup test which states 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.

=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.

<BR><BR>

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>.

<BR><BR>

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.

<BR><BR>

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>.

<BR><BR>

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>

<BR><BR>

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.

<BR><BR>

Thus <math>H\,</math> is a subgroup of <math>G\,</math>.

[[Category:Mathematical theorems]]

[[Category:Articles containing proofs]]