Inverse map is involutive - Revision history

Vipul at 19:39, 31 March 2022

2022-03-31T19:39:59Z

← Older revision		Revision as of 19:39, 31 March 2022
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	==Proof==		==Proof==

	~~<center>{{#widget:YouTube\|id=Ha9SUEeZzkQ}}</center>~~

	===Proof of reversal law===		===Proof of reversal law===

Vipul at 09:37, 4 April 2013

2013-04-04T09:37:42Z

← Older revision		Revision as of 09:37, 4 April 2013
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	==Proof==		==Proof==

			<center>{{#widget:YouTube\|id=Ha9SUEeZzkQ}}</center>

	===Proof of reversal law===		===Proof of reversal law===

Vipul: /* Proof of reversal law */

2012-06-24T15:09:34Z

Proof of reversal law

← Older revision		Revision as of 15:09, 24 June 2012
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	===Proof of reversal law===		===Proof of reversal law===

	~~This follows from associativity~~. Consider:		In order to show that the element <math>a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}</math> is a two-sided inverse of <math>a_1a_2 \ldots a_n</math>, it suffices to show that their product both ways is the identity element. Consider first the product:

	<math>a_1a_2 \ldots ~~a_na_n~~^{-1}a_{n-1}^{-1}\~~ldots~~ a_1^{-1}</math>		<math>(a_1a_2 \ldots a_{n-1}a_n)(a_n^{-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1})</math>

	~~We first cancel <math>a_n</math> and <math>a_n^{-1}</math>~~, and ~~keep proceeding in this way, till the whole word reduces to the identity element.~~		Due to associativity, we can drop the parentheses and we get:

	~~A similar argument follows for the product:~~		<math>a_1a_2 \ldots a_{n-1}a_na_n^{-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1}</math>

	<math>a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}a_1a_2 \ldots a_n</math>		Now, consider the middle product <math>a_na_n^{-1}</math>. This is the identity element, and since the identity element has no effect on the remaining product, it can be removed, giving the product:

			<math>a_1a_2 \ldots a_{n-1}a_{n-1}^{-1}\dots a_2^{-1}a_1^{-1}</math>

			We now repeat the argument with the middle product <math>a_{n-1}a_{n-1}^{-1}</math> and ''cancel'' them. Proceeding this way, we are able to cancel all terms and eventually get the identity element.

			A similar argument follows for the product the other way around:

			<math>(a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1})(a_1a_2 \ldots a_n)</math>

	Thus, the elements are two-sided inverses of each other.		Thus, the elements are two-sided inverses of each other.

			'''Note''': In fact, it suffices to check only one of the two inverse conditions, i.e., check only that the first product is the identity element. This is because, in a group, every element has a two-sided inverse. Further, [[equality of left and right inverses in monoid]] forces any ''one-sided'' (left ''or'' right) inverse to be equal to the two-sided inverse.

	===Proof for applying it twice===		===Proof for applying it twice===

Vipul: /* Proof for applying it twice */

2012-06-21T18:00:18Z

Proof for applying it twice

← Older revision		Revision as of 18:00, 21 June 2012
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	This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.		This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.

	More explicitly, if <math>b = a^{-1}</math>, that means that <math>ab = ba = e</math>. But this is precisely the condition for stating that <math>b ~~= a~~^{-1}</math>.		More explicitly, if <math>b = a^{-1}</math>, that means that <math>ab = ba = e</math>. But this is precisely the condition for stating that <math>a = b^{-1}</math>.
	<section end="main"/>		<section end="main"/>

Vipul: /* Proof for applying it twice */

2012-06-21T17:55:40Z

Proof for applying it twice

← Older revision		Revision as of 17:55, 21 June 2012
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	This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.		This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.

			More explicitly, if <math>b = a^{-1}</math>, that means that <math>ab = ba = e</math>. But this is precisely the condition for stating that <math>b = a^{-1}</math>.
	<section end="main"/>		<section end="main"/>

Vipul at 16:54, 10 June 2008

2008-06-10T16:54:21Z

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			{{basic fact}}

			<section begin="main"/>
	==Statement==		==Statement==

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	This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.		This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.
			<section end="main"/>

			==References==

			===Textbook references===

			* {{booklink-proved\|DummitFoote}}, Page 18, Proposition 1

Vipul: 1 revision

2008-05-07T23:45:23Z

1 revision

← Older revision	Revision as of 23:45, 7 May 2008
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Vipul: New page: ==Statement== The inverse map in a group, i.e. the map sending any element of the group, to its inverse element, is an ''involutive map'', in the sense that it has the following t...

2008-05-03T13:44:07Z

New page: ==Statement== The inverse map in a group, i.e. the map sending any element of the group, to its inverse element, is an ''involutive map'', in the sense that it has the following t...

New page

==Statement==

The inverse map in a group, i.e. the map sending any element of the [[group]], to its [[inverse element]], is an ''involutive map'', in the sense that it has the following two properties:

* It satisfies the reversal law:

<math>(a_1a_2 \ldots a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}</math>

* Applying it twice sends an element to itself:

<math>(a^{-1})^{-1} = a</math>

It fact, both these are true in the greater generality of a [[monoid]], under the condition that all the <math>a_i</math>s have two-sided inverse <math>a_i^{-1}</math> (note: we still need a monoid to guarantee that two-sided inverses, when they exist, are unique).

==Proof==

===Proof of reversal law===

This follows from associativity. Consider:

<math>a_1a_2 \ldots a_na_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}</math>

We first cancel <math>a_n</math> and <math>a_n^{-1}</math>, and keep proceeding in this way, till the whole word reduces to the identity element.

A similar argument follows for the product:

<math>a_n^{-1}a_{n-1}^{-1}\ldots a_1^{-1}a_1a_2 \ldots a_n</math>

Thus, the elements are two-sided inverses of each other.

===Proof for applying it twice===

This is direct from the definition. let <math>b = a^{-1}</math>. Then, by the inherent symmetry in the definition of [[inverse element]], we see that <math>a = b^{-1}</math>.