Notes on Elementary Properties of Group, Uniqueness of identity element,Uniqueness of inverse element,Cancellation law | Grade 12 > Mathematics > Elementary Group Theory | KULLABS.COM

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Elementary properties of a group:

We know the definition of a group. It says that a group must have an identity element and every element must have an inverse. But there is no mention of the number of identities or the number of inverses of an element. For this investigation about these matters and others, we should prove some theorems: These theorems such as the cancellation law, uniqueness of identity element, uniqueness of inverse element and others reveal some properties of groups. Whihch means hat is G is a group with binary operation * then G satisfies the following properties:

Uniqueness of Identity Element (Theorem I)

In a group, there is one and only one identity element. (or in a group the identity element is unique).

Proof:

$$Let\;e\;be\;an\;identity\;element\;in\;a\;group\;(\;G,\circ\;).\;If\;possible\;let\;e’\;be\;another\;identity\;element.\;Then\;we\;have\;$$

$$\;e\circ\;e’=e’\circ\;e=e’\;\;\;considering\;e\;as\;the\;identity\;element.\;$$

$$\;e\circ\;e’=e’\circ\;e=e\;\;\;considering\;e’\;as\;the\;identity\;element.\;$$

Hence e=e’, i.e. there is one and only one identity element.

Unique Inverse law (Theorem ii)

$$Every\;element\;in\;the\;group\;(\;G,\circ\;)\;has\;unique\;inverse.\;$$

Proof:

We know that an element a is a group has inverse a-1 such that,

$$a^{-1}\circ\;a=e=a\circ\;a^{-1}$$

Suppose a’ is another inverse of a. So, we have

$$a’\circ\;a=e=a\circ\;a’$$

$$Now,\;\;\;a’=e\circ\;a’\;\;\;\;\;\;as\;e\;is\;the\;indentity\;element$$

$$\;=\;(a^{-1}\circ\;a\;)\;\circ\;a’\;\;\;\;\;\;as\;a=a^{-1}\circ\;a$$

$$\;=a^{-1}\;\circ\;(a\circ\;a’\;)\;\;\;\;\;by\;associativity\;law\;$$

$$\;=a^{-1}\circ\;e\;\;\;\;\;\;as\;a\circ\;a’=e\;$$

$$\;=\;a^{-1}\;\;\;\;\;\;\;\;\;as\;e\;is\;the\;identity\;element.\;$$

This shows that the inverse element is unique.

Cancellation law (Theorem iii)

$$If\;a,\;b,\;c\;are\;the\;elements\;of\;group\;(\;G,\;\circ\;)\;and\;a\circ\;b=a\circ\;c,\;then\;b=c.\;Also\;if\;b\circ\;c=c\circ\;a,\;then\;b=c\;$$

Proof:

$$As\;a\in\;(G,\;\circ\;)\;,\;a\;has\;the\;inverse\;a^{-1}\;such\;that\;a^{-1}\;\circ\;a=\;a\circ\;a^{-1}=e\;$$

We have,

$$a\circ\;b=b\circ\;c$$

Multiplying the above equation by a­­-1 on the left. Then,

$$a^{-1}\circ\;(a\circ\;b)\;=a^{-1}\circ\;(a\circ\;c\;)\;$$

$$or,\;a^{-1}\circ\;(a\circ\;b)=a^{-1}\circ\;(a\circ\;c)\;$$

$$or,\;(a^{-1}\circ\;a)\circ\;b=(a^{-1}\circ\;a)\circ\;c\;\;\;by\;associativity\;law\;$$

$$or,\;\;e\circ\;b=e\;\circ\;c\;\;\;\;\;\;\;\;\;\;\;\;\;because\;a^{-1}\circ\;a=e\;$$

$$or,\;\;\;\;b=c\;\;\;\;\;\;\;\;\;\;\;\;\;\;as\;e\;is\;the\;identity\;element.\;$$

$$Similarly,\;we\;can\;show\;that\;if\;b\circ\;a\;=\;c\circ\;a,\;then\;b=c.\;$$

Theorem iv

$$If\;a,b\in\;(G,\circ\;),\;then,\;$$

$$i)\;(a\circ\;b)^{-1}\;=b^{-1}\circ\;a^{-1}$$

$$ii)\;(a^{-1})^{-1}=a$$

Proof:

We have,

$$(a\circ\;b)\circ\;(b^{-1}\circ\;a^{-1})\;$$

$$\;=\;((a\circ\;b)\circ\;b^{-1})\;\circ\;a^{-1}\;\;\;\;by\;associativity\;law$$

$$\;=\;(a\circ\;(b\circ\;b^{-1}))\;\circ\;a^{-1}$$

$$\;=\;(a\circ\;e)\;\circ\;a^{-1}$$

$$\;=e\;$$

$$Similarly,\;we\;can\;show\;that\;(b^{-1}\circ\;a^{-1})\;\circ\;(a\circ\;b)\;=e\;$$

$$\therefore\;a\circ\;b\;is\;the\;inverse\;of\;b^{-1}\circ\;a^{-1}.\;$$

$$i.e.\;\;(a\;\circ\;b)^{-1}\;=b^{-1}\circ\;a^{-1}$$

Again,

$$We\;have\;a^{-1}\circ\;a=e$$

$$Multiplying\;both\;sides\;on\;the\;left\;by\;(a^{-1})^{-1}\;we\;get,\;$$

$$\;(a^{-1})^{-1}\;\circ\;a^{-1}\circ\;a=(a^{-1})^{-1}\;e\;$$

$$\;or,\;((a^{-1}\circ\;a^{-1})\circ\;a=(a^{-1})^{-1}\;\;\;\;by\;associativity\;law\;$$

$$\;or,\;\;\;\;e\circ\;a\;=(a^{-1})^{-1}\;$$

Theorem V:

$$If\;elemenets\;a\;and\;b\;are\;of\;the\;group\;(G,\;\circ\;),\;then\;$$

$$a\circ\;x=b\;and\;x\circ\;a=b$$

$$have\;unique\;solutions\;in\;(G,\circ\;).\;$$

Proof:

$$Let\;a\;be\;an\;element\;of\;the\;group\;(G,\circ\;).\;So\;a^{-1},\;the\;inverse\;element\;of\;a$$

$$\;is\;also\;in\;(G,\circ)\;and\;a^{-1}\;\circ\;a=a\circ\;a^{-1}=e$$

$$Now,a\circ\;x=b$$

$$a^{-1}\circ\;(a\circ\;x)= a^{-1}\circ\;b\;\;\;Multiplying\;both\;sides\;by\;a^{-1}\;on\;the\;left\;$$

$$\;or\;(a^{-1}\circ\;a)\circ\;x= a^{-1}\circ\;b\;\;\;Associativity\;law\;$$

$$\;or\;e\circ\;x=\;a^{-1}\circ\;b\;$$

$$\;or\;x\;=a^{-1}\circ\;b$$

This is the required equation.

To show the uniqueness of the solution we suppose that x­ and x­2­ are the solutions, then

$$a\circ\;x_1=b\;and\;a\circ\;x_2=b\;$$

$$\therefore\;a\circ\;x_1\;=a\circ\;x_2\;\;\;\;each\;being\;equal\;to\;b$$

$$or,a^{-1}\circ\;(a\circ\;x_1)=a^{-1}\circ\;(a\circ\;x_2)$$

$$or,(a^{-1}\circ\;a)\circ\;x_1=(a^{-1}\circ\;a)\circ\;x_2$$

$$\;e\circ\;x_1\;=e\circ\;x_2$$

$$or,\;x_1=x_2\;$$

So, the solution is unique.

$$Similarly\;we\;can\;show\;that\;x\circ\;a=b\;has\;a\;unique\;solution\;in\;(G,\circ\;).\;$$

Example:

$$If\;a,\;b\;\in\;G\;and\;G\;is\;abelian,\;show\;that\;(ab)^2=a^2b^2$$

Solution:

Since G is abelian so,

Now,

$$(ab)^2=(ab)(ab)=a(ba)b$$

$$\;=a(ab)b\;\;\;=(aa)(bb)$$

$$\;=a^2b^2$$

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )

• In a group, there is one and only one identity element. (or in a group the identity element is unique).
• $$If\;a,\;b,\;c\;are\;the\;elements\;of\;group\;(\;G,\;\circ\;)\;and\;a\circ\;b=a\circ\;c,\;then\;b=c.\;Also\;if\;b\circ\;c=c\circ\;a,\;then\;b=c\;$$
• $$Every\;element\;in\;the\;group\;(\;G,\circ\;)\;has\;unique\;inverse.\;$$
• $$If\;a,b\in\;(G,\circ\;),\;then,\;$$

$$i)\;(a\circ\;b)^{-1}\;=b^{-1}\circ\;a^{-1}$$

$$ii)\;(a^{-1})^{-1}=a$$

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